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Matrix exponential

In mathematics, the matrix exponential is a matrix function on square matrices analogous to the ordinary exponential function. It is used to solve systems of linear differential equations. In the theory of Lie groups, the matrix exponential gives the exponential map between a matrix Lie algebra and the corresponding Lie group.

Let $X$ be an $n×n$ real or complex matrix. The exponential of $X$, denoted by $e^{X}$ or $exp(X)$, is the $n×n$ matrix given by the power series

$$e^X = \sum_{k=0}^\infty \frac{1}{k!} X^k$$

where $$X^0$$ is defined to be the identity matrix $$I$$ with the same dimensions as $$X$$. The series always converges, so the exponential of $X$ is well-defined.

Equivalently, $$e^X = \lim_{k \rightarrow \infty} \left(I + \frac{X}{k} \right)^k$$

where $I$ is the $n×n$ identity matrix.

When $X$ is an $n×n$ diagonal matrix then $exp(X)$ will be an $n×n$ diagonal matrix with each diagonal element equal to the ordinary exponential applied to the corresponding diagonal element of $X$.

Elementary properties
Let $X$ and $Y$ be $n×n$ complex matrices and let $a$ and $b$ be arbitrary complex numbers. We denote the $n×n$ identity matrix by $I$ and the zero matrix by 0. The matrix exponential satisfies the following properties.

We begin with the properties that are immediate consequences of the definition as a power series:
 * $e^{0} = I$, where $exp(X^{T}) = (exp X)^{T}$ denotes the transpose of $X^{T}$.
 * $X$, where $exp(X^{∗}) = (exp X)^{∗}$ denotes the conjugate transpose of $X^{∗}$.
 * If $X$ is invertible then $Y$
 * If $e^{YXY^{−1}} = Ye^{X}Y^{−1}.|undefined$ is invertible then $e^{aX}e^{bX} = e^{(a + b)X}$

The next key result is this one:
 * If $$XY=YX$$ then $$e^Xe^Y=e^{X+Y}$$.

The proof of this identity is the same as the standard power-series argument for the corresponding identity for the exponential of real numbers. That is to say, as long as $$X$$ and $$Y$$ commute, it makes no difference to the argument whether $$X$$ and $$Y$$ are numbers or matrices. It is important to note that this identity typically does not hold if $$X$$ and $$Y$$ do not commute (see Golden-Thompson inequality below).

Consequences of the preceding identity are the following:

Using the above results, we can easily verify the following claims. If $e^{X}e^{−X} = I$ is symmetric then $X$ is also symmetric, and if $e^{X}$ is skew-symmetric then $X$ is orthogonal. If $e^{X}$ is Hermitian then $X$ is also Hermitian, and if $e^{X}$ is skew-Hermitian then $X$ is unitary.

Finally, a Laplace transform of matrix exponentials amounts to the resolvent, $$\int_0^\infty e^{-ts}e^{tX}\,dt = (sI - X)^{-1}$$ for all sufficiently large positive values of $s$.

Linear differential equation systems
One of the reasons for the importance of the matrix exponential is that it can be used to solve systems of linear ordinary differential equations. The solution of $$ \frac{d}{dt} y(t) = Ay(t), \quad y(0) = y_0, $$ where $A$ is a constant matrix and y is a column vector, is given by $$ y(t) = e^{At} y_0. $$

The matrix exponential can also be used to solve the inhomogeneous equation $$ \frac{d}{dt} y(t) = Ay(t) + z(t), \quad y(0) = y_0. $$ See the section on applications below for examples.

There is no closed-form solution for differential equations of the form $$ \frac{d}{dt} y(t) = A(t) \, y(t), \quad y(0) = y_0, $$ where $A$ is not constant, but the Magnus series gives the solution as an infinite sum.

The determinant of the matrix exponential
By Jacobi's formula, for any complex square matrix the following trace identity holds:

In addition to providing a computational tool, this formula demonstrates that a matrix exponential is always an invertible matrix. This follows from the fact that the right hand side of the above equation is always non-zero, and so $e^{X}$, which implies that $det(e^{A}) ≠ 0$ must be invertible.

In the real-valued case, the formula also exhibits the map $$\exp \colon M_n(\R) \to \mathrm{GL}(n, \R)$$ to not be surjective, in contrast to the complex case mentioned earlier. This follows from the fact that, for real-valued matrices, the right-hand side of the formula is always positive, while there exist invertible matrices with a negative determinant.

Real symmetric matrices
The matrix exponential of a real symmetric matrix is positive definite. Let $$S$$ be an $e^{A}$ real symmetric matrix and $$x \in \R^n$$ a column vector. Using the elementary properties of the matrix exponential and of symmetric matrices, we have:

$$x^Te^Sx=x^Te^{S/2}e^{S/2}x=x^T(e^{S/2})^Te^{S/2}x =(e^{S/2}x)^Te^{S/2}x=\lVert e^{S/2}x\rVert^2\geq 0.$$

Since $$e^{S/2}$$ is invertible, the equality only holds for $$x=0$$, and we have $$x^Te^Sx > 0$$ for all non-zero $$x$$. Hence $$e^S$$ is positive definite.

The exponential of sums
For any real numbers (scalars) $x$ and $y$ we know that the exponential function satisfies $n×n$. The same is true for commuting matrices. If matrices $X$ and $Y$ commute (meaning that $e^{x+y} = e^{x} e^{y}$), then, $$e^{X+Y} = e^Xe^Y.$$

However, for matrices that do not commute the above equality does not necessarily hold.

The Lie product formula
Even if $X$ and $Y$ do not commute, the exponential $XY = YX$ can be computed by the Lie product formula $$e^{X+Y} = \lim_{k\to\infty} \left(e^{\frac{1}{k}X}e^{\frac{1}{k}Y}\right)^k.$$

Using a large finite $k$ to approximate the above is basis of the Suzuki-Trotter expansion, often used in numerical time evolution.

The Baker–Campbell–Hausdorff formula
In the other direction, if $X$ and $Y$ are sufficiently small (but not necessarily commuting) matrices, we have $$e^Xe^Y = e^Z,$$ where $Z$ may be computed as a series in commutators of $X$ and $Y$ by means of the Baker–Campbell–Hausdorff formula: $$Z = X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]]+ \cdots,$$ where the remaining terms are all iterated commutators involving $X$ and $Y$. If $X$ and $Y$ commute, then all the commutators are zero and we have simply $e^{X + Y}$.

Inequalities for exponentials of Hermitian matrices
For Hermitian matrices there is a notable theorem related to the trace of matrix exponentials.

If $A$ and $B$ are Hermitian matrices, then $$\operatorname{tr}\exp(A + B) \leq \operatorname{tr}\left[\exp(A)\exp(B)\right].$$

There is no requirement of commutativity. There are counterexamples to show that the Golden–Thompson inequality cannot be extended to three matrices – and, in any event, $Z = X + Y$ is not guaranteed to be real for Hermitian $tr(exp(A)exp(B)exp(C))$, $A$, $B$. However, Lieb proved that it can be generalized to three matrices if we modify the expression as follows $$\operatorname{tr}\exp(A + B + C) \leq \int_0^\infty \mathrm{d}t\, \operatorname{tr}\left[e^A\left(e^{-B} + t\right)^{-1}e^C \left(e^{-B} + t\right)^{-1}\right].$$

The exponential map
The exponential of a matrix is always an invertible matrix. The inverse matrix of $C$ is given by $e^{X}$. This is analogous to the fact that the exponential of a complex number is always nonzero. The matrix exponential then gives us a map $$\exp \colon M_n(\Complex) \to \mathrm{GL}(n, \Complex)$$ from the space of all n×n matrices to the general linear group of degree $n$, i.e. the group of all n×n invertible matrices. In fact, this map is surjective which means that every invertible matrix can be written as the exponential of some other matrix (for this, it is essential to consider the field C of complex numbers and not R).

For any two matrices $X$ and $Y$, $$\left\| e^{X+Y} - e^X\right\| \le \|Y\| e^{\|X\|} e^{\|Y\|}, $$

where $e^{−X}$ denotes an arbitrary matrix norm. It follows that the exponential map is continuous and Lipschitz continuous on compact subsets of $‖ · ‖$.

The map $$t \mapsto e^{tX}, \qquad t \in \R$$ defines a smooth curve in the general linear group which passes through the identity element at $M_{n}(C)$.

In fact, this gives a one-parameter subgroup of the general linear group since $$e^{tX}e^{sX} = e^{(t + s)X}.$$

The derivative of this curve (or tangent vector) at a point t is given by

The derivative at $t = 0$ is just the matrix X, which is to say that X generates this one-parameter subgroup.

More generally, for a generic $$-dependent exponent, $t = 0$,

Taking the above expression $X(t)$ outside the integral sign and expanding the integrand with the help of the Hadamard lemma one can obtain the following useful expression for the derivative of the matrix exponent, $$\left(\frac{d}{dt}e^{X(t)}\right)e^{-X(t)} = \frac{d}{dt}X(t) + \frac{1}{2!} \left[X(t), \frac{d}{dt}X(t)\right] + \frac{1}{3!} \left[X(t), \left[X(t), \frac{d}{dt}X(t)\right]\right] + \cdots $$

The coefficients in the expression above are different from what appears in the exponential. For a closed form, see derivative of the exponential map.

Directional derivatives when restricted to Hermitian matrices
Let $$X$$ be a $$n \times n$$ Hermitian matrix with distinct eigenvalues. Let $$X = E \textrm{diag}(\Lambda) E^*$$ be its eigen-decomposition where $$E$$ is a unitary matrix whose columns are the eigenvectors of $$X$$, $$E^*$$ is its conjugate transpose, and $$\Lambda = \left(\lambda_1, \ldots, \lambda_n\right)$$ the vector of corresponding eigenvalues. Then, for any $$n \times n$$ Hermitian matrix $$V$$, the directional derivative of $$\exp: X \to e^X$$ at $$X$$ in the direction $$V$$ is $$ D \exp (X) [V] \triangleq \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left(\displaystyle e^{X + \epsilon V} - e^{X} \right) = E(G \odot \bar{V}) E^* $$ where $$\bar{V} = E^* V E$$, the operator $$\odot$$ denotes the Hadamard product, and, for all $$1 \leq i, j \leq n$$, the matrix $$G$$ is defined as $$ G_{i, j} = \left\{\begin{align} & \frac{e^{\lambda_i} - e^{\lambda_j}}{\lambda_i - \lambda_j} & \text{ if } i \neq j,\\ & e^{\lambda_i} & \text{ otherwise}.\\ \end{align}\right. $$ In addition, for any $$n \times n$$ Hermitian matrix $$U$$, the second directional derivative in directions $$U$$ and $$V$$ is $$ D^2 \exp (X) [U, V] \triangleq \lim_{\epsilon_u \to 0} \lim_{\epsilon_v \to 0} \frac{1}{4 \epsilon_u \epsilon_v} \left(\displaystyle e^{X + \epsilon_u U + \epsilon_v V} - e^{X - \epsilon_u U + \epsilon_v V} - e^{X + \epsilon_u U - \epsilon_v V} + e^{X - \epsilon_u U - \epsilon_v V} \right) = E F(U, V) E^* $$ where the matrix-valued function $$F$$ is defined, for all $$1 \leq i, j \leq n$$, as $$ F(U, V)_{i,j} = \sum_{k=1}^n \phi_{i,j,k}(\bar{U}_{ik}\bar{V}_{jk}^* + \bar{V}_{ik}\bar{U}_{jk}^*) $$ with $$ \phi_{i,j,k} = \left\{\begin{align} & \frac{G_{ik} - G_{jk}}{\lambda_i - \lambda_j} & \text{ if } i \ne j,\\ & \frac{G_{ii} - G_{ik}}{\lambda_i - \lambda_k} & \text{ if } i = j \text{ and } k \ne i,\\ & \frac{G_{ii}}{2} & \text{ if } i = j = k.\\ \end{align}\right. $$

Computing the matrix exponential
Finding reliable and accurate methods to compute the matrix exponential is difficult, and this is still a topic of considerable current research in mathematics and numerical analysis. Matlab, GNU Octave, R, and SciPy all use the Padé approximant. In this section, we discuss methods that are applicable in principle to any matrix, and which can be carried out explicitly for small matrices. Subsequent sections describe methods suitable for numerical evaluation on large matrices.

