My apologies if this isn't a well-enough-posed question, I think I'm partly unsure of what exact question to even ask.

There are many different ways in which we can take a function of a matrix.

  1. We can apply a holomorphic function to any matrix, using power series, Cauchy's integral formula, or even just applying the known formula to the Jordan blocks.
  2. We can apply any continuous (or up to measure zero, measurable) function to a normal matrix by approximating with Stone-Weierstrass, or applying the function to the eigenvalues.
  3. We can apply a smooth (or even $C^k$, where $k$ is the size of the largest Jordan block) function to any matrix with real eigenvalues, by either using the formula on Jordan blocks or by approximating with Stone-Weierstrass in the $C^k$ norm.

My general question is therefore:

What happens when we try to apply general smooth functions of $\mathbb{C}$ to general matrices with complex eigenvalues?

If a matrix is diagonalizable, we can easily just diagonalize it, and apply any function to the eigenvalues. Even if it isn't diagonalizable, it seems we should still be able to apply the same formula to the Jordan blocks. While $f$ is not holomorphic, it is still totally legal to apply the differential operator $\partial_z=\frac{1}{2}(\partial_x-i\partial_y)$ to a smooth function.

Specifically, I am wondering:

  1. Is there something that goes wrong when we try to do this? Are there discontinuities, or some other taken-for-granted property that breaks down? And if so, is this true even in the diagonalizable case?
  2. In all the other cases of matrix functions, there are methods of computing them that don't involve actually taking a Jordan decomposition. Is there such a method for smooth functions of complex matrices? If not, is there any rigorous sense in which there can't be?
  3. Absent answers to the previous two questions, is there some other reason why I haven't ever heard of this before?

1 Answer 1


This depends of which amount of functional calculus you wish for the map $(f,A)\mapsto f(A)$. What seems reasonable is that $f\mapsto f(A)$ be a morphism of algebra, and $\sigma(f(A))=f(\sigma(A))$. At least, this is what happens when $A$ is normal, or if we restrict to holomorphic functions $f$. But this does not work in a more general setting, because functions commute but matrices don't. Let me give an example within ${\bf M}_n({\mathbb C})$.

On the one hand, you need a definition of $c(A)$ where $c:z\mapsto\bar z$ is complex conjugation. I can accept either $c(A)=\bar A$ or $c(A)=A^*$, the Hermitian conjugate. But there are many matrices for which $A\bar A\ne\bar A A$ and $AA^*\ne A^*A$, so how will you define $f(A)$ when $f(z)=|z|^2=zc(z)$ ?

  • $\begingroup$ Under what I proposed in the question, $c(A)$ would not equal to $\overline{A}$ or $A^*$. Instead, (for a diagonalizable matrix at least) it would correspond to diagonalizing the matrix, and replacing the diagonal matrix with its complex conjugate. This would necessarily commute with $A$ because diagonal matrices commute, and in fact equals $A^*$ precisely for normal matrices. $\endgroup$ Commented Jan 11, 2022 at 20:38

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