How expressive is $e^A$ in the sense of universal approximation?

Question

For any real matrix $B\in\mathbb{R}^{n\times n},n\ge 2$ and precision $\varepsilon$, is there a real matrix $A\in\mathbb{R}^{n\times n}$ such that $\|e^A-B\|_F<\varepsilon$? ($F$ refers to Frobenius norm)

What if we add some restrictions to $B$?

Answer 1

[EDIT: added and then removed a stronger argument that did not work.]

A partial answer providing a starting point and expanding on the comment:

• if $B$ has no real negative eigenvalues, the answer is yes.
• if $B$ has a simple negative eigenvalue, the answer is no.

Theorem 11.1 on Higham's book Functions of matrices states:

For $B\in\mathbb{C}^{n\times n}$ with no eigenvalues on $\mathbb{R}^{-}$, $$ \log B = \int_0^1 (B-I)(t(B-I)+I)^{-1} dt. $$ Here $\log B$ is a matrix such that $e^{A}=B$. Hence if $B$ has no eigenvalues with negative real part you can reach $B$ exactly, it is in the image of the exponential.

A converse holds: suppose $B$ has a simple real negative eigenvalue $\lambda$; then it has no real logarithm $A$. Indeed, note that we can find a real eigenvector $v$ for $B$. Any matrix logarithm must have an eigenvalue $\log \lambda \not\in\mathbb{R}$ with an associated real eigenvector $v$, as can be seen with a Jordan form. That is impossible, since $Av$ must be real.

To conclude, we must note that having a negative real eigenvalue with odd multiplicity is an open property: if $B$ has a such an eigenvalue $\lambda$, then there is a $\varepsilon >0$ such that any real matrix $C\in\mathbb{R}^{n\times n}$ with $\|B-C\|\leq \varepsilon$ has the same property. Indeed, the characteristic polynomial $p_B(t) = \det (B-tI)$ is continuous in the entries of $B$ and has a sign change at $\lambda$. So we can find $a<b<0$ such that $p_B(a)$ and $p_B(b)$ have opposite signs. A sufficiently close matrix $C$ will have a sufficiently close characteristic polynomial, hence $p_C(a)$ and $p_C(B)$ have opposite signs, and $C$ has an eigenvalue in $[a,b]$ with odd multiplicity, too.

Answer 2

Edit 2 (in fact a complete rewrite): The condition from Noam's comment is almost, but not quite, the right condition. That lies somewhere between Noam's condition and condition (C) below, but lies strictly between the two.

I tried for a bit to find it, but I'm running out of steam now, and perhaps there is circumstantial evidence that the situation is actually quite messy at the level of the fine details (see below).

A result of Culver from 1966 (see Theorem 1 there) implies that an invertible $B$ is a real matrix exponential $B=e^A$ if and only if condition (C) holds.

(C) Every Jordan block corresponding to a negative eigenvalue (if any) occurs an even number of times.

So a matrix $B$ can be approximated by real matrix exponentials if every neighborhood of $B$ contains an invertible matrix satisfying (C).

This clearly holds when $B$ itself satisfies (C) (even when $B$ is not invertible).

Conversely, if there is a negative eigenvalue of odd multiplicity, then a small perturbation can not get rid of it (though it could remove degeneracies), and we won't reach a matrix satisfying (C).

Another type of matrix that can not be approximated is $B=\bigl( \begin{smallmatrix} -1 & 1\\ 0 & -1 \end{smallmatrix} \bigr)$. Again, a small perturbation will either keep the matrix non-diagonalizable with a single Jordan block, or it will give us two distinct negative eigenvalues, and (C) will fail in both cases.

On the other hand, $$ B=\begin{pmatrix} -1 & 1 & 0 & 0\\ 0 & -1 & 0 &0 \\ 0 & 0 &-1&0 \\0&0&0&-1\end{pmatrix} $$ can be approximated by matrix exponentials: now we can remove the degeneracy in the first block (for example by a small perturbation in the $(2,1)$ entry) and then slightly adjust the third and fourth eigenvalues to obtain a diagonalizable matrix with two double negative eigenvalues.