Diagonalizable case
If a matrix is diagonal: $$A = \begin{bmatrix} a_1 & 0 & \cdots & 0 \\ 0 & a_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_n \end{bmatrix} ,$$ then its exponential can be obtained by exponentiating each entry on the main diagonal: $$e^A = \begin{bmatrix} e^{a_1} & 0 & \cdots & 0 \\ 0 & e^{a_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & e^{a_n} \end{bmatrix} .$$

This result also allows one to exponentiate diagonalizable matrices. If

and $e^{X(t)}$ is diagonal, then

Application of Sylvester's formula yields the same result. (To see this, note that addition and multiplication, hence also exponentiation, of diagonal matrices is equivalent to element-wise addition and multiplication, and hence exponentiation; in particular, the "one-dimensional" exponentiation is felt element-wise for the diagonal case.)

Example : Diagonalizable
For example, the matrix $$ A = \begin{bmatrix} 1 & 4\\ 1 & 1\\ \end{bmatrix}$$ can be diagonalized as $$\begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}\begin{bmatrix} -1 & 0\\ 0 & 3\\ \end{bmatrix}\begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}^{-1}.$$

Thus, $$e^A = \begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}e^\begin{bmatrix} -1 & 0\\ 0 & 3\\ \end{bmatrix}\begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}^{-1}=\begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}\begin{bmatrix} \frac{1}{e} & 0\\ 0 & e^3\\ \end{bmatrix}\begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}^{-1} = \begin{bmatrix} \frac{e^4+1}{2e} & \frac{e^4-1}{e}\\ \frac{e^4-1}{4e} & \frac{e^4+1}{2e}\\ \end{bmatrix}.$$

Nilpotent case
A matrix $t$ is nilpotent if $A = UDU^{−1}$ for some integer q. In this case, the matrix exponential $D$ can be computed directly from the series expansion, as the series terminates after a finite number of terms:

$$e^N = I + N + \frac{1}{2}N^2 + \frac{1}{6}N^3 + \cdots + \frac{1}{(q - 1)!}N^{q-1} ~.$$

Since the series has a finite number of steps, it is a matrix polynomial, which can be computed efficiently.

Using the Jordan–Chevalley decomposition
By the Jordan–Chevalley decomposition, any $$n \times n$$ matrix X with complex entries can be expressed as $$X = A + N $$ where
 * A is diagonalizable
 * N is nilpotent
 * A commutes with N

This means that we can compute the exponential of X by reducing to the previous two cases: $$e^X = e^{A+N} = e^A e^N. $$

Note that we need the commutativity of A and N for the last step to work.

Using the Jordan canonical form
A closely related method is, if the field is algebraically closed, to work with the Jordan form of $N$. Suppose that $e^{A} = Ue^{D}U^{−1}$ where $X$ is the Jordan form of $J$. Then $$e^{X} = Pe^{J}P^{-1}.$$

Also, since $$\begin{align} J &= J_{a_1}(\lambda_1) \oplus J_{a_2}(\lambda_2) \oplus \cdots \oplus J_{a_n}(\lambda_n), \\ e^J &= \exp \big( J_{a_1}(\lambda_1) \oplus J_{a_2}(\lambda_2) \oplus \cdots \oplus J_{a_n}(\lambda_n) \big) \\ &= \exp \big( J_{a_1}(\lambda_1) \big) \oplus \exp \big( J_{a_2}(\lambda_2) \big) \oplus \cdots \oplus \exp \big( J_{a_n}(\lambda_n) \big). \end{align}$$

Therefore, we need only know how to compute the matrix exponential of a Jordan block. But each Jordan block is of the form $$\begin{align} &           &     J_a(\lambda) &= \lambda I + N \\ &\Rightarrow & e^{J_a(\lambda)} &= e^{\lambda I + N} = e^\lambda e^N. \end{align}$$

where $X$ is a special nilpotent matrix. The matrix exponential of $N$ is then given by $$e^J = e^{\lambda_1} e^{N_{a_1}} \oplus e^{\lambda_2} e^{N_{a_2}} \oplus \cdots \oplus e^{\lambda_n} e^{N_{a_n}}$$

Projection case
If $J$ is a projection matrix (i.e. is idempotent: $N^{q} = 0$), its matrix exponential is:

Deriving this by expansion of the exponential function, each power of $P$ reduces to $P$ which becomes a common factor of the sum: $$e^P = \sum_{k=0}^{\infty} \frac{P^k}{k!} = I + \left(\sum_{k=1}^{\infty} \frac{1}{k!}\right)P = I + (e - 1)P ~.$$

Rotation case
For a simple rotation in which the perpendicular unit vectors $e^{N}$ and $X = PJP$ specify a plane, the rotation matrix $P$ can be expressed in terms of a similar exponential function involving a generator $R$ and angle $G$. $$\begin{align} G &= \mathbf{ba}^\mathsf{T} - \mathbf{ab}^\mathsf{T} & P &= -G^2 = \mathbf{aa}^\mathsf{T} + \mathbf{bb}^\mathsf{T} \\ P^2 &= P & PG &= G = GP ~, \end{align}$$ $$\begin{align} R\left( \theta \right) = e^{G\theta} &= I + G\sin (\theta) + G^2(1 - \cos(\theta)) \\ &= I - P + P\cos (\theta) + G\sin (\theta ) ~.\\ \end{align}$$

The formula for the exponential results from reducing the powers of $θ$ in the series expansion and identifying the respective series coefficients of $P^{2} = P$ and $G$ with $e^{P} = I + (e − 1)P$ and $a$ respectively. The second expression here for $b$ is the same as the expression for $G^{2}$ in the article containing the derivation of the generator, $−cos(θ)$.

In two dimensions, if $$a = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$$ and $$b = \left[ \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right]$$, then $$G = \left[ \begin{smallmatrix} 0 & -1 \\ 1 & 0\end{smallmatrix} \right]$$, $$G^2 = \left[ \begin{smallmatrix}-1 & 0 \\ 0 & -1\end{smallmatrix} \right]$$, and $$R(\theta) = \begin{bmatrix}\cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{bmatrix} = I \cos(\theta) + G \sin(\theta)$$ reduces to the standard matrix for a plane rotation.

The matrix $sin(θ)$ projects a vector onto the $e^{Gθ}$-plane and the rotation only affects this part of the vector. An example illustrating this is a rotation of $R(θ)$ in the plane spanned by $R(θ) = e^{Gθ}$ and $P = −G^{2}$,

$$\begin{align} \mathbf{a} &= \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} & \mathbf{b} &= \frac{1}{\sqrt{5}}\begin{bmatrix} 0 \\ 1 \\ 2 \\ \end{bmatrix} \end{align}$$ $$\begin{align} G = \frac{1}{\sqrt{5}}&\begin{bmatrix} 0 & -1 & -2 \\    1 &  0 &  0 \\     2 &  0 &  0 \\  \end{bmatrix} & P = -G^2 &= \frac{1}{5}\begin{bmatrix} 5 & 0 & 0 \\    0 & 1 & 2 \\     0 & 2 & 4 \\  \end{bmatrix} \\ P\begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix} = \frac{1}{5}&\begin{bmatrix} 5 \\ 8 \\ 16 \\ \end{bmatrix} = \mathbf{a} + \frac{8}{\sqrt{5}}\mathbf{b} & R\left(\frac{\pi}{6}\right) &= \frac{1}{10}\begin{bmatrix} 5\sqrt{3} &     -\sqrt{5} &     -2\sqrt{5} \\ \sqrt{5} &  8 + \sqrt{3} & -4 + 2\sqrt{3} \\ 2\sqrt{5} & -4 + 2\sqrt{3} & 2 + 4\sqrt{3} \\ \end{bmatrix} \\ \end{align}$$

Let $ab$, so $30° = π/6$ and its products with $a$ and $b$ are zero. This will allow us to evaluate powers of $N = I - P$.

$$\begin{align} R\left( \frac{\pi}{6} \right) &= N + P\frac{\sqrt{3}}{2} + G\frac{1}{2} \\ R\left( \frac{\pi}{6} \right)^2 &= N + P\frac{1}{2} + G\frac{\sqrt{3}}{2} \\ R\left( \frac{\pi}{6} \right)^3 &= N + G \\ R\left( \frac{\pi}{6} \right)^6 &= N - P \\ R\left( \frac{\pi}{6} \right)^{12} &= N + P = I \\ \end{align}$$

Evaluation by Laurent series
By virtue of the Cayley–Hamilton theorem the matrix exponential is expressible as a polynomial of order $G$−1.

If $n$ and $N^{2} = N$ are nonzero polynomials in one variable, such that $P$, and if the meromorphic function $$f(z)=\frac{e^{t z}-Q_t(z)}{P(z)}$$ is entire, then $$e^{t A} = Q_t(A).$$ To prove this, multiply the first of the two above equalities by $G$ and replace $P$ by $z$.

Such a polynomial $R$ can be found as follows−see Sylvester's formula. Letting $A$ be a root of $a$, $Q_{t}$ is solved from the product of $P$ by the principal part of the Laurent series of $P$ at $f$: It is proportional to the relevant Frobenius covariant. Then the sum St of the Qa,t, where $a$ runs over all the roots of $a$, can be taken as a particular $P(A) = 0$. All the other Qt will be obtained by adding a multiple of $P$ to $P(z)$. In particular, $Q_{t}(z)$, the Lagrange-Sylvester polynomial, is the only $Q_{a,t}(z)$ whose degree is less than that of $P$.

Example: Consider the case of an arbitrary 2×2 matrix, $$A := \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$

The exponential matrix $Q_{t}$, by virtue of the Cayley–Hamilton theorem, must be of the form $$e^{tA} = s_0(t)\, I + s_1(t)\,A.$$

(For any complex number $P$ and any C-algebra $z$, we denote again by $B$ the product of $z$ by the unit of $z$.)

