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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.

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up vote 3 down vote accepted

The following was explained to me by Jean-Pierre Ferrier.

For any matrix $a$ in $A:=M_n(\mathbb C)$, write $\Lambda(a)$ for the set of eigenvalues of $a$. Let $U$ be an open subset of $\mathbb C$, and 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, and let $\mathcal O(U)$ be the $\mathbb C$-algebra of holomorphic functions on $U$. 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. (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) There is an $r>0$ and a neighborhood $N$ of $a$ in $A$ such that $$ f(b)=\frac{1}{2\pi i}\ \sum_{\lambda\in\Lambda(a)}\ \int_{|z-\lambda|=r}\ \frac{f(z)}{z-b}\ dz $$ 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. 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 (i) is clear. To prove (ii) we can keep on assuming $\mathbb C[a]\simeq\mathbb C[X]/(X-\lambda)^m$. On replacing $a$ with $a-\lambda$, we can even assume $a^n=0$. Choose $r>0$ so that $U$ contains the closed disk of radius $r$ centered at $0$, let $N$ be the set of those $b$ in $A$ whose eigenvalues $\lambda$ satisfy $|\lambda|<r/2$, and let $b$ be in $N$. Replacing $a$ with $b$ in the above argument, we can assume $b^n=0$. Now (ii) follows from Cauchy's Integral Formula and the equalities $$ f(b)=\sum_{k=0}^{n-1}\ \frac{f^{(k)}(0)}{k!}\ b^k,\quad \frac{1}{z-b}=\sum_{k=0}^{n-1}\ \frac{b^k}{z^{k+1}}\quad. $$

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