Multiplicity of an Eigenvalue of the Exponential of an Operator

Question

I am trying to prove the following:

Let $T:\mathbf R^n\to\mathbf R^n$ be a linear operator. Then $$\det e^T=e^{\text{Trace}(T)}$$

To do this I took the following apporach.

Let $\lambda\in\mathbf R$ be an eigenvalue of $T$. Then $Tv=\lambda v$ for some nonzero vector $v\in \mathbf R^n$.

Noting that $e^Tv=e^\lambda v$, we get that $e^\lambda$ is an eigenvalue of $e^T$.

So I now try to show that the multiplicity of $\lambda$ for $T$ is same as the multiplicity of $e^\lambda$ for $e^T$.

To do that I need to show that $\dim \ker(T-\lambda I)^n=\dim \ker(e^T-e^\lambda I)^n$.

But I have no idea how to go further.

Can anybody help?

(I have seen a proof of the above mentioned result using a completely different approach but still it would be of some importance to explicitly establish the fact that, if it is true, the multiplicity of $\lambda$ for $T$ is same as the multiplicity of $e^\lambda$ for $e^T$).

Answer 1

First assume $\lambda=0$. Take $v\in ker(T)^n$. We have to show $v\in ker(T-I)^n$. Then $$ (e^T-I)v =\sum_{k=1}^\infty \frac1{k!}T^kv=\sum_{k=1}^{n-1} \frac1{k!}T^kv. $$ We prove per Induction that $$ (e^T-I)^j v = T^j p_j(T)v, $$ where $p_j$ is a polynomial. Assume that this claim is true for all $j$ with $1\le j\le n-1$. Then $$ (e^T-I)^{j+1} =(e^T-I) T^j p_j(T)v = T^j p_j(T) (e^T-I)v = T^{j+1} p_j(T)p_1(T)v. $$ Setting $p_{j+1}:=p_1p_j$ finishes the induction. Hence it follows $(e^T-I)^nv=0$. This proves $ker(T)^n \subset ker(e^T-I)^n$.

Second, let $\lambda\ne0$. Set $\tilde T:=T-\lambda I$. Then $e^{\tilde T} = e^{T-\lambda I} = e^{-\lambda}e^T$ as $T$ and $I$ commute. From the first step, it follows $$ ker(T-\lambda I)^n = ker(\tilde T)^n\subset ker(e^{\tilde T}-I )^n = ker(e^{-\lambda}e^T-I)^n = ker(e^T-e^\lambda I)^n. $$

Answer 2

@ caffeinemachine , your required equality (concerning the kernels) is true if $spectrum(T)\subset\mathbb{R}$ (because $\exp$ is injective over $\mathbb{R}$) but, otherwise, is false. A priori, $T$ admits non-real eigenvalues ; how do you proceed with them ?

To show (*): $\det(e^T)=e^{tr(T)}$, there are $2$ methods.

1. Staying over $\mathbb{R}$. Show that $\dfrac{d}{dt}\log(\det(e^{tT}))=tr(T)$. You must know the derivative of "$\det$".

2. Going up over $\mathbb{C}$. Use the fact that $T$ is similar to a triangular matrix $U$ and calculate the diagonal of $e^U$. Thus we obtain the eigenvalues of $e^T$ with multiplicity and therefore we deduce (*). Moreover this method is valid for any matrix function $f$ instead of $\exp$ (cf. Matrix functions, Higham).