Let $B$ and $α$ be the roots of the characteristic polynomial of $β$, $$P(z) = z^2 - (a + d)\ z + ad - bc = (z - \alpha)(z - \beta) ~ .$$

Then we have $$S_t(z) = e^{\alpha t} \frac{z - \beta}{\alpha - \beta} + e^{\beta t} \frac{z - \alpha}{\beta - \alpha}~,$$ hence $$\begin{align} s_0(t) &= \frac{\alpha\,e^{\beta t} - \beta\,e^{\alpha t}}{\alpha - \beta}, & s_1(t) &= \frac{e^{\alpha t} - e^{\beta t}}{\alpha - \beta} \end{align}$$

if $S_{t}(z)$; while, if $S_{t}(z)$, $$S_t(z) = e^{\alpha t} (1 + t (z - \alpha)) ~,$$

so that $$\begin{align} s_0(t) &= (1 - \alpha\,t)\,e^{\alpha t},& s_1(t) &= t\,e^{\alpha t}~. \end{align}$$

Defining $$\begin{align} s &\equiv \frac{\alpha + \beta}{2} = \frac{\operatorname{tr} A}{2}~, & q &\equiv \frac{\alpha - \beta}{2} = \pm\sqrt{-\det\left(A - sI\right)}, \end{align}$$

we have $$\begin{align} s_0(t) &= e^{st}\left(\cosh(qt) - s\frac{\sinh(qt)}{q}\right), & s_1(t) &= e^{st}\frac{\sinh(qt)}{q}, \end{align}$$

where $Q_{t}$ is 0 if $e^{tA}$, and $A$ if $α ≠ β$.

Thus,

Thus, as indicated above, the matrix $t$ having decomposed into the sum of two mutually commuting pieces, the traceful piece and the traceless piece, $$A = sI + (A-sI)~,$$

the matrix exponential reduces to a plain product of the exponentials of the two respective pieces. This is a formula often used in physics, as it amounts to the analog of Euler's formula for Pauli spin matrices, that is rotations of the doublet representation of the group SU(2).

The polynomial $α = β$ can also be given the following "interpolation" characterization. Define $sin(qt)/q$, and $t = 0$. Then $q = 0$ is the unique degree $S_{t}$ polynomial which satisfies $e_{t}(z) ≡ e^{tz}$ whenever $A$ is less than the multiplicity of $k$ as a root of $a$. We assume, as we obviously can, that $P$ is the minimal polynomial of $P$. We further assume that $A$ is a diagonalizable matrix. In particular, the roots of $A$ are simple, and the "interpolation" characterization indicates that $n ≡ deg P$ is given by the Lagrange interpolation formula, so it is the Lagrange−Sylvester polynomial.

At the other extreme, if $S_{t}(z)$, then $$S_t = e^{at}\ \sum_{k=0}^{n-1}\ \frac{t^k}{k!}\ (z - a)^k ~.$$

The simplest case not covered by the above observations is when $$P = (z - a)^2\,(z - b)$$ with $< n$, which yields $$S_t = e^{at}\ \frac{z - b}{a - b}\ \left(1 + \left(t + \frac{1}{b - a}\right)(z - a)\right) + e^{bt}\ \frac{(z - a)^2}{(b - a)^2}.$$

Evaluation by implementation of Sylvester's formula
A practical, expedited computation of the above reduces to the following rapid steps. Recall from above that an n×n matrix $S_{t}^{(k)}(a) = e_{t}^{(k)}(a)$ amounts to a linear combination of the first $P$−1 powers of $n$ by the Cayley–Hamilton theorem. For diagonalizable matrices, as illustrated above, e.g. in the 2×2 case, Sylvester's formula yields $S_{t}$, where the $A$s are the Frobenius covariants of $B$.

It is easiest, however, to simply solve for these $A$s directly, by evaluating this expression and its first derivative at $P = (z - a)^{n}$, in terms of $B$ and $A$, to find the same answer as above.

But this simple procedure also works for defective matrices, in a generalization due to Buchheim. This is illustrated here for a 4×4 example of a matrix which is not diagonalizable, and the $I$s are not projection matrices.

Consider $$A = \begin{bmatrix} 1 & 1 &          0 &            0 \\  0 & 1 &           1 &            0 \\  0 & 0 &           1 & -\frac{1}{8} \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{bmatrix} ~,$$ with eigenvalues $a ≠ b$ and $exp(tA)$, each with a multiplicity of two.

Consider the exponential of each eigenvalue multiplied by $B$, $exp(tA) = B_{α} exp(tα) + B_{β} exp(tβ)$. Multiply each exponentiated eigenvalue by the corresponding undetermined coefficient matrix $t = 0$. If the eigenvalues have an algebraic multiplicity greater than 1, then repeat the process, but now multiplying by an extra factor of $t$ for each repetition, to ensure linear independence.

(If one eigenvalue had a multiplicity of three, then there would be the three terms: $$B_{i_1} e^{\lambda_i t}, ~ B_{i_2} t e^{\lambda_i t}, ~ B_{i_3} t^2 e^{\lambda_i t} $$. By contrast, when all eigenvalues are distinct, the $t$s are just the Frobenius covariants, and solving for them as below just amounts to the inversion of the Vandermonde matrix of these 4 eigenvalues.)

Sum all such terms, here four such, $$\begin{align} e^{At} &= B_{1_1} e^{\lambda_1 t} + B_{1_2} t e^{\lambda_1 t} + B_{2_1} e^{\lambda_2 t} + B_{2_2} t e^{\lambda_2 t}, \\ e^{At} &= B_{1_1} e^{\frac{3}{4} t} + B_{1_2} t e^{\frac{3}{4} t} + B_{2_1} e^{1 t} + B_{2_2} t e^{1 t} ~. \end{align}$$

To solve for all of the unknown matrices $B$ in terms of the first three powers of $B$ and the identity, one needs four equations, the above one providing one such at $A$ = 0. Further, differentiate it with respect to $t$, $$A e^{A t} = \frac{3}{4} B_{1_1} e^{\frac{3}{4} t} + \left( \frac{3}{4} t + 1 \right) B_{1_2} e^{\frac{3}{4} t} + 1 B_{2_1} e^{1 t} + \left(1 t + 1 \right) B_{2_2} e^{1 t} ~,$$

and again, $$\begin{align} A^2 e^{At} &= \left(\frac{3}{4}\right)^2 B_{1_1} e^{\frac{3}{4} t} + \left( \left(\frac{3}{4}\right)^2 t + \left( \frac{3}{4} + 1 \cdot \frac{3}{4}\right) \right) B_{1_2} e^{\frac{3}{4} t} + B_{2_1} e^{1 t} + \left(1^2 t + (1 + 1 \cdot 1 )\right) B_{2_2} e^{1 t} \\ &= \left(\frac{3}{4}\right)^2 B_{1_1} e^{\frac{3}{4} t} + \left( \left(\frac{3}{4}\right)^2 t + \frac{3}{2} \right) B_{1_2} e^{\frac{3}{4} t} + B_{2_1} e^{t} + \left(t + 2\right) B_{2_2} e^{t} ~, \end{align}$$

and once more, $$\begin{align} A^3 e^{At} &= \left(\frac{3}{4}\right)^3 B_{1_1} e^{\frac{3}{4} t} + \left( \left(\frac{3}{4}\right)^3 t + \left( \left(\frac{3}{4}\right)^2 + \left(\frac{3}{2}\right) \cdot \frac{3}{4}\right) \right) B_{1_2} e^{\frac{3}{4} t} + B_{2_1} e^{1 t} + \left(1^3 t + (1 + 2) \cdot 1 \right) B_{2_2} e^{1 t} \\ &= \left(\frac{3}{4}\right)^3 B_{1_1} e^{\frac{3}{4} t}\! + \left( \left(\frac{3}{4}\right)^3 t\! + \frac{27}{16} \right) B_{1_2} e^{\frac{3}{4} t}\! + B_{2_1} e^{t}\! + \left(t + 3\cdot 1\right) B_{2_2} e^{t} ~. \end{align}$$

(In the general case, $t$−1 derivatives need be taken.)

Setting $n$ = 0 in these four equations, the four coefficient matrices $t$s may now be solved for, $$\begin{align} I &= B_{1_1} + B_{2_1} \\ A &= \frac{3}{4} B_{1_1} + B_{1_2} + B_{2_1} + B_{2_2} \\ A^2 &= \left(\frac{3}{4}\right)^2 B_{1_1} + \frac{3}{2} B_{1_2} + B_{2_1} + 2 B_{2_2} \\ A^3 &= \left(\frac{3}{4}\right)^3 B_{1_1} + \frac{27}{16} B_{1_2} + B_{2_1} + 3 B_{2_2} ~, \end{align} $$

to yield $$\begin{align} B_{1_1} &= 128 A^3 - 366 A^2 + 288 A - 80 I \\ B_{1_2} &=  16 A^3 -  44 A^2 +  40 A - 12 I \\ B_{2_1} &= -128 A^3 + 366 A^2 - 288 A + 80 I \\ B_{2_2} &=  16 A^3 -  40 A^2 +  33 A -  9 I ~. \end{align}$$

Substituting with the value for $B$ yields the coefficient matrices $$\begin{align} B_{1_1} &= \begin{bmatrix}0 & 0 & 48 & -16\\ 0 & 0 & -8 & 2\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}\\ B_{1_2} &= \begin{bmatrix}0 & 0 &  4 &  -2\\ 0 & 0 & -1 & \frac{1}{2}\\ 0 & 0 & \frac{1}{4} & -\frac{1}{8}\\ 0 & 0 & \frac{1}{2} & -\frac{1}{4} \end{bmatrix}\\ B_{2_1} &= \begin{bmatrix}1 & 0 & -48 & 16\\ 0 & 1 & 8 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}\\ B_{2_2} &= \begin{bmatrix}0 & 1 &  8 &  -2\\ 0 & 0 & 0 &  0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix} \end{align}$$

so the final answer is $$e^{tA} = \begin{bmatrix} e^t & te^t & \left(8t - 48\right) e^t\! + \left(4t + 48\right)e^{\frac{3}{4}t} & \left(16 - 2\,t\right)e^t\! + \left(-2t - 16\right)e^{\frac{3}{4}t}\\ 0 & e^t & 8e^t\! + \left(-t - 8\right) e^{\frac{3}{4}t} & -2e^t + \frac{t + 4}{2}e^{\frac{3}{4}t}\\ 0 & 0 & \frac{t + 4}{4}e^{\frac{3}{4}t} & -\frac{t}{8}e^{\frac{3}{4}t}\\ 0 & 0 & \frac{t}{2}e^{\frac{3}{4}t} & -\frac{t - 4}{4}e^{\frac{3}{4}t} ~. \end{bmatrix} $$

The procedure is much shorter than Putzer's algorithm sometimes utilized in such cases.

Illustrations
Suppose that we want to compute the exponential of $$B = \begin{bmatrix} 21 & 17 & 6 \\  -5 & -1 & -6 \\   4 &  4 & 16 \end{bmatrix}.$$

Its Jordan form is $$J = P^{-1}BP = \begin{bmatrix} 4 & 0 &  0 \\  0 & 16 &  1 \\  0 &  0 & 16 \end{bmatrix},$$ where the matrix P is given by $$P = \begin{bmatrix} -\frac14 & 2 &  \frac54 \\ \frac14 & -2 & -\frac14 \\ 0 & 4 &        0 \end{bmatrix}.$$

Let us first calculate exp(J). We have $$J = J_1(4) \oplus J_2(16) $$

The exponential of a 1×1 matrix is just the exponential of the one entry of the matrix, so $λ_{1} = 3/4$. The exponential of J2(16) can be calculated by the formula $λ_{2} = 1$ mentioned above; this yields

$$\begin{align} &\exp \left( \begin{bmatrix} 16 & 1 \\ 0 & 16 \end{bmatrix} \right) = e^{16} \exp \left( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right) = \\[6pt] {}={} &e^{16} \left(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + {1 \over 2!}\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} + \cdots {} \right) = \begin{bmatrix} e^{16} & e^{16} \\ 0 & e^{16} \end{bmatrix}. \end{align} $$

Therefore, the exponential of the original matrix $A$ is $$\begin{align} \exp(B) &= P \exp(J) P^{-1} = P \begin{bmatrix} e^4 & 0 & 0 \\ 0 & e^{16} & e^{16} \\ 0 & 0 & e^{16} \end{bmatrix} P^{-1} \\[6pt] &= {1 \over 4} \begin{bmatrix} 13e^{16} - e^4 & 13e^{16} - 5e^4 & 2e^{16} - 2e^4 \\ -9e^{16} + e^4 & -9e^{16} + 5e^4 & -2e^{16} + 2e^4 \\ 16e^{16}      & 16e^{16}        &  4e^{16} \end{bmatrix}. \end{align}$$

Linear differential equations
The matrix exponential has applications to systems of linear differential equations. (See also matrix differential equation.) Recall from earlier in this article that a homogeneous differential equation of the form $$ \mathbf{y}' = A\mathbf{y} $$ has solution $exp(λ_{i}t)$.

