Is the map $z \mapsto (A^z)_{i,j}$ analytic?

Question

Let $A \in M_n(\mathbb{C})$ be a matrix satisfying $A=A^*$ and $x \ne 0 \implies \langle Ax,x\rangle > 0$. For such a matrix, we can define $A^z$ where $z \in \mathbb{C}$ (I suppose, for example using the continuous functional calculus on the $C^*$-algebra $M_n(\mathbb{C})$). Is it true that the map $$\mathbb{C}\to \mathbb{C}: z \mapsto (A^z)_{i,j}$$ is analytic (i.e. complex differentiable)?

Answer 1

Following the hint of @Abdelmalek Abdesselam in the comment above, let me give an answer.

Since $A$ is a self-adjoint, there is a unitary $U \in M_n(\mathbb{C})$ such that $A = U^* DU$ where $$D= \operatorname{Diag}(d_1, \dots, d_n)$$ is a diagonal matrix.

Given $z \in \mathbb{C}$, consider the function $$f_z: \sigma(A) = \sigma(D) \subseteq [0, \infty) \to \mathbb{C}: x \mapsto x^z$$

choose a sequence of polynomials $\{p_n^z\}_{n=1}^\infty$ such that $\|p_n^z-f^z\|_\infty \to 0$ when $n \to \infty$. Then by functional calculus $$A^z = f_z(A) = \lim_n p_n^z(A) = \lim_n p_n^z(U^* DU) =\lim_n U^*p_n^z(D)U = U^*D^z U.$$

Note that $D^z = \operatorname{Diag}(d_1^z, \dots, d_n^z)$. To see this, simply observe that $$p_n^z(D) = \operatorname{Diag}(p_n^z(d_1), \dots, p_n^z(d_n))$$ The left hand side converges to $D^z$ when $n \to \infty$, while the right hand-side converges to $$\operatorname{Diag}(d_1^z, \dots, d_n^z)$$ using the fact that convergence in $M_n(\mathbb{C})$ is entrywise. It thus follows that $$(A^z)_{ij}= (U^* D^zU)_{ij}= (U^* \operatorname{Diag}(d_1^z, \dots, d_n^z) U)_{i,j}= P(d_1^z, \dots, d_n^z)$$ for some polynomial $P$ in $n$-variables (not-depending on $z\in \mathbb{C}$). Hence, it is clear that $$\mathbb{C}\to \mathbb{C}: z \mapsto (A^z)_{ij}$$ is an analytic function, as desired.