Arctan of Jordan Matrix I am trying to find a formula for the arctan of a Jordan matrix. The answer would be a 2x2 matrix of lambda values.
I was thinking of using the Taylor series somehow but Im really unsure about it. Please let me know if you have any ideas.
 A: HINT: As usual, write your matrix $A=\lambda I + J$, where $J=\left[\begin{matrix} 0&1\\0&0\end{matrix}\right]$. Expand $\arctan(x)$ in power series about $x=\lambda$. What happens when you do that and substitute $x=A$?
A: It can be guessed that
$\arctan[J]=\left[\begin{array}{cc}
\arctan(\lambda)&\frac{1}{1+\lambda^2}\\0&\arctan(\lambda)
\end{array}\right]$,
for
$ 
J=\left[\begin{array}{cc}
\lambda&1\\0&\lambda
\end{array}\right]$ and with $|\lambda|<1$.
This is by numerical experiments.
A: Let
$$ A = \left[\begin{array}{cc}\lambda & 1\\0&\lambda\end{array}\right]$$
It can be decomposed
$$ A= D+N$$
where $$ D= \left[\begin{array}{cc}\lambda & 0\\0&\lambda\end{array}\right],\qquad  N = \left[\begin{array}{cc}0 & 1\\0&0\end{array}\right]$$
Let's assume that $f$ is a function that can be expanded into a Taylor series:
$$ f(x) = \sum_{n=0}^\infty a_n x^n$$
Let us note that
$$ f'(x) = \sum_{n=0}^\infty a_n nx^{n-1}$$
We then define
$$ f(A) = \sum_{n=0}^\infty a_n A^n$$
(assuming this series is convergent in some sense). Since matrices $D$ and $N$ commute, we can use the Newton's binomial formula to get
$$ A^n = (D+N)^n = \sum_{k=0}^n \left(\begin{array}{c}n\\k\end{array}\right) D^{n-k} N^k$$
We have $N^0 = 1$ and $N^k = 0$ for $k\ge 2$, so
$$ A^n = D^n + nD^{n-1}N$$
$$ f(A) = \sum_{n=0}^\infty \left(a_n D^n + a_n n D^{n-1}N\right) = f(D) + f'(D) N$$
That is
$$ f\left(\left[\begin{array}{cc}\lambda & 1\\0&\lambda\end{array}\right]\right) = \left[\begin{array}{cc}f(\lambda) & f'(\lambda)\\0&f(\lambda)\end{array}\right]$$
This can be used also for Jordan matrices of bigger dimension, but then we need to keep more terms in the expansion, as in bigger dimension we need a higher power $k$ so that $N^k=0$.