If we consider the vector $$ \mathbf{y}(t) = \begin{bmatrix} y_1(t) \\ \vdots \\y_n(t) \end{bmatrix}  ~,$$ we can express a system of inhomogeneous coupled linear differential equations as $$ \mathbf{y}'(t) = A\mathbf{y}(t)+\mathbf{b}(t).$$ Making an ansatz to use an integrating factor of $B_{i}$ and multiplying throughout, yields $$\begin{align} &           &       e^{-At}\mathbf{y}'-e^{-At}A\mathbf{y} &= e^{-At}\mathbf{b} \\ &\Rightarrow &      e^{-At}\mathbf{y}'-Ae^{-At}\mathbf{y} &= e^{-At}\mathbf{b} \\ &\Rightarrow & \frac{d}{dt} \left(e^{-At}\mathbf{y}\right) &= e^{-At}\mathbf{b}~. \end{align}$$

The second step is possible due to the fact that, if $exp(J_{1}(4)) = [e^{4}]$, then $e^{(λI + N)} = e^{λ} e^{N}$. So, calculating $J_{n}(a)$ leads to the solution to the system, by simply integrating the third step with respect to $B$.

A solution to this can be obtained by integrating and multiplying by $$e^{\textbf{A}t}$$ to eliminate the exponent in the LHS. Notice that while $$e^{\textbf{A}t}$$ is a matrix, given that it is a matrix exponential, we can say that $$e^{\textbf{A}t} e^{-\textbf{A}t} = I$$. In other words, $$\exp{\textbf{A}t} = \exp{{(-\textbf{A}t)}^{-1}}$$.

Example (homogeneous)
Consider the system $$\begin{matrix} x' &=& 2x & -y & +z \\ y' &=&   & 3y & -1z \\ z' &=& 2x & +y & +3z \end{matrix}~.$$

The associated defective matrix is $$A = \begin{bmatrix} 2 & -1 & 1 \\  0 &  3 & -1 \\  2 &  1 &  3 \end{bmatrix}~.$$

The matrix exponential is $$e^{tA} = \frac{1}{2}\begin{bmatrix} e^{2t}\left( 1 + e^{2t} - 2t\right) & -2te^{2t}      &  e^{2t}\left(-1 + e^{2t}\right) \\ -e^{2t}\left(-1 + e^{2t} - 2t\right) & 2(t + 1)e^{2t} & -e^{2t}\left(-1 + e^{2t}\right) \\ e^{2t}\left(-1 + e^{2t} + 2t\right) & 2te^{2t}       &  e^{2t}\left( 1 + e^{2t}\right) \end{bmatrix}~,$$

so that the general solution of the homogeneous system is $$\begin{bmatrix}x \\y \\ z\end{bmatrix} = \frac{x(0)}{2}\begin{bmatrix}e^{2t}\left(1 + e^{2t} - 2t\right) \\ -e^{2t}\left(-1 + e^{2t} - 2t\right) \\ e^{2t}\left(-1 + e^{2t} + 2t\right)\end{bmatrix} + \frac{y(0)}{2}\begin{bmatrix}-2te^{2t} \\ 2(t + 1)e^{2t} \\ 2te^{2t}\end{bmatrix} + \frac{z(0)}{2}\begin{bmatrix}e^{2t}\left(-1 + e^{2t}\right) \\ -e^{2t}\left(-1 + e^{2t}\right) \\ e^{2t}\left(1 + e^{2t}\right)\end{bmatrix} ~, $$

amounting to $$\begin{align} 2x &= x(0)e^{2t}\left(1 + e^{2t} - 2t\right) + y(0)\left(-2te^{2t}\right) + z(0)e^{2t}\left(-1 + e^{2t}\right) \\[2pt] 2y &= x(0)\left(-e^{2t}\right)\left(-1 + e^{2t} - 2t\right) + y(0)2(t + 1)e^{2t} + z(0)\left(-e^{2t}\right)\left(-1 + e^{2t}\right) \\[2pt] 2z &= x(0)e^{2t}\left(-1 + e^{2t} + 2t\right) + y(0)2te^{2t} + z(0)e^{2t}\left(1 + e^{2t}\right) ~. \end{align}$$

Example (inhomogeneous)
Consider now the inhomogeneous system $$\begin{matrix} x' &=& 2x & - & y & + & z & + & e^{2t} \\ y' &=&   &   & 3y& - & z & \\ z' &=& 2x & + & y & + & 3z & + & e^{2t} \end{matrix} ~.$$

We again have $$A = \left[\begin{array}{rrr} 2 & -1 & 1 \\  0 &  3 & -1 \\  2 &  1 &  3 \end{array}\right] ~,$$

and $$\mathbf{b} = e^{2t}\begin{bmatrix}1 \\0\\1\end{bmatrix}.$$

From before, we already have the general solution to the homogeneous equation. Since the sum of the homogeneous and particular solutions give the general solution to the inhomogeneous problem, we now only need find the particular solution.

We have, by above, $$\begin{align} \mathbf{y}_p &= e^{tA}\int_0^t e^{(-u)A}\begin{bmatrix}e^{2u} \\0\\e^{2u}\end{bmatrix}\,du+e^{tA}\mathbf{c} \\[6pt] &= e^{tA}\int_0^t \begin{bmatrix} 2e^u - 2ue^{2u} & -2ue^{2u}   & 0 \\ -2e^u + 2(u+1)e^{2u} & 2(u+1)e^{2u} & 0 \\ 2ue^{2u} & 2ue^{2u}    & 2e^u \end{bmatrix}\begin{bmatrix}e^{2u} \\0 \\e^{2u}\end{bmatrix}\,du + e^{tA}\mathbf{c} \\[6pt] &= e^{tA}\int_0^t \begin{bmatrix} e^{2u}\left( 2e^u - 2ue^{2u}\right) \\ e^{2u}\left(-2e^u + 2(1 + u)e^{2u}\right) \\ 2e^{3u} + 2ue^{4u} \end{bmatrix}\,du + e^{tA}\mathbf{c} \\[6pt] &= e^{tA}\begin{bmatrix} -{1 \over 24}e^{3t}\left(3e^t(4t - 1) - 16\right) \\ {1 \over 24}e^{3t}\left(3e^t(4t + 4) - 16\right) \\ {1 \over 24}e^{3t}\left(3e^t(4t - 1) - 16\right) \end{bmatrix} + \begin{bmatrix} 2e^t - 2te^{2t} &     -2te^{2t} & 0 \\ -2e^t + 2(t + 1)e^{2t} & 2(t + 1)e^{2t} & 0 \\ 2te^{2t} &      2te^{2t} & 2e^t \end{bmatrix}\begin{bmatrix}c_1 \\c_2 \\c_3\end{bmatrix} ~, \end{align}$$ which could be further simplified to get the requisite particular solution determined through variation of parameters. Note c = yp(0). For more rigor, see the following generalization.

Inhomogeneous case generalization: variation of parameters
For the inhomogeneous case, we can use integrating factors (a method akin to variation of parameters). We seek a particular solution of the form $e^{a}/0!$, $$\begin{align} \mathbf{y}_p'(t) & = \left(e^{tA}\right)'\mathbf{z}(t) + e^{tA}\mathbf{z}'(t) \\[6pt] & = Ae^{tA}\mathbf{z}(t) + e^{tA}\mathbf{z}'(t) \\[6pt] & = A\mathbf{y}_p(t) + e^{tA}\mathbf{z}'(t)~. \end{align}$$

For $e^{a}/1!$ to be a solution, $$\begin{align} e^{tA}\mathbf{z}'(t) &= \mathbf{b}(t) \\[6pt] \mathbf{z}'(t) &= \left(e^{tA}\right)^{-1}\mathbf{b}(t) \\[6pt] \mathbf{z}(t) &= \int_0^t e^{-uA}\mathbf{b}(u)\,du + \mathbf{c} ~. \end{align}$$

Thus, $$\begin{align} \mathbf{y}_p(t) & = e^{tA}\int_0^t e^{-uA}\mathbf{b}(u)\,du + e^{tA}\mathbf{c} \\ & = \int_0^t e^{(t - u)A}\mathbf{b}(u)\,du + e^{tA}\mathbf{c}~, \end{align}$$ where $e^{a}/2!$ is determined by the initial conditions of the problem.

More precisely, consider the equation $$Y' - A\ Y = F(t)$$

with the initial condition $e^{At} y(0)$, where
 * $t$ is an $A$ by $n$ complex matrix,
 * $n$ is a continuous function from some open interval $F$ to $e^{−At}$,
 * $$t_0$$ is a point of $I$, and
 * $$Y_0$$ is a vector of $AB = BA$.

Left-multiplying the above displayed equality by $e^{At}B = Be^{At}$ yields $$Y(t) = e^{(t - t_0)A}\ Y_0 + \int_{t_0}^t e^{(t - x)A}\ F(x)\ dx ~.$$

We claim that the solution to the equation $$P(d/dt)\ y = f(t)$$

with the initial conditions $$y^{(k)}(t_0) = y_k$$ for $e^{At}$ is $$y(t) = \sum_{k=0}^{n-1}\ y_k\ s_k(t - t_0) + \int_{t_0}^t s_{n-1}(t - x)\ f(x)\ dx ~,$$

where the notation is as follows:
 * $$P\in\mathbb{C}[X]$$ is a monic polynomial of degree $y_{p}(t) = exp(tA) z(t)$,
 * $I$ is a continuous complex valued function defined on some open interval $f$,
 * $$t_0$$ is a point of $I$,
 * $$y_k$$ is a complex number, and

$y_{p}$ is the coefficient of $$X^k$$ in the polynomial denoted by $$S_t\in\mathbb{C}[X]$$ in Subsection Evaluation by Laurent series above.

To justify this claim, we transform our order $I$ scalar equation into an order one vector equation by the usual reduction to a first order system. Our vector equation takes the form $$\frac{dY}{dt} - A\ Y = F(t),\quad Y(t_0) = Y_0,$$ where $n$ is the transpose companion matrix of $A$. We solve this equation as explained above, computing the matrix exponentials by the observation made in Subsection Evaluation by implementation of Sylvester's formula above.

In the case $P$ = 2 we get the following statement. The solution to $$ y'' - (\alpha + \beta)\ y' + \alpha\,\beta\ y = f(t),\quad y(t_0) = y_0,\quad y'(t_0) = y_1 $$

is $$y(t) = y_0\ s_0(t - t_0) + y_1\ s_1(t - t_0) + \int_{t_0}^t s_1(t - x)\,f(x)\ dx,$$

where the functions $c$ and $Y(t_{0}) = Y_{0}$ are as in Subsection Evaluation by Laurent series above.

