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Let $f\colon \mathbb R\to\mathbb R$ be a function and $A\in M_n:=Hom(\mathbb R^n,\mathbb R^n)$ a linear operator with real eigenvalues and diagonal Jordan form. Then one naturally defines an operator $f(A)\in M_n$ by diagonalizing $A$ and applying $f$ to each eigenvalue. This yields a map from a subset of $M_n$ to $M_n$. Let me denote this map by $F$ rather than $f$ to avoid confusion.

In many cases $F$ naturally extends to all operators. For example, if $f$ is polynomial then one can substitute any matrix into the polynomial, if $f(x)=e^x$ then there is a matrix exponent, and so on. This gets more complicated with functions like square root, sign, etc, but they still can be applied to large sets of operators.

Is there a general theory for this type of constructions, when $f$ is not necessarily analytic? More precisely,

Question: If $f$ is smooth ($C^1$, $C^\infty$, etc), is $F$ smooth as well?

By the the smoothness of $F$, which is defined on some strange subset of $M_n$, I mean that it has a smooth extension to an open subset of $M_n$. Also, hopefully the smooth extension can be constructed naturally so that it is $GL(n)$-equivariant.

More generally, $f$ may be defined on an open subset of $\mathbb R$ or $\mathbb C$, then $F$ is defined on the set of diagonalizable operators whose eigenvalues all belong to the domain of $f$, and one asks the same question. (In the complex case, $f$ is not assumed analytic, only smooth in the sense of real analysis).

My primary motivation is curiosity, but recently I've run across some special cases of the question in differential geometry and data analysis. I think I can prove that the answer is affirmative if "smooth" is replaced by "locally Lipschitz" and construct a counter-example if "smooth" is replaced by "continuous". From some old papers I got the impression that perhaps this kind of results were well-known long ago, but I could not find any references.

Update. I was wrong about Lipschitz continuity. If $f(x)=|x|$ and $A=\begin{pmatrix} 0 & \varepsilon \\ 0 & 0 \end{pmatrix}$, then $A$ can be approximated by diagonalizable operators with real eigenvalues of either sign, so by continuity one would have to define $F(A)=A$ and $F(A)=-A$ at the same time. This probably means that $C^1$ should be removed from the wishlist and the question makes sense only for $C^\infty$ smoothness.

A type of application that I have in mind is like this: given a smooth family of operators, one wants to apply a function to all of them and keep control over the derivatives. Some interesting functions are $f(z)=\frac{z}{|z|}$ for $z\in\mathbb C\setminus\{0\}$ and non-degenerate operators, or $f$ being a bump function which preserves eigenvalues near some distinguished point and annihilates those that are far away.

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    $\begingroup$ In case you don't know the buzzword yet: this is called "functional calculus" en.wikipedia.org/wiki/Functional_calculus $\endgroup$ Commented Apr 30 at 8:38
  • $\begingroup$ I haven't tried to check the details, but one idea is to use variants of Sylvester's formula for non-diagonalizable matrices. I think if the Jordan form is diagonal then those formulas degenerate into the usual Sylvester's formula, and that is clearly smooth if $f$ is. $\endgroup$ Commented Apr 30 at 14:53
  • $\begingroup$ Thank you for the pointers. @Paul, why is the Sylvester formula smooth? I don't see this from its appearance. When a matrix has eigenvalues with multiplicities, even the number of terms in the formula may be different for nearby matrices. $\endgroup$ Commented Apr 30 at 16:12
  • $\begingroup$ Yes that's true, the argument has to use the more general formula throughout if it's going to work. When I get the time I'll try to write out what I have in mind as an answer, at which point I will find out for sure whether the idea works. I'll leave a comment if it doesn't. $\endgroup$ Commented Apr 30 at 18:57

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If $A = V \operatorname{diag}(J_1, \dots, J_s) V^{-1}$ is a Jordan decomposition, then $f(A) = V \operatorname{diag}(f(J_1), \dots, f(J_s)) V^{-1}$, where functions of Jordan blocks are defined as $$ J_i = \begin{bmatrix} \lambda_i & 1\\ & \lambda_i & \ddots\\ & & \ddots & 1\\ & & & \lambda_i \end{bmatrix} \in \mathbb{C}^{k_i\times k_i}, \quad f(J_i) := \begin{bmatrix} f(\lambda_i) & f'(\lambda_i) & \dots & \frac{f^{(k_i-1)}(\lambda_i)}{(k_i-1)!}\\ & f(\lambda_i) & \ddots & \vdots\\ & & \ddots & f'(\lambda_i)\\ & & & f(\lambda_i) \end{bmatrix}. $$ The presence of higher derivatives is surprising at first, but this is the only natural way to define functions of non-diagonalizable matrices. You can convince yourself of the validity of this formula for polynomials by computing powers and polynomials of Jordan blocks. This classical formula for a matrix function is equivalent to the other natural definitions based on power series, Cauchy integrals, Hermite interpolation, or diagonalization + continuity.

This formula shows that in general you "lose" $k-1$ orders of differentiability, where $k$ is the dimension of the largest Jordan block of the matrix you are applying your function to. The formula also shows that one needs sufficient regularity in the function to define $f(A)$ on a non-diagonalizable $A$; it is not sufficient that $f(\lambda_i)$ is defined for all eigenvalues.

A good reference for functions of (finite) matrices is Higham's book Functions of matrices.

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    $\begingroup$ Yes, but I think this procedure is known to the OP (it's outlined in the first paragraph of OP's question) and it's not quite what the question is about. Rather the question asks if the so obtained function is smooth in $A$, which now boils down to whether and when the eigenvalues of $A$ depend smoothly on $A$. This is discussed in one of the related MO questions: Conditions for smooth dependence of the eigenvalues and eigenvectors of a matrix on a set of parameters. $\endgroup$ Commented May 1 at 10:37
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    $\begingroup$ @M.G. OP mentions he knows how to define matrix functions for matrices with diagonal Jordan form, though. $\endgroup$ Commented May 1 at 11:33
  • $\begingroup$ @M.G. But you're right that what I wrote doesn't fully solve the question whether the matrix-valued function is smooth in $A$. I don't think it is something that can be settled just by looking at continuity of eigenvectors and eigenvalues: eigenvectors are typically not differentiable when there are nontrivial Jordan forms, but matrix functions can be. $\endgroup$ Commented May 1 at 11:38
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    $\begingroup$ Yes, apologies, you are right, OP explicitly says diagonal Jordan form rather than just Jordan form. $\endgroup$ Commented May 1 at 12:53
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    $\begingroup$ @Federico Thank you, the book is helpful and inspiring. There is Cauchy integral formula there, it implies smoothness of the matrix function if $f$ is analytic. If $f$ is not analytic, then it might be the case that $f$ cannot be extended to a neighborhood of a matrix with non-single eigenvalues. Even if all eigenvalues are real, non-real eigenvalues probably always occur in a neighborhood, and then one needs complex differentiablity, which implies analyticity. But it is so strange that even a simple cutoff function does not work. I have to think more about it. $\endgroup$ Commented May 1 at 15:59

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