Arctan of Jordan Matrix

Question

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.

Answer 1

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.

Answer 2

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 $x=A$?

Answer 3

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 $N^k=0$.