Matrix-matrix exponentials
The matrix exponential of another matrix (matrix-matrix exponential), is defined as $$X^Y = e^{\log(X) \cdot Y}$$ $$^Y\!X = e^{Y \cdot \log(X)}$$ for any normal and non-singular $C^{n}$ matrix $n$, and any complex $C^{n}$ matrix $X$.

For matrix-matrix exponentials, there is a distinction between the left exponential $Y$ and the right exponential $^{Y}X$, because the multiplication operator for matrix-to-matrix is not commutative. Moreover,
 * If $X^{Y}$ is normal and non-singular, then $X$ and $X^{Y}$ have the same set of eigenvalues.
 * If $^{Y}X$ is normal and non-singular, $X$ is normal, and $e^{−tA}$, then $0 ≤ k < n$.
 * If $Y$ is normal and non-singular, and $X$, $X$, $Y$ commute with each other, then $n > 0$ and $s_{k}(t)$.

title: Matrix exponential


text:
Mikhael (film)

Mikhael is a 2019 Indian Malayalam-language action thriller film directed by Haneef Adeni and produced by Anto Joseph. It stars Nivin Pauly in the title role, alongside an ensemble cast of Unni Mukundan, Manjima Mohan, Siddique, JD Chakravarthy, Suraj Venjaramoodu, K.P.A.C Lalitha, Ashokan, Jayaprakash, Sudev Nair, Kalabhavan Shajohn, Sijoy Varghese, Babu Antony and Baiju. The music was composed by Gopi Sundar, while Mahesh Narayanan and Vishnu Panicker handled the cinematography and editing.

Mikhael was released on 18 January 2019 and received mixed reviews from critics and audiences, with criticism on the screenplay, plot, and Nivin Pauly's performance. On the other hand, Unni Mukundan's performance, casting, background scores, and climax fight received critical acclaim. The film was a commercial success.

Plot
George Peter, an influential gold smuggler and don in Kerala, and his driver are killed under mysterious circumstances. ACP Muhammed Easa and his colleague CI Isaac John arrive at the spot and discover that the murder was carried out with a surgical blade, deducing that a doctor was behind it. They start investigating medical stores. At the same time, George's cold-blooded younger brother Marco meets George's partners, William Davis and his elder son, Abraham, to ask whether it was their work. When Abraham speaks in a disgusting manner, an agitated Marco shoots him to death. During his funeral, Easa asks William and his younger son Francy about Abraham's murder, but they don't reveal anything.

One night, a mysterious person shoots Patrick, Davis's secretary. Easa and Isaac take him to Star Hospital, where he gets murdered despite being protected with maximum security. This solidifies Easa's and Isaac's theory that the murderer is a doctor. A contract killer interrogates a pharmacist to learn the truth about the murders. Both gangs start believing that someone is after them. Isaac becomes suspicious of Easa since he revealed that he had an informant, but did not reveal anything about him. Isaac investigates on his own, discovering that Dr. Michael John, a young surgeon of Star Hospital is both the killer and the informant. Issac also finds that Easa is helping Mikhael since they have contacted each other several times. Issac forces Easa to reveal everything to prevent Mikhael's arrest.

Past: When Mikhael was 10 years old, he persuaded his father, John, to let him drive his car, which resulted in an accident that killed him, a martial arts trainer. Michael's mother Ancy, blamed him for his father's death. Meanwhile, George's son, Gerald, challenged Jenny, Mikhael's sister, to a kung-fu fight in front of the entire school. Jenny defeats him, causing his friends and other students to tease him for losing to a girl. Depressed over the matter, Gerald commits suicide by jumping from the school terrace. George visits Jenny and asks her to jump from the school terrace in the same way Gerald had jumped or he will kill her entire family.

After George continuously disturbs and harasses her family, Jenny decides to jump off the school terrace. Before she jumps, Mikhael arrives and saves her. After learning about the harassment, Mikhael files a complaint to the police, but the SI thrashes and warns him, as they are allied with George. Ancy asks him to visit them in Christmas, but he refuses. George later kills Ancy, causing Mikhael to burn Peter's Audi Q7 car. George and his henchmen catch Mikhael, beat him brutally, and leave him on a railroad track to die. He manages to escape and eventually catches up to George and his driver and kills them.

Present: Marco learns about Mikhael and Jenny and vows to finish them. Easa also reveals that he ordered Mikhael to kill Patrick since he was a common connection between both teams and killing him will provoke a gang war. Meanwhile, Jenny gets kidnapped by William and Francy as they learn about Mikhael's involvement in Patrick's death, but Michael saves Jenny. William and Francy are arrested by Easa and he also arrests Marco for killing Abraham, but Marco is released on bail. He kidnaps Jenny and takes her to the helipad of a skyscraper and challenges Michael to a fight. Michael arrives at the helipad and fights with Marco. Using a hidden surgical blade in his pocket, Michael kills Marco and illegally donates his organs, thus saving Jenny and his family.

Production
Mikhael is the second directorial venture of Haneef Adeni, produced by Anto Joseph under the banner Anto Joseph Film Company. The film was announced on 12 July 2018. The pooja function was held in Kochi on 3 September, with principal photography beginning the same day.

Principal photography ended on November 26, 2018.

Music
The film features songs composed by Gopi Sundar.

Reception
Mikhael received mixed reviews from critics, with criticism on screenplay, plot and Nivin Pauly's performance, but praise on Unni Mukundan's performance, casting, background scores and climax fight.

Critical response
Sajin Shrijith of Cinema Express gave 4/5 stars and wrote "Mikhael sees Nivin successfully accomplishing something he failed to do in films like Richie and Kayamkulam Kochunni. With a role tailor-made for Nivin's personality, rides solely on the actor's shoulders, and he carries the film effortlessly." Renukha Padmanabhan of The Week gave 3/5 stars and wrote "The narrative takes recourse to frequent flashbacks, but the director makes sure that the audience do not lose track of the happenings as he switches between the past and the present. Gopi Sundar's background score fits well into the proceedings; however, there isn't much novelty and you are left with a sense of deja vu while listening to the score."

Deepa Soman of The Times of India gave 2.5/5 stars and wrote "By the first half itself, the reasons for the events yet to unfold are almost clear and those waiting to be revealed post intermission hardly evoke any thrill as they arrive. The long second half is all about waiting to see what all the hero gets to do to the villains, to punish them." Behindwoods gave 2.5/5 stars and wrote "Mikhael is the kind of film which will entertain you if you do not expect everything to be logical. To enjoy the movie to the fullest you should let the film go in whichever direction it feels like going, and not think too much about anything. There are many characters and stories revolving around Mike's and his sister's story. Some characters are interesting while some are cliches in new covers."

Sowmya Rajendran from The News Minute gave 2/5 stars and wrote "Even if you watch the film knowing what to expect of Haneef Adeni, 'Mikhael' is disappointing." Manoj Kumar of The Indian Express gave 1.5/5 stars and wrote "Adeni could have named this film The Great Brother, instead of Mikhael" and also termed the film as a "migraine-inducing incoherent mess, which kills our desire to be entertained." Navamy Sudheesh of The Hindu called the film an average action fare, wrote that "while the film is engaging at some levels, it lacks that raw energy and intensity of a thorough thriller." Litty Simon of Malayala Manorama wrote "The slow-paced thriller won't really blow your mind, but it comes with enough masala for a Nivin Pauly movie."

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Mikhael (film)


text:
South African Premier Division

The South African Premier Division, officially referred to as the Betway Premiership for sponsorship purposes, is the premier men's professional soccer league and the highest division in the league system of South Africa, organised since 1996.

As the top-level division in South Africa, it is commonly referred to as the Premier Soccer League (PSL) – the administrator of association football in the country.

History
The league was founded in 1996 after an agreement between the National Soccer League and the remnants of the National Professional Soccer League. It is stated on its website that the former name still exists today, but for legal purposes, with the Premier Soccer League being its trade name.

The league was reduced from 18 to 16 teams after the end of the 2001–02 season to avoid fixture congestion, causing the disbandment of two teams; Ria Stars and Free State Stars, though the latter was reformed and competing as of the 2020–21 season. In 2004, a match-fixing scandal rocked the football/soccer landscape in South Africa. An investigation codenamed "Operation Dribble" was launched by the police system of South Africa police in June 2004. More than 40 arrests were made, ranging from club bosses to match commissioners, referees and their assistants. Kaizer Chiefs successfully defended their league title in 2005 after they won the prestigious trophy in 2004 for the first time in a decade. The 2005–06 season saw Mamelodi Sundowns capturing the title for a fourth time. Since the 2017–18 season, they have dominated the league with winning league title after league title.

In 2007, the PSL signed a television deal with SuperSport worth R 1.6 billion, which was the biggest sporting deal in the history of the country and ranked the league into the top 15 in the world in terms of commercial broadcast deals. In the same year, ABSA replaced Castle Lager as the title sponsor, with the latter having sponsored the league since its inception. SuperSport United won 3 league titles in a row between 2008 and 2010.

As of May 2019, the league is rated as the 6th best in Africa according to the CAF 5-year ranking system.

For the 2018–19 season, the PSL gives each club a monthly grant of R 2 million, with funds coming from the television broadcasting rights and national sponsorships, with the league champion earning R 10 million.

In 2019, ABSA increased their sponsorship deal to the PSL to R 39.9 million, which meant that the league winners would secure R 15 million. ABSA would however announce its cancellation of its sponsorship with the PSL a year later on 4 June 2020.

On 24 September 2020, Dr. Irvin Khoza and MultiChoice announced on live television that the new sponsor would be DStv, effectively rebranding the league as the DStv Premiership. MultiChoice also announced that Showmax would replace DStv as the jersey sponsor of SuperSport United F.C. so the latter would be independent to sponsor the league outright.

CEOs

 * 🏴 Trevor Phillips: 1996–1998
 * Joe Ndlela: 1998–2000
 * Robin Petersen: 2000–2001 (resigned)
 * 🏴 Trevor Phillips: 2002–2007
 * 🇳🇴 Kjetil Siem: 2007–2011
 * Zola Majavu: 2011 (resigned)
 * Stanley Matthews: 2012 (resigned)
 * Cambridge Mokanyane: 2013 (acting)
 * Brand de Villers: 2013–2015
 * Mato Madlala: 2016 – present (acting)

Association ranking for 2021–22 CAF competitions
Association ranking for the 2021–22 CAF Champions League and the 2021–22 CAF Confederation Cup will be based on results from each CAF tournament from 2017 to 2020–21.

 * Legend
 * CL: CAF Champions League
 * CC: CAF Confederation Cup
 * {{legend|#FFFFBB|Associations ranked 1–12 and eligible to enter two teams in each CAF tournament}}

Sponsorship
The league has had title sponsorships since its inception. Like the EFL Cup and the English Football League divisions in England, the league takes its title sponsor's name to determine its official common name:

 * 1996–2007: Castle Lager (Castle Premiership)
 * 2007–2020: ABSA (ABSA Premiership)
 * 2020–present: DStv (DStv Premiership)

Format
As of the 2021–22 season, the league is composed of 16 teams competing from August to May each season, similar to the format of most European football leagues. Each team plays the other teams twice in a double round-robin format using the three points for a win system.

