Holomorphic function of a matrix

A statement is made below. The questions are:

(a) Is the statement true?

(b) If it is, does it appear in the literature?

Here is the statement.

For any matrix $A$ in $M_n(\mathbb C)$, write $\Lambda(A)$ for the set of eigenvalues of $A$.

Recall that there is a unique continuous $\mathbb C[X]$-algebra morphism $$\mathcal O(\Lambda(A))\to M_n(\mathbb C),$$ where $\mathcal O(\Lambda(A))$ is the algebra of those functions which are holomorphic on (some open neighborhood of) $\Lambda(A)$. Recall also that this morphism is usually denoted by $f\mapsto f(A)$. (Here $X$ is an indeterminate.)

Let $U$ be an open subset of $\mathbb C$, let $U'$ be the open subset of $M_n(\mathbb C)$ defined by the condition $$\Lambda(A)\subset U,$$ and let $f$ be holomorphic on $U$. (The fact the $U'$ is open follows from Rouché's Theorem.)

STATEMENT. The map $A\mapsto f(A)$ from $U'$ to $M_n(\mathbb C)$ is holomorphic.

-

The following was explained to me by Jean-Pierre Ferrier, and Ahmed Jeddi made useful comments.

For any matrix $a$ in $A:=M_n(\mathbb C)$, write $\Lambda(a)$ for the set of eigenvalues of $a$. For each $\lambda$ in $\Lambda(a)$, write $1_\lambda\in\mathbb C[a]$ for the projector onto the $\lambda$-generalized eigenspace parallel to the other generalized eigenspaces, and put $a_\lambda:=a1_\lambda$, and $z_\lambda:=z1_\lambda$ for $z$ in $\mathbb C$.

Let $U$ be an open subset of $\mathbb C$, let $U'$ be the subset of $A$, which is open by Rouché's Theorem, defined by the condition $\Lambda(a)\subset U$, let $a$ be in $U'$, let $X$ be an indeterminate, let $\mathcal O(U)$ be the $\mathbb C$-algebra of holomorphic functions on $U$, and equip $\mathcal O(U)$ and $\mathbb C[a]$ with the $\mathbb C[X]$-algebra structures associated respectively with the element $z\mapsto z$ of $\mathcal O(U)$ and the element $a$ of $\mathbb C[a]$.

Theorem 1. (i) There is a unique $\mathbb C[X]$-algebra morphism from $\mathcal O(U)$ to $\mathbb C[a]$. We denote this morphism by $f\mapsto f(a)$.

(ii) The map $U'\ni a\mapsto f(a)\in A$ is holomorphic.

(iii) For any $a$ in $U'$ we have $$f(a)=\sum_{\lambda\in\Lambda(a),0\le k<n}\frac{f^{(k)}(\lambda)_\lambda}{k!}\ (a-\lambda)^k.$$

Proof of (i) and (iii). By the Chinese Remainder Theorem, $\mathbb C[a]$ is isomorphic to the product of $\mathbb C[X]$-algebras of the form $\mathbb C[X]/(X-\lambda)^m$, with $\lambda\in\mathbb C$. So we can assume that $\mathbb C[a]$ is of this form, and the lemma follows from the fact that, for any $f$ in $\mathcal O(U)$, there is unique $g$ in $\mathcal O(U)$ such that $$f(z)=\sum_{k=0}^{m-1}\ \frac{f^{(k)}(\lambda)}{k!}\ (z-\lambda)^k+(z-\lambda)^mg(z)$$ for all $z$ in $U$. q.e.d.

Say that a cycle is a formal finite sum of smooth closed curves. Let $\gamma$ be a cycle in $U\setminus\Lambda(a)$ such that $I(\gamma,\lambda)=1$ for all $\lambda\in\Lambda(a)$ (where $I(\gamma,\lambda)$ is the winding number of $\gamma$ around $\lambda$), and let $N$ be the set of those $b$ in $A$ such that $\Lambda(b)\subset U$, and that $\gamma$ is a cycle in $U\setminus\Lambda(b)$ satisfying $I(\gamma,\lambda)=1$ for all $\lambda\in\Lambda(b)$. As already observed, Rouché's Theorem implies that $N$ is an open neighborhood of $a$ in $A$. Theorem 2 below will imply Part (ii) of Theorem 1.

Theorem 2. We have $$f(b)=\frac{1}{2\pi i}\int_\gamma\ \frac{f(z)}{z-b}\ dz$$ for all $f$ in $\mathcal O(U)$ and all $b$ in $N$. In particular the map $b\mapsto f(b)$ from $U'$ to $A$ is holomorphic.

Proof. We have $$\frac{1}{2\pi i}\int_\gamma\ \frac{f(z)}{z-b}\ dz$$ $$=\frac{1}{2\pi i}\int_\gamma\ \frac{f(z)}{z-b}\ \sum_{\lambda\in\Lambda(b)}1_\lambda\ dz$$ $$=\sum_{\lambda\in\Lambda(b)}\frac{1_\lambda}{2\pi i}\int_\gamma\ \frac{f(z)}{z-b}\ dz$$ $$=\sum_{\lambda\in\Lambda(b)}\frac{1_\lambda}{2\pi i}\int_\gamma\ \frac{f(z)\ dz}{(z-\lambda)-(b-\lambda)}$$ $$=\sum_{\lambda\in\Lambda(b)}\frac{1_\lambda}{2\pi i}\int_\gamma\ f(z)\ \sum_{k=0}^{n-1}\ \frac{(b-\lambda)^k}{(z-\lambda)^{k+1}}\ dz$$ $$=\sum_{\lambda\in\Lambda(b),0\le k<n}\frac{1_\lambda}{2\pi i}\int_\gamma\ \frac{f(z)\ dz}{(z-\lambda)^{k+1}}\ (b-\lambda)^k$$ $$\overset{(*)}{=}\sum_{\lambda\in\Lambda(b),0\le k<n}I(\gamma,\lambda)\ \frac{f^{(k)}(\lambda)_\lambda}{k!}\ (b-\lambda)^k$$ $$=\sum_{\lambda\in\Lambda(b),0\le k<n}\frac{f^{(k)}(\lambda)_\lambda}{k!}\ (b-\lambda)^k$$ $$\overset{(**)}{=}f(b),$$ where Equality $(*)$ follows from the Residue Theorem, and Equality $(**)$ from Part (iii) of Theorem 1. q.e.d.

-
@ Pierre-Yves Gaillard , I just read your post. We can find your theorem 2, as 6.2.28 p. 427 (or problem 33, p.445) in Horn, Johnson, Topics in matrix analysis. They only say that the function $A\rightarrow f(A)$ is continuous ! – loup blanc Jan 17 at 13:25
@loupblanc - Thank you very much for your interesting comment! – Pierre-Yves Gaillard Jan 17 at 13:37