Prove that $e^{iAt}$ is unitary if $A$ is Hermitian

Question

I would like to prove that if $A$ is Hermitian matrix, then exponential $e^{iAt}$ for $t \in \mathbb{R}$ is unitary. This means I need to show that $e^{iAt}[e^{iAt}]^\dagger=[e^{iAt}]^\dagger e^{iAt} = I$. I have a proof (below) but I would like to verify if my reasoning is correct.

I started with spectral decomposition of $A$ and definition of a matrix exponential: $$ [e^{iAt}]^\dagger = \Big[\sum_{j=1}^n e^{i \lambda_j t} u_j^\dagger u_j \Big]^\dagger = \sum_{j=1}^n (e^{i \lambda_j t})^* [u_j^\dagger u_j]^\dagger = \sum_{j=1}^n e^{-i \lambda_j t} [u_j^\dagger u_j] = e^{-iAt}, $$ where $\lambda_j$ is eigenvalue of $A$ and $u_j$ is a respective eigenvector.

This is possible because firstly eigenvalues $\lambda_j \in \mathbb{R}$ thanks to hemiticity of $A$, hence complex conugate to $e^{i \lambda_j t}$ is $e^{-i \lambda_j t}$, and secondly $[u_j^\dagger u_j]^\dagger = u_j^\dagger (u_j^\dagger)^\dagger = u_j^\dagger u_j$.

So, now we have $e^{iAt}[e^{iAt}]^\dagger = e^{iAt}e^{-iAt}= e^{itA-itA}$ which is possible because $A$ apparently commutes with itself. Finnaly $e^{itA-itA} = e^O = I$ and similarly for the matrices switched. So, the $e^{iAt}$ is unitary.

Is