At the conclusion of each season, the champion and runner-up of the Premier Division qualify for the CAF Champions League, while the 3rd-place team and the Nedbank Cup champions qualify for the CAF Confederation Cup. The bottom team in the league is automatically relegated and replaced by the winner of the Motsepe Foundation Championship, whiles the team finishing immediately above the bottom team enters a mini-league playoff series with the 2nd and 3rd-placed league-finished teams of the National First Division, with the playoff winners earning their place in the Premiership for the following season.

Broadcasting
The league's broadcaster is SuperSport, who in turn sub-leases the broadcasting rights to the SABC, so as to broadcast some matches on public television in South Africa. SuperSport was awarded an initial $277 million 5-year broadcast/TV rights deal in 2007 by the Premier Soccer League to help commercialize the league globally beyond South Africa. matches on sunset times between Tuesdays and Fridays and on weekends whereas the SABC telecast matches played on Wednesday nights and on weekends.

Since 2016, as SuperSport is a brand owned by the MultiChoice Group, the league is available to watch live and on demand via their specified services; the DStv App (formerly DStv Mobile and DStv Now) and Showmax Pro. Both are accessible via the DStv website for PC and Mac and as mobile apps on Android and iOS/iPadOS for mobile phones, tablets and Smart TVs.

Most successful managers

 * 🇿🇦 Gordon Igesund is the only manager to have won the league with 4 clubs; Manning Rangers 1996-97, Orlando Pirates 2000-01, Santos 2001-02, Mamelodi Sundowns 2006-07.
 * 🇿🇦 Rulani Mokwena become the only manager to have retained the league title 4 times in a row (4) Mamelodi Sundowns 2020–21, 2021–22, 2022–23, 2023-24; breaking previous record of 🇿🇦 Gavin Hunt and 🇿🇦 Pitso Mosimane of the managers that have retained the title the most times; (3) SuperSport United 2007–08, 2008–09, 2009–10, (3) Mamelodi Sundowns 2017–18, 2018–19, 2019–20 respectively.
 * Seven foreign managers have won the league, with 🇷🇴 Ted Dumitru having won the most titles, 4.
 * Since the founding of the Premier Soccer League (PSL) in 1996, the following two records Ernst Middendorp holds remain unsurpassed: he has coached eight different top-flight clubs in the PSL, and has been appointed on twelve occasions as the permanent head coach of a top-flight team competing in the PSL.

League records

 * Ever presents (714 matches): Kaizer Chiefs, Mamelodi Sundowns, Orlando Pirates, SuperSport United
 * Most Premier Division titles: 14 – Mamelodi Sundowns (1997/98), (1998/99), (1999/00), (2005/06), (2006/07), (2013/14), (2015/16), (2017/18), (2018/19), (2019/20), (2020/21), (2021/22), (2022/23), (2023/24)
 * Biggest Premier Division win: 8–1 – SuperSport United vs Zulu Royals (2003/04)
 * Most wins: 373– Mamelodi Sundowns
 * Most draws: 154 – Orlando Pirates
 * Most defeats: 184 – Ajax Cape Town
 * Most goals scored in a season: 73 – Kaizer Chiefs (1998/99)
 * Most goals conceded in a season: 85 – Mother City (1999/00)
 * Most points in a season: 75 – Mamelodi Sundowns (1998/99) & (1999/00) and Kaizer Chiefs (1998/99)
 * Most points in a season (30 games): 71 – Mamelodi Sundowns (2015/16)

League participants
As of the 2023–24 season.

All-time top goalscorers

 * 123 goals: Siyabonga Nomvete
 * 111 goals: Daniel Mudau
 * 104 goals: Manuel Bucuane
 * 104 goals: Mabhuti Khenyeza
 * 101 goals: Siphiwe Tshabalala
 * 101 goals: Collins Mbesuma

NB: list includes all players who have scored at least 100 goals in the PSL era. Also includes cup competitions.

Source:

Founding members
Bold entries are participating in the current season.

 * AmaZulu
 * Bloemfontein Celtic
 * Cape Town Spurs
 * Hellenic
 * Jomo Cosmos
 * Kaizer Chiefs
 * Manning Rangers
 * Michau Warriors
 * Moroka Swallows

 * Orlando Pirates
 * QwaQwa Stars
 * Real Rovers
 * Mamelodi Sundowns
 * SuperSport United
 * Bush Bucks
 * Vaal Professionals
 * Witbank Aces
 * Wits University

title: South African Premier Division


text:
Kitchen Stories

Kitchen Stories (Salmer fra Kjøkkenet) is a 2003 Norwegian film by Bent Hamer, director of Eggs and O' Horten.

Plot
Swedish efficiency researchers come to Norway to study Norwegian men, in an effort to help optimize their use of their kitchens. Folke Nilsson (Tomas Norström) is assigned to study the habits of Isak Bjørvik (Joachim Calmeyer). By the rules of the research institute, Folke has to sit on an umpire's chair in Isak's kitchen and observe him from there, but never talk to him. Isak volunteered for the program with the promise of a horse, but he only receives a dala horse, a little painted wooden statue.

Isak stops using his kitchen and observes Folke through a hole in the ceiling instead. The two lonely men, observer and observed, slowly overcome the initial Norwegian-Swede and subject-observer distrust and become friends. Isak's friend Grant visits him often. Grant is a concentration camp survivor and feels Folke is stealing his friend.

The friendship between Folke and Isak costs Folke his job during an inspection. He is forced to leave and drive up to the Swedish border, but then he returns, only to find Isak has died of a broken heart. Folke, now alone, occupies Isak's home and takes up Isak's friendship with Grant.

Inspiration for the film
Bent Hamer was amused after perusing post-World War II research books on the efficiency of the Swedish housewife, and pondered on the idea of research being done on men. This led him to make the film Kitchen Stories with Swedish researchers and a Norwegian man as the main character.

Cast

 * Joachim Calmeyer as Isak Bjørvik, the man in the kitchen
 * Tomas Norström as Folke Nilsson, the Swedish researcher
 * Bjørn Floberg as Grant, the friend of Isak
 * Reine Brynolfsson as Malmberg
 * Sverre Anker Ousdal as Dr. Jack Zac. Benjaminsen, Swedish researcher
 * Leif Andrée as Dr. Ljungberg, Swedish researcher
 * Gard B. Eidsvold ground crew
 * Lennart Jähkel as Green
 * Trond Brænne as spokesman
 * Bjørn Jenseg as caretaker
 * Jan Gunnar Røise as assistant caretaker
 * Karin Lunden as female corporate assistant

Awards

 * Best Film, Amanda Award, Norway 2003
 * Best Director, Copenhagen International Film Festival 2003
 * Best Director, São Paulo International Film Festival 2003
 * FIPRESCI Prize, Tromsø International Film Festival 2003

title: Kitchen Stories


text:
Protoavis

Protoavis (meaning "first bird") is a problematic taxon known from fragmentary remains from Late Triassic Norian stage deposits near Post, Texas. The animal's true classification has been the subject of much controversy, and there are many different interpretations of what the taxon actually is. When it was first described, the fossils were described as being from a primitive bird which, if the identification is valid, would push back avian origins some 60-75 million years.

The original describer of Protoavis texensis, Sankar Chatterjee of Texas Tech University, interpreted the type specimen to have come from a single animal, specifically a 35 cm tall bird that lived in what is now Texas, USA, around 210 million years ago. Though it existed far earlier than Archaeopteryx, its skeletal structure is more bird-like. Protoavis has been reconstructed as a carnivorous bird that had teeth on the tip of its jaws and eyes located at the front of the skull, suggesting a nocturnal or crepuscular lifestyle. Reconstructions usually depict it with feathers, as Chatterjee originally interpreted structures on the arm to be quill knobs, the attachment point for flight feathers found in some modern birds and non-avian dinosaurs. However, re-evaluation of the fossil material by subsequent authors such as Lawrence Witmer have been inconclusive regarding whether or not these structures are actual quill knobs.

However, this description of Protoavis assumes that Protoavis has been correctly interpreted as a bird. Many palaeontologists doubt that Protoavis is a bird, or that all remains assigned to it even come from a single species, because of the circumstances of its discovery and unconvincing avian synapomorphies in its fragmentary material. When they were found at the Tecovas and Bull Canyon Formations in the Texas panhandle in 1973, in a sedimentary strata of a Triassic river delta, the fossils were a jumbled cache of disarticulated bones that may reflect an incident of mass mortality following a flash flood.

Description
Protoavis is usually depicted as being a bipedal archosaur, similar to several poposaurids and rauisuchids that lived during roughly the same time as Protoavis. In a description published by Sankar Chatterjee, structures were identified as quill knobs, although there has been debate as to whether these are actually quill knobs or not.

Skull and braincase
The braincase of Protoavis is similar in some respects to Troodon, with an enlarged cerebellum that shifted the optic lobes ventrolaterally, and also has a large floccular lobe. The inner ear is also pretty similar and bird-like in both taxa. The canalicular systems and the cochlear process differ in both taxa, and the vestibular region is relatively small and located in a ventral position to most of the anterior and posterior semicircular canals. The anterior semicircular canal is significantly longer than the others, and the cochlear process is a relatively long, vertically oriented tube. However, Protoavis is also remarkedly non-bird like in that it possess only a single exit for the trigeminal. However, these characters are not robust enough to identify Protoavis as a bird.

The skull has an extremely narrow parietal with block like dorsal aspect, very broad, T-shaped frontals that form the "lateral wings" that Chatterjee applies to the lack of postorbitals. There are short curved ulnae with olecranon processes, and a possible scapula with bent shaft, and the cervicals have profiles and aspects to their exterior that are very similar to the Megalancosaurus cervical series. All the cervicals but the most posterior and axis/atlas have hypapophyses and those triangular neural spines; all characteristics that have been described in Megalancosaurus. This suggests that portions of Protoavis may be drepanosaurid in nature.

Chatterjee presents the skull of Protoavis as complete, although only the caudal aspect of the cranium is represented in the available fossils. Chatterjee argues that the temporal region displays a streptostylic quadrate with orbital process for attachment of the M. protractor pterygoidei et quadrati, with associated confluence of the orbits with the temporal fenestrae, thus facilitating prokinesis. He further asserts that the braincase of Protoavis bears a number of characters seen in Ornithurae, including the structure of the otic capsule, the widespread pneumatization of the braincase elements, a full complement of tympanic recesses, and the presence of an epiotic.

Of this material, only the quadrate and orbital roof, in addition to limited portions of the braincase are preserved with enough fidelity to permit any definitive interpretation. The quadrates of TTU P 9200 and TTU P 9201 are not particularly alike; a fact not easily explained away if the material is conspecific, as Chatterjee insists. There does not appear to be an orbital process present on either bone, and the modifications of the proximal condyle permitting wide range of motion against the squamosal, are not readily apparent. Furthermore, the quadratojugal and jugal appear far more robust in the Protoavis specimens themselves, than represented by Chatterjee. The size and development of the quadratojugal seems to contradict Chatterjee's assertion that this bone contacted the quadrate via a highly mobile pin joint. These data render the assertion of prokinesis in the skull of Protoavis questionable at best, and it seems most parsimonious to conclude that the specimen displays a conventional opisthostylic quadrate.

