matrix wise tangent inverse (arctan) Given a matrix $X$, an expression for the matrix cosine and sine are given by
$$
\textrm{cos}(X) = \frac{e^{iX} + e^{-iX}}{2}\\
\textrm{sin}(X) = \frac{e^{iX} - e^{-iX}}{2i}
$$
I have been trying to find a convenient expression for the arctan with no luck. Is there any expression for the arctan?
$$
\textrm{tan}^{-1}(X) = \text{?}
$$
 A: Using Taylors formula (or term-by-term differentiation and geometric series), one can show that the Taylor series for $\tan^{-1}(x)$ or $\arctan(x)$ is the following 
$$\arctan(x)=\sum\limits_{k=0}^{\infty}\frac{(-1)^k}{2k+1}x^{2k+1}=x-\frac{x^3}{3}+\frac{x^5}{5}+\cdots,$$
which is absolutely convergent for $|x|<1$. This can be shown using the Ratio Test.
Using similar reasoning, we can identify a possible candidate for infinite series expansion of $\arctan(A)$ as
$$\arctan(A)=\sum\limits_{k=0}^\infty \frac{(-1)^k}{2k+1}A^{2k+1}=A-\frac{A^3}{3}+\frac{A^5}{5}+\cdots$$
However, we will also need to be very careful about when this series converges. It will converge if the matrix is small enough in the sense of the matrix norm being less than one $||A||<1$.
This results in a very nice situation if $A$ is a diagonalizable matrix, i.e., if $A=P^{-1}\Lambda P$ where $\Lambda=\left[\begin{array}{cccc}\lambda_1&0&\cdots&0\\0&\lambda_2&0\cdots&0\\
\vdots&0_\vdots&\ddots&\vdots\\
0&0&\cdots0&\lambda_n\end{array}\right]$, $P=[{\bf v}_1,...,{\bf v}_n]$, and $\lambda_k,{\bf v}_k$ is the eigenvalue/vector pair coming from diagonalization.
Specifically, consider the following sum. If we assume that $|\lambda_k|<1$ for each eigenvalue then the following computations are valid.
$$\begin{align*}
\sum\limits_{k=0}^\infty \frac{(-1)^k}{2k+1}A^{2k+1}&=A-\frac{A^3}{3}+\frac{A^5}{5}+...\\
&=P^{-1}\Lambda P-\frac{(P^{-1}\Lambda P)^3}{3}+\frac{(P^{-1}\Lambda P)^5}{5}+...\\
&=P^{-1}\Lambda P-\frac{(P^{-1}\Lambda^3 P)}{3}+\frac{(P^{-1}\Lambda^5 P)}{5}+...\\
&=P^{-1}(\Lambda-\frac{\Lambda^3}{3}+\frac{\Lambda^5}{5}+...)P\\
&=P^{-1}\left[\begin{array}{cccc}\arctan(\lambda_1)&0&\cdots&0\\0&\arctan(\lambda_2)&0\cdots&0\\
\vdots&0_\vdots&\ddots&\vdots\\
0&0&\cdots0&\arctan(\lambda_n)\end{array}\right]P
\end{align*}
$$
I hope this helps. Please let me know if you have any questions.
