Analytical function of a matrix - equivalence of the definitions

Question

It may seem obvious but actually I have not found any FORMAL proof that the definition of the analytical function via power series and via Jordan form are equivalent. Can anybody help me and explain this one?

Answer 1

To fix a context, let us say $f: \mathbb{D} \to \mathbb{C}$ is analytic on the unit disk, with power series $f(z) = \sum_{k=0}^\infty c_kz^k$. For a matrix $A \in \mathbb{C}^{n\times n}$, let $f(A) = \sum_{k=0}^\infty c_kA^k$, which converges if $||A|| <1$ for some matrix norm $||\cdot||$.

Claim: If $B$ is an invertible matrix such that $||BAB^{-1}|| < 1$, then $$f(BAB^{-1}) = Bf(A)B^{-1}.$$

Proof: $(BAB^{-1})^k = BA^kB^{-1}$ for all $k \geq 0$, so $$\sum_{k=0}^N c_k(BAB^{-1})^k = B\left(\sum_{k=0}^N c_kA^k\right)B^{-1}.$$ Then take the limit as $N \to \infty$, observing that conjugation by $B$ is continuous.

Therefore, it suffices to compute $f$ of a matrix in Jordan form. It further suffices to compute $f$ of a Jordan block, since for a block diagonal matrix $$ \begin{bmatrix} A & 0 \\ 0 & A'\end{bmatrix}^k = \begin{bmatrix}A^k & 0 \\ 0 & A'^k\end{bmatrix}.$$ If $J_{n+1}(\lambda)$ is the Jordan block of dimension $n+1$ with eigenvalue $\lambda$, it can be proven that $$ J_{n+1}(\lambda)^k = \begin{bmatrix} \lambda^k & \binom{k}{1}\lambda^{k-1} & \binom{k}{2}\lambda^{k-2} &\cdots & \binom{k}{n}\lambda^{k-n} \\ 0 & \lambda^k & \binom{k}{1}\lambda^{k-1} & \cdots & \binom{k}{n-1}\lambda^{k-n+1} \\ 0 & 0 & \lambda^k & \cdots & \binom{k}{n-2}\lambda^{k-n+2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda^k \end{bmatrix}$$ by induction and the binomial formula. Hence, the matrix series $f(J_{n+1}(\lambda))$ has entries $$ \sum_{k=0}^\infty c_k \binom{k}{d} \lambda^{k-d} = \sum_{k=d}^\infty c_k \frac{k!}{d!(k-d)!} \lambda^{k-d} = \frac{1}{d!} \sum_{k=d}^\infty c_k (k\cdot (k-1) \cdots (k-d+1))\lambda^{k-d} $$ which is $1/d!$ times the term-by-term $k$th derivative of $f$ at $\lambda$. Taking the derivative term-by-term is justified since $f$ is analytic on the disk.

This shows that $$ f(J_{n+1}(\lambda)) = \begin{bmatrix} f(\lambda) & f'(\lambda) & \frac{1}{2!}f''(\lambda) & \cdots & \frac{1}{n!}f^{(n)}(\lambda) \\ 0 & f(\lambda) & f'(\lambda) & \cdots & \frac{1}{(n-1)!}f^{(n-1)}(\lambda) \\ 0 & 0 & f(\lambda) & \cdots & \frac{1}{(n-2)!}f^{(n-2)}(\lambda) \\ 0 & 0 & 0 & \cdots & \frac{1}{(n-3)!}f^{(n-3)}(\lambda) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & f(\lambda) \end{bmatrix},$$ as desired.