The braincase is where Protoavis comes close to being as avian as Chatterjee has maintained. The otic capsule is allegedly organized in avian fashion, with three distinct foramina arranged as such: fenestra ovalis, fenestra pseudorotunda, and the caudal tympanic recess, with a bony metotic strut positioned between the fenestra pseudorotunda and caudal tympanic recess. The claim that the full complement of tympanic recesses seen in ornithurines, are similarly observed in Protoavis is questionable, as the preservation of the braincase is not adequate to permit concrete observations on the matter. Chatterjee omits in his 1987 account of the braincase, the presence of a substantial post-temporal fenestra, which in all Aves (including Archaeopteryx), is reduced or absent altogether, and the lack of a pneumatic sinus on the paroccipital. Furthermore, the braincase possesses multiple characters symplesiomorphic of Coelurosauria, including an expanded cerebellar auricular fossa, and a vagal canal opening into the occiput. What is preserved of the preorbital skull curiously lacks apomorphic characters to be expected in a specimen, which is allegedly more closely allied to Pygostylia than is Archaeopteryx lithographica. Most telling is the complete absence of accessory fenestrae in the antorbital fossa, leading to maxillary sinuses.

Post-cranial anatomy
The post-cranial remains are as badly preserved, or worse, than the cranial elements, and their interpretation by Chatterjee are in many cases unsubstantiated or speculative. Of the postcranial skeleton, Chatterjee has isolated the axial skeleton as displaying a suite of avian characters, including heterocoelus centra, hypapophyses and reduction of the neural spines. First and foremost, the preservation quality of the vertebrae is poor. While the centra are modified, they do not appear to be truly heterocoelus. The presence of incipient hypapophyses in and of itself might be considered indicative of avian affinity, but their poor development and presence on vertebrae otherwise thoroughly non-avian, is most parsimoniously regarded as mild convergence until further material should be brought to light. The reduction of the neural spines is questionable.

Curiously, Gregory Paul has noted that the cervicals of Protoavis and drepanosaurs are astonishingly similar, such they are hardly distinguishable from one another. Considering the modification of the drepanosaur neck for the purposes of snap-action predation, it becomes more likely that superficial similarities in the cervicals of both taxa are in fact only convergent with Aves. Chatterjee does not identify the remaining vertebrae as particularly avian in their osteology.

Pectoral girdle
The pectoral girdle is discussed by Chatterjee as being highly derived in Protoavis, displaying synapomorphies of avialans more derived than Archaeopteryx, including the presence of a hypocleidium-bearing furcula, and a hypertrophied, carinate sternum. Chatterjee's interpretation of the fossils identified as such in his reviews of the Protoavis material are open to question due to the preservation quality of the elements and as of this time, it is not clear whether either character was in fact present in Protoavis. The glenoid appears to be oriented dorsolaterally permitting a wide range of humeral movement. Chatterjee implies that this is a highly derived trait which allies Protoavis to Aves, but why this should be so is not clearly discussed in the descriptions of the animal. In and of itself, the orientation of the glenoid is not a sufficient basis for placing Protoavis within Aves. The scapular blade is far broader than illustrated by Chatterjee in his 1997 account, and not particularly avian in its gross form. The coracoid, identified by Chatterjee as strut-like and retroverted, is, like the supposed furcula and sternum, too poorly preserved to permit accurate identification. Moreover, the original spatial relationship of the alleged coracoid to the scapula is entirely unknown. Uncinate processes and sternal ribs are missing.

Pelvic girdle
Chatterjee asserts that the pelvic girdle is apomorphic comparative to archaic birds and displays a retroverted pubis, fusion of the ischium and ilium, an antitrochanter, and the presence of a renal fossa. The pubis does appear to display opisthopuby, although this has yet to be verified. The alleged fusion of the ischium and ilium into an ilioischiadic plate is currently not substantiated by the fossils at hand, despite Chatterjee's auspicious illustration to the contrary in The Rise of Birds. At this time the pelvic girdle is not sufficiently well preserved to ascertain whether or not a renal fossa was present, although as no known avian from the Mesozoic displays a renal fossa, it is not clear why Protoavis should, even if it is more derived than Archaeopteryx. Similarly, it is unclear if the alleged antitrochanter has been correctly identified as such.

Arms and legs
The manus and carpus are among the few areas of the Protoavis material which are well preserved, and they are astonishingly non-avian. The distal carpals, while long, are in no way similar to those observed in the urvogel or other archaic birds. There is no semilunate element, and the structure of the radiale and ulnare would have limited the flexibility of the wrist joint. The manus is not tridactyl, and metacarpal V is present. In even the most basal avialian, Archaeopteryx, there is no vestige of the fifth metacarpal and its presence in Protoavis seems incongruous with the claim that it is a bird, let alone one more derived than Archaeopteryx.

Chatterjee claims that the humerus of Protoavis is "remarkably avian", but as in all matters with the fossils referred to this taxon, accurate identification of the elaborate trochanters, ridges, etc., attributed to the humerus by Chatterjee is impossible at this time. The expanded distal condyles, which appear to be present in the humerus of Protoavis and enlarged deltopectoral crest (a ridge for the attachment of chest and shoulder muscles), are congruent with the morphology of ceratosaur humeri, as is the apparent presence of a distal brachial depression.

The femur of Protoavis is astonishingly similar to non-tetanurans, namely coelophysoids. The proximal femur displays a trochanteric shelf caudal to the lesser and greater trochanters, a feature distinguishing non-tetanurans theropods from Tetanurae. Further similarities between the proximal humerus of Protoavis and that of non-tetanuran theropods are found in the shared presence of an enlarged obturator ridge, whose morphology in Protoavis is again, uncannily like that observed in robust basal theropods, e.g., "Syntarus" kayentakatae. The resemblance between the femur of Protoavis and that of a non-tetanuran theropod becomes ever more pronounced at the distal end of the bone. Both share a crista tibiofibularis groove, a feature of a non-tetanuran theropod separating the medial and lateral condyles.

The tibia of Protoavis allegedly possesses both a lateral and cranial cnemial crest, though the validity of this claim is subject to question due to the preservation quality of the material. The fibula is continuous to the astragalocalcaneal unit. A tibiotarsus is absent, unusual considering Chatterjee's claims for the pygostylian affinity of Protoavis, as is a tarsometatarsus. The ascending process of the astragalus is reduced, a character entirely incongruous with a highly derived status for Protoavis. Curiously, such abbreviation of the ascending process is found in ceratosaurs, and in its general osteology, the Protoavis tarsus and pes, is quite similar to those of non-tetanuran theropods. Chatterjee's restoration of the hallux as reversed is nothing more than speculation, as the original spatial relationships of the pedal elements are impossible to ascertain at this time.

Quill knobs
Reconstructions usually depict Protoavis with feathers, as Chatterjee originally interpreted structures on the arm to be quill knobs, the attachment point for flight feathers found in some modern birds and non-avian dinosaurs. However, re-evaluation of the fossil material by subsequent authors such as Lawrence Witmer have been inconclusive regarding whether or not these structures are actual quill knobs.

In his 1997 account, Chatterjee infers the presence of feathers from alleged quill knobs on the badly smashed ulna and metacarpals III and IV, and infers the presence of remiges from such structures (though he does caution that this is uncertain). As is the case with the alleged quill knobs on the ulna, the metacarpal structures appear to be attributable to post-mortem damage. Moreover, the thumb, unlike the case in all birds, is not medially divergent. Considering how poorly preserved the ulna is, it is entirely premature to make any definitive conclusions as to the presence of quill knobs until such time as more adequate material becomes available. Upon further examination of the material no structures were isolated that could be deemed as homologous to remigial papillae.

Classification and taxonomy
The taxonomy of Protoavis is controversial, with several palaeornithologists considering it to be an early ancestor of modern birds, while most others in the palaeontological community regard it as a chimaera, a mixture of several specimens. American palaeontologist Gregory Paul suggested that Protoavis is a herrerasaur. In a paper by Phil Currie and X.J. Zhao discussing a braincase of a Troodon formosus, they compared the bird-like characters of Troodon and Protoavis. In the paper, they made a number of corrections involving both Chatterjee's and Currie's own misinterpretations of parts of Troodon cranial anatomy before the particular braincase being described was found. At least a couple of the corrections (the anterior tympanic recess, and the relatively kinetic quadrate-squamosal contact) made Troodon more bird-like then Chatterjee made out in his Protoavis paper, but overall these particular corrections seemed to have little bearing on the avian features of Protoavis. Currie and Zhao did not explicitly state whether or not they considered Protoavis to be a theropod, however they suggested that although Protoavis has characters suggesting avian affinities, most of these are also found in theropod dinosaurs.

Protoavis is a bird
"The most remarkable thing about Protoavis is that, although it predates Archaeopteryx by 75 million years, it is considerably more advanced than Archaeopteryx...Protoavis is more closely related to modern birds than is Archaeopteryx."

- Sankar Chatterjee

Sankar Chatterjee and a few other palaeornithologists claimed that this material documents a Triassic origin of birds and the presence of a bird more advanced than Archaeopteryx. Though it existed approximately 75 million years before the oldest known bird, its skeletal structure is allegedly more bird-like. Protoavis has been reconstructed as a carnivorous bird that had teeth on the tip of its jaws and eyes located at the front of the skull, suggesting a nocturnal or crepuscular lifestyle. The fossil bones are too badly preserved to allow an estimate of flying ability; although reconstructions usually show feathers, judging from thorough study of the fossil material there is no indication that these were present.

However, this description of Protoavis assumes that Protoavis has been correctly interpreted as a bird. Almost all palaeontologists doubt that Protoavis is a bird, or that all remains assigned to it even come from a single species, because of the circumstances of its discovery and weak avialan synapomorphies in its fragmentary material. When they were found at a Dockum Group quarry in the Texas panhandle in 1984, in a sedimentary stratum of a Triassic river delta, the fossils were a jumbled cache of disarticulated bones reflecting an incident of mass mortality following a flash flood.

Protoavis is a chimaera
"Except for a few elements, the available material of Protoavis is extremely fragmentary. Chatterjee’s interpretations of certain bones are questionable, and even the association of elements into specimens and then into a single taxon seems difficult to support."

- Luis Chiappe

Chatterjee was convinced that some of these crushed bones belonged to two individuals – one old, one young – of the same species. However, only a few parts were found, primarily a skull and some limb bones which moreover do not well agree in their proportions respective to each other, and this has led many to believe that the Protoavis fossil is chimaeric, made up of more than one organism: the pieces of skull appear like those of a coelurosaur, while the femur and ankle bone catalogued under TTU P-9200 and TTU P-9201 respectively suggest affinities to non-tetanuran theropods and at least some vertebrae are most similar to those of Megalancosaurus, a drepanosaurid. However, those supposed similarities between the cervicals of Protoavis and drepanosaurids were the same similarities that Feduccia and Wild (1993) used to argue for an affinity between Archaeopteryx and drepanosaurids.

"'Everywhere one turns; the very fossils ascribed thereto challenge the validity of Protoavis. The most parsimonious conclusion to be inferred from these data is that Chatterjee's contentious find is nothing more than a chimera, a morass of long-dead archosaurs.'"

If it really is a single animal and not a chimera, Protoavis would raise questions about when birds began to diverge from other theropods, if they are a lineage of theropod dinosaurs at all, but until better evidence is produced, the animal's status currently remains uncertain. Furthermore, paleobiogeography suggests that true birds did not colonize the Americas until the Cretaceous; the most primitive undisputed bird-like maniraptorans found to date are all Eurasian. Certainly, the fossils are most parsimoniously attributed to primitive dinosaurian and other reptiles as outlined above. However, coelurosaurs and ceratosaurs are in any case not too distantly related to the ancestors of birds and in some aspects of the skeleton not unlike them, explaining how their fossils could be mistaken as avian. Palaeontologist Zhonghe Zhou stated:

"'[Protoavis] has neither been widely accepted nor seriously considered as a Triassic bird ... [Witmer], who has examined the material and is one of the few workers to have seriously considered Chatterjee’s proposal, argued that the avian status of P. texensis is probably not as clear as generally portrayed by Chatterjee, and further recommended minimization of the role that Protoavis plays in the discussion of avian ancestry.'"

