TITLE: $\sin(A)$, where $A$ is a matrix
QUESTION [23 upvotes]: If $A$ is an $n\times n$ matrix with elements $a_{ij}$ $i=$i'th row, $j=$j'th column. Then  $e^A$ is also a matrix as can be seen by expanding it in a power series.Is $e^A$ always convergent and defined for any $n\times n$ matrix? What are the elements $e^A_{ij}$ in terms of $a_{ij}$ ?
$\sin(x)$ for real numbers $x$ can be interpreted easily geometrically by looking at the unit circle. Is there any geometrical interpretation of $\sin(A)$ when $A$ is a matrix?
What applications does it have?
Slight addition:
Is $\sin(A)$ periodic in any sense, i.e. is there a matrix $B$ not $0$ such that $\sin(A+B)=\sin(A)$ for all matrices $A$?
Does $\sin(A)^2+\cos(A)^2=I$ hold?
And do all the regular rules from algebra transfer, i.e. $e^Ae^B=e^{(A+B)}$ ?
Is there a consistent definition of $M/N$ for matrices $M,N$, such that $e^A/e^B=e^{(A-B)}$ holds?

REPLY [2 votes]: First version of the answer
Let $A$ be an $n$ by $n$ complex matrix, let 
$$
M(X)=\prod\ (X-\lambda_j)^{m_j}\ \in\mathbb C[X]
$$ 
be its minimal polynomial (the $\lambda_j$ being distinct and the $m_j$ positive), let $U\subset\mathbb C$ be an open set containing the $\lambda_j$, let $B$ be the $\mathbb C[X]$-algebra of holomorphic functions on $U$. 
Then there is a unique continuous $\mathbb C[X]$-algebra morphism from $B$ to $\mathbb C[A]$. 
Denote it by $f\mapsto f(A)$. 
For each $f$ in $B$, there is a unique polynomial $P(X)\in\mathbb C[X]$ of degree less than $\deg M(X)$ which satisfies $P(A)=f(A)$. 
Moreover $P(X)$ is given by the formula 
$$
P(X)=\sum_j\ T_j\left(T_j(f)\ \frac{(X-\lambda_j)^{m_j}}{P(X)}\right)\frac{P(X)}{(X-\lambda_j)^{m_j}}\quad, 
$$ 
where $T_j(g)$ or $T_j(g(X))$ denotes the degree less than $m_j$ Taylor polynomial of $g$ at $\lambda_j$, viewed as an element of $\mathbb C[X]$. 
The matrix $A$ is diagonalizable if and only if all the $m_j$ are equal to $1$. In this case, we have 
$$ 
P(X)=\sum_j\ \prod_{k\neq j}\ \frac{X-\lambda_k}{\lambda_j-\lambda_k}\quad.
$$
Same answer with less indices
Let $A$ be an $n$ by $n$ complex matrix, let 
$$
M(X)=\prod_{\lambda\in\Lambda}\ (X-\lambda)^{m_\lambda}\ \in\mathbb C[X]
$$ 
be its minimal polynomial (the $m_\lambda$ being positive), let $U\subset\mathbb C$ be an open set containing $\Lambda$, let $B$ be the $\mathbb C[X]$-algebra of holomorphic functions on $U$. 
Then there is a unique continuous $\mathbb C[X]$-algebra morphism from $B$ to $\mathbb C[A]$. 
Denote it by $f\mapsto f(A)$. 
For each $f$ in $B$, there is a unique polynomial $P(X)\in\mathbb C[X]$ of degree less than $\deg M(X)$ which satisfies $P(A)=f(A)$. 
Moreover $P(X)$ is given by the formula 
$$
P(X)=\sum_{\lambda\in\Lambda}\ T_\lambda\left(T_\lambda(f)\ \frac{(X-\lambda)^{m_\lambda}}{P(X)}\right)\frac{P(X)}{(X-\lambda)^{m_\lambda}}\quad, 
$$ 
where $T_\lambda(g)$ or $T_\lambda(g(X))$ denotes the degree less than $m_\lambda$ Taylor polynomial of $g$ at $\lambda$, viewed as an element of $\mathbb C[X]$. 
The matrix $A$ is diagonalizable if and only if all the $m_\lambda$ are equal to $1$. In this case, we have 
$$
P(X)=\sum_{\lambda\in\Lambda}\ \ \prod_{\mu\in\Lambda\setminus\{\lambda\}}\ \frac{X-\mu}{\lambda-\mu}\quad.
$$