Properties of the per-element exponential (Hadamard exponential) for matrices

I'm asking this question mostly out of curiosity, though I do also have a potential application.

In linear algebra we usually define the matrix emponential as $e^A = I + A + \frac{1}{2}A^2 + \frac{1}{6}A^3 + \dots$, which has lots of nice properties. However, we could also define a different kind of "matrix exponentiation", which I'll write $e^{\circ A}$, where $(e^{\circ A})_{ij} = e^{A_{ij}}$, i.e. we just apply the exponential function to each element independently.

After writing this question I guessed that the name of this operation would be "Hadamard exponential." An internet search revealed that it's mentioned by this name in a few textbooks and research papers, but in general I can find very little written about its properties from a linear algebra point of view. (I've edited this post to use what seems to be standard notation for the Hadamard exponential.)

One obvious thing is that it inherits all the usual properties of exponentiation, as long as we use the Hadamrd product $(\circ)$ (i.e. per-element multiplication) instead of the usual matrix product. Then we can immediately apply results like the Schur product theorem to conclude that if $e^{\circ A}$ and $e^{\circ B}$ are both positive definite then so is $e^{\circ (A+B)}$. Another obvious property is that for real matrices, the elements of $e^{\circ A}$ are all positive, and hence the Perron-Frobenius theorem applies.

However, what I would particularly like to know is whether anything can be said about the eigenvalues and eigenvectors of $e^{\circ A}$ in terms of the eigendecomposition of $A$. I suspect that there is no straightforward relationship in general, but I would expect there to be inequality constraints.

In short, my question is, has the operation I've called $\operatorname{eexp}$ been studied in linear algebra, and what is known about its properties?

• For the Hadamard exponential, we can use the fact that $\exp(x) > x$ to state that if $A$ is positive semi-definite, then $\exp(\circ A)$ is positive definite. – EA304GT Oct 25 '18 at 2:15

1 Answer

There are some wonderful theorems regarding entrywise functions of matrices, especially regarding positive definite matrices. In particular:

• If $A,B$ are positive semidefinite, so is $A \circ B$
• If $A$ is positive semidefinite, then so is $e^{\circ A}$.
• Define $f[A]$ to be an entrywise function of (real) square matrices (of arbitrary size). Then $f$ takes positive definite matrices to positive definite matrices if and only if it is an analytic function whose power series has only non-negative coefficients.

These results are apparently important in the context of numerical analysis, especially when it comes to thresholding (i.e. rounding values to zero while keeping the resulting error to within certain bounds). See also this question on MO.

Another quick result is, if $\|\cdot\|$ denotes the Frobenius norm, then $$\|A\circ B\| \leq \|A\| \cdot \|B\| \\ \left\| e^{\circ A}\right\| \leq e^{\|A\|}$$ One last quick and useful result: if $u,v$ are column vectors, then $$A \circ (uv^T) = \operatorname{diag}(u) A \operatorname{diag}(v)$$

• Thank you, this is helpful. A couple of questions: (1) can you recommend a good resource that covers this, for further reading? (2) do you happen to know if those positive semidefiniteness results are restricted to Hermitian/symmetric matrices, or if they extend to the more general definition that applies to any complex/real matrix? – Nathaniel Jan 1 '16 at 5:23
• There's a chapter of Horn and Johnson with some of this in it, but I ran across what I know at a colloquium. In this context, postive semi definite means "Hermitian and positive semidefinite". – Omnomnomnom Jan 1 '16 at 13:09
• For the record, both definitions are general. One definition requires that $xAx^*$ be positive (non-negative) for all $x\in \Bbb C^n$, and the other just requires it be positive for $x\in \Bbb R^n$. – Omnomnomnom Jan 1 '16 at 13:13
• Ah, that's too bad, I'm mostly interested in non-Hermitian matrices. I've seen parts of the Horn and Johnson chapter in a Google books preview - I'll try and get a hard copy and read the rest. – Nathaniel Jan 1 '16 at 13:17
• I don't know how worth the investment that would be, honestly; it's just a part of chapter 7 that covers what you're looking for. Also, H and J only ever talks about Hermitian positive semidefinite matrices. – Omnomnomnom Jan 1 '16 at 13:23