Welman has argued that the quadrate of Protoavis displays synapomorphies of Theropoda. Paul has demonstrated the drepanosaur affinities of the cervical vertebrae. Gauthier & Rowe, and Dingus & Rowe have argued convincingly for identifying the hind limb of Protoavis as belonging to a ceratosaur. Feduccia has argued that Protoavis represents an arboreal "thecodont". In a study of early ornithischian dinosaurs, Sterling Nesbitt and others determined some of the partial remains of Protoavis to be a non-tetanuran theropod. The entire skull and neck are considered to be most likely from a drepanosaurid because the skull and neck are too big compared to the dorsal vertebrae of Protoavis.

In discussions of evolution
Scientists such as Alan Feduccia have cited Protoavis in an attempt to refute the hypothesis that birds evolved from dinosaurs. However, some scientists have claimed the only consequence would be to push the point of bird divergence further back in time. At the time when such claims were originally made, the affiliation of birds and maniraptoran theropods which today is well-supported and generally accepted by most ornithologists was much more contentious; most Mesozoic birds have only been discovered since then. Chatterjee himself has since used Protoavis to support a close relationship between dinosaurs and birds.

"'As there remains no compelling data to support the avian status of Protoavis or taxonomic validity thereof, it seems mystifying that the matter should be so contentious. The author very much agrees with Chiappe in arguing that at present, Protoavis is irrelevant to the phylogenetic reconstruction of Aves. While further material from the Dockum beds may vindicate this peculiar archosaur, for the time being, the case for Protoavis is non-existent.'"

Phylogenetic implications
It has been argued that if valid, Protoavis will represent the death knell to the theropod descent of birds. Palaeontologists counter that if valid, Protoavis in no way falsifies the theropod origin of birds. The very fact that Chatterjee used his putative bird to defend theropod origin seems to contradict the argument of Alan Feduccia that a true bird from the Triassic would bring about the collapse of the theropod "dogma".

Discovery and history
"Smushed and mashed and broken."

- Jacques Gauthier informally describing the holotype specimen.

Archosaur discoveries are comparatively abundant in Texas, and have been recovered in some quantity since E. D. Cope worked the redbeds of the panhandle over a century ago. The holotype specimen of Protoavis (TTU P 9200), the paratype (TTU P 9201), and all referred materials, were discovered in the Dockum Group, from the panhandle of Texas. The Dockum dates from the Carnian through the early Norian, in the terminal Triassic and is composed of four units of decreasing age: the Santa Rose Formation, the Tecovas Formation, the Trujillo Formation, the Cooper Canyon Formation, and the Bull Canyon Formation. Many skeletal elements and partial elements of Protoavis were collected from the Post (Miller) Quarry of the Bull Canyon Formation in the 1980s and other specimens referred to Protoavis were collected from the underlying Kirkpatrick Quarry of the Tecovas Formation. The specimens altogether consists of a partial skull and postcranial remains belonging to possibly several large individuals. The bones were completely freed of the surrounding matrix, and some were heavily reconstructed and the identification of some of the elements have been questioned by other palaeornithologists and palaeontologists.

The type material was collected from mudstone deposits in June 1973 and initially identified as a juvenile Coelophysis bauri. The level of the Dockum group from which the Protoavis material was recovered, was most likely deposited in a deltaic river system. The bone bed excavated by Sankar Chatterjee and his students of Texas Tech University, in which Protoavis was discovered, likely reflects an incident of mass mortality following a flash flood. Chatterjee, who first described Protoavis, has assigned the binomial Protoavis texensis ("first bird from Texas") to the small cache of bones, allegedly conspecific. He interpreted the type specimen to have come from a single animal, specifically a 35 cm tall bird that lived in what is now Texas, USA, between 225 and 210 million years ago.

Due to the nature of the bones being jumbled into sandstone nodules, and completely disarticulated, it has been suggested that Protoavis was reworked from later sediments. However, a basic stratigraphic principle, the "principle of inclusions", is a special case of the principle of cross-cutting relationships. It states that rock has to exist before it can be included in other sedimentary rock. Reworking is the process of weathering fossils or rock containing fossils out of rocks already present, transporting them, and redepositing them in sediments which are later lithified as new sedimentary rocks. Since the Jurassic rocks occurred after the Triassic sediments of the Dockum Group, they could not have been reworked into the Dockum sediments as inclusions.

Palaeoenvironment
The inferred palaeoclimate of the Dockum Group would have been subtropical and governed by a distinct dry/wet season pattern, with the latter marked by monsoonal rains. The botanical evidence indicates that the area was densely forested, and the abundance of both invertebrate and vertebrate material from the site suggests that the locale was in general richly populated by a wide variety of species. Dinosaurs were still fairly rare in the Dockum group, and only some ceratosaurs and other basal forms are well documented. The principal carnivores of the locale would have been poposaurids such as Postosuchus, a species well represented in the Triassic redbeds of Texas. Other archaic archosaurs, such as rhynchosaurs and aetosaurs, were also fairly common.

Taphonomy
Both the holotype and paratype were recovered from disparate locations, both disarticulated and unassociated. Consequently, spatial relationships are impossible to determine. No record of the original orientation of the material even as recovered, exists. Further material assigned to the taxon has been recovered in isolation with no apparent spatial relationships to each other, and more or less has been referred to Protoavis spuriously. Thus, the presentation of the holotype and paratype as coherent skeletons by Chatterjee is fallacious. Such representations are ad hoc conglomerations of bone whose status as conspecific is not apparent from their taphonomy.

Not only were the remains recovered disarticulated and unassociated, there are glaring morphometric differences in the various components of the holotype and paratype. For instance, the scapulae and coracoids are so reduced, that the association with the axial skeleton is extremely difficult to support. Juvenile ontogeny cannot be invoked credibly to explain this discrepancy. Furthermore, the degree of morphometric variation in the holotype and paratype seems incongruent with the component material representing a conspecific assemblage of bones.

The fossils themselves display significant postmortem damage, and are in some cases so badly crushed and distorted at the hand of geological processes, that accurate interpretation thereof is impossible. Consequently, the lucid analyses offered by Chatterjee   are in many cases more artistic creativeness than an accurate description.

In his definitive analysis of the material, The Rise of Birds (1997), Chatterjee failed to illustrate the Protoavis fossils via pictures or sketches of the fossils proper, and instead offers the reader artistic reconstructions. For this, Chatterjee has been sharply criticized. Such an approach in science is entirely intolerable in that it idealizes the material at hand, and obscures the very fragmentary nature of the fossils, and their poor state of preservation.

title: Protoavis


text:
L'Oudon

L'Oudon is a former commune in the Calvados department in the Normandy region in northwestern France. On 1 January 2017, it was merged into the new commune Saint-Pierre-en-Auge.

title: L'Oudon


text:
Baltimora

Baltimora was an Italian music project from Milan, active from 1984 to 1987. They are best known for their 1985 single "Tarzan Boy" and are often considered a one-hit wonder in the United Kingdom and the United States. In other European countries, including their native Italy, Baltimora scored a follow-up hit "Woody Boogie" the same year.

Early years
In early 1984, Maurizio Bassi, a music producer and musician from Milan, met Jimmy McShane, a native of Derry, Northern Ireland. They decided to form an act fronted by McShane, a trained singer, dancer and actor, who had previously tried to break into the West End theater scene and toured with Dee D. Jackson. McShane and Bassi chose the name Baltimora when, one evening together, McShane took a map of the United States, closed his eyes and happened to place his finger on Baltimore. They changed the final letter to an 'a' to make the name more in keeping with the act's Italian roots.

Bassi recruited prominent Italian sessionmen to record their first album, including Giorgio Cocilovo on guitar and Gabriele "Lele" Melotti on drums. Fellow Italo disco producer Tom Hooker has claimed that Baltimora's lead vocals were performed by Maurizio Bassi, as he'd done with Carrara, with McShane sometimes providing the backing vocals. The bulk of Baltimora's song writing fell on Bassi and American lyricist Naimy Hackett, though McShane wrote the lyrics to their track "Survivor in Love".

Success with "Tarzan Boy"
Baltimora's first single, "Tarzan Boy", was released in April 1985, and became a huge European success, peaking at No. 6 on the Italian single chart and entering the top 5 in numerous European countries, including Germany, Switzerland, Austria, Sweden, France, the Netherlands and Norway. The song eventually made it to the United Kingdom charts, where it reached No. 3 in August 1985.

"Tarzan Boy" was released in Canada in October 1985 and peaked on the Canadian charts at No. 5 by the end of the year. However, it took a while for the single to enter the Hot 100 single chart in the United States (where it was released on EMI). When it did, it remained on the Billboard Hot 100 chart for six months, peaking at No. 13 in February 1986. Baltimora performed on the American TV show Solid Gold, which helped further their success in America. The second single, "Woody Boogie", which also gained notable success, entered the top 20 in Germany, Switzerland and Sweden.

Baltimora's first album, Living in the Background, was released in Europe at the end of 1985 and in United States in 1986. Despite the success of "Tarzan Boy", Living in the Background only charted moderately, entering the top 20 album charts only in a few countries in Europe, including Sweden, reaching the no. 49 spot on the US charts.

When Baltimora's second album, Survivor in Love (1987) and the single "Key Key Karimba" failed to chart, Baltimora lost label support and Bassi decided to disband.

Jimmy McShane died of an AIDS-related illness in March 1995.

Studio personnel

 * Maurizio Bassi: session arranger, musician, producer, and songwriter from Milan, Italy.
 * Gaetano Leandro is an Italian pop keyboardist known for his collaboration with American singer Amii Stewart. He joined the group in the studio for the Survivor in Love album, supplementing Bassi's keyboards.
 * Giorgio Cocilovo (born 1956) is an Italian jazz fusion guitarist and music teacher from Milan.
 * Claudio Bazzari (Milan, 12 September 1949) is an Italian blues and country rhythm guitarist, he has played in various groups and as a session musician and has collaborated with many artists, including Fabrizio De André and Mario Lavezzi.
 * Pier Michelatti is an Italian session bassist, known for his collaboration with Pink Project, Fabrizio and Cristiano De André and Baltimora.
 * Dino D'Autorio (born 1 February 1954) is an Italian pop, rock, jazz, funk and blues bassist, and music teacher/arranger from Penne, Italy. He replaced Pier Michelatti as Baltimora's bassist in 1987.
 * Gabriele "Lele" Melotti (born 20 September 1953) is an Italian session drummer and percussionist from Bologna, Italy.

Style
Baltimora's music style is mainly described as Italo disco   and sometimes as new wave, including their first album, Living in the Background.

title: Baltimora


text:
Ouffières

Ouffières is a commune in the Calvados department in the Normandy region in northwestern France.

Geography
The commune is part of the area known as Suisse Normande.

The commune is made up of the following collection of villages and hamlets, Neumer and Ouffières.

The River Orne plus six streams The Neumer, The Flagy, The Aunay Douffieres, The Trois Cours, The Pisseux and The Val de Cropton are the seven watercourses running through the commune.

title: Ouffières