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Is there anything noteworthy about the following set?

$$ \Omega_k ≔ \left\{e^A \mid \operatorname{rank}(A)=k \right\} $$

Are there any special algebraic properties to $e^A$ in this case? Does the manifold have any significant properties? What's its dimensionality?

Obviously $Ω_0 = \{\}$. In the case $k=1$, we have $A = u v^\top$ for some vectors $u,v$, hence

$$ \Omega_1 = \Big\{+\frac{e^{u^⊤v}-1}{u^\top v} uv^\top \mid u, v \in \Bbb R^n, u v^\top \neq 0 \Big\} $$

So $Ω_1$ consists of a subset of the set of rank-1 perturbations of the identity matrix. Note that $\frac{e^{u^⊤v}-1}{u^⊤v}$ is a shorthand for $φ_1(u^⊤v)=∑_{k∈ℕ} \frac{(u^⊤v)^k}{(k+1)!}$ which is defined even when $u⟂v$.

Rodrigo de Azevedo
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Hyperplane
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  • In general, because ${A \mid \operatorname{rank}(A) = k}$ is not a manifold, we shouldn't expect its image under the exponential map to be a manifold – Ben Grossmann Sep 06 '22 at 20:44
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    @BenGrossmann Are you sure? My understanding of differential geometry is pedestrian, but I have seen quite a few articles saying this is a manifold. Are you maybe thinking of ${A∣\operatorname{rank}(A)≤k}$? – Hyperplane Sep 06 '22 at 21:13
  • You're right, that is what I had in mind, and some more careful googling confirms your claim – Ben Grossmann Sep 06 '22 at 21:22
  • Ignoring the case where things aren't diagonalizable (because I haven't thought deeply enough about it), if $Av=\lambda v$, then $e^Av=e^{\lambda}v$, and so $A$ being rank $k$ means $\ker A$ is $n-k$ dimensional, means that $\ker e^A -I$ is $n-k$ dimensional. Whether you view this to be an algebraic property is perhaps up to you. But I would expect that you're looking at invertible matrices $B$ where $B-I$ has specified rank. Except that e^${2\pi i}=1$, so this doesn't quite work. Hmmm.... – Aaron Sep 07 '22 at 15:59

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Since matrix exponentiation commutes with conjugation, and because of the rank-nullity theorem, we can reduce the problem to understanding the image of Jordan blocks.

If $N$ is nilpotent, then $N$ and $kI$ commute, so $e^{kI+N}=e^{kI}e^N=e^ke^N$, and so we can further reduce to understanding $e^N$.

Lemma: if $N^m\neq 0$ but $N^{m+1}=0$, then $(e^{N}-I)^m\neq 0$ but $(e^{N}-I)^{m+1}= 0$

Proof: $e^N-I=N(I+N/2+N^2/6+\ldots)$, so $(e^N-I)^{\ell}$ is a a multiple of $N^{\ell}$, which is zero if $\ell>m$. On the other hand, fully expanding out the power series and using $N^{m+1}=0$ we get $(e^N-I)^m=N^m\neq 0$.

With this, one can show that if $J$ is a single Jordan block, then $e^J$ is conjugate to a single Jordan block.

As as consequence of this observation, one can show that $\ker(e^A-I)$ is the direct sum of the eigenspaces of $A$ with eigenvalue $2\pi i k, k\in \mathbb Z$. Further, every matrix $B$ such that $B$ is invertible and $\dim\ker B-I \geq n-k$ can be written as $e^A$ for some $A$ with rank $k$.

Finally, appealing to the rank nullity theorem, we conclude

$$\Omega_k=\{B\in GL_n(\mathbb C) \mid \operatorname{rank}(B-I)\leq k\}$$

With this final result in mind, there is actually a simple argument to get half of it:

Because $\operatorname{im}(AB)\subset \operatorname{im}(A)$, we have that $\operatorname{im}(e^A-I)=\operatorname{im}(A(I+A/2+A^2/6+\ldots))\subset \operatorname{im}(A)$, and so the rank is non-increasing. This gives us inclusion one way.

Unfortunately, I don't yet see a way to get the other inclusion without something akin to JNF.

Aaron
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  • Of that is very nice, so in fact, in some sense the matrix exponential indeed maps low rank matrices to matrices "close" to the identity. – Hyperplane Sep 08 '22 at 11:42
  • In your argument, you allowed complex valued matrices, but I am currently only interested in real matrices, where my example shows that $Ω_1 ⊆{B∈_n(ℝ)∣\operatorname{rank}(B-)=1}$. In fact, I believe if we restrict $A$ to be real, we get $Ω_k = {B∈_n(ℝ)∣\operatorname{rank}(B-)=k}$. – Hyperplane Sep 19 '22 at 10:57
  • @Hyperplane No, the issue is eigenvalues of $2\pi i k$, which can still happen even if the matrix is real. For example, if $A$ is $2\pi$ times the $90^{\circ}$ rotation matrix, then $e^A=I$. However, what I think will happen is that because complex eigenvalues/eigenvectors will pair up under conjugation, the rank must drop by an even number. – Aaron Sep 19 '22 at 11:58
  • You're right, so in the end we should get $Ω_k = {B∈_n(ℝ)∣\operatorname{rank}(B−)∈S_k}$ where $S_k = {k, k-2, k-4, …}$. – Hyperplane Sep 19 '22 at 13:03
  • @Hyperplane You have containment, but I don't know that you have equality. The image of the matrix exponential map is apparently not all of $GL_n(\mathbb R)$, so I would have to put in more thought before I had a coherent picture of what is going on. – Aaron Sep 19 '22 at 22:30
  • This gives a clue https://math.stackexchange.com/q/191228/99220 – Hyperplane Sep 19 '22 at 22:40