Matrix exponent eigenvalues

Question

I was thinking about this for quite some time and didn't find it nowhere. So I hope someone can help me.

let $e:\frak{gl}$$(n,R)$ $\to Gl(n,R)$ be the exponential application.

I know that if $v$ is a eigenvetor of $A$ with eigenvalue $\lambda$, then $$e^Av=e^{\lambda}v.$$

My question is:

if $e^Av=v$ for a nonzero $v$ (i.e. 1 is a eigenvalue of $e^A$), can I conclude that $A$ necessarily has $0$ as one of its eigenvalue?

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I'm specially interested in the case that A is $$ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ I want to prove that if $e^Av=v$ then $v=0$ (because A doesn't have $0$ as it eigenvalue).

Answer 1

if $e^Av=v$ for a nonzero $v$ (i.e. 1 is a eigenvalue of $e^A$), can I conclude that $A$ necessarily has $0$ as one of its eigenvalue?

No. For instance,

$$\begin{pmatrix} 0 & - 2 \pi \\ 2 \pi & 0 \end{pmatrix}$$

does not have 0 as an eigenvalue, yet its matrix exponential is the identity matrix, which does have 1 as an eigenvalue.

I'm specially interested in the case that A is $$ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$

The exponential of this matrix is $$ \begin{pmatrix} \cos(1) & -\sin(1) \\ \sin(1) & \cos(1) \end{pmatrix}$$

which has only complex eigenvalues and so does not have any fixed vectors.

I want to prove that if $e^Av=v$ then $v=0$ (because A doesn't have $0$ as it eigenvalue).

For this particular $e^A$, the statement is true, but not for the reason you claim, but rather from an explicit computation of the nullspace of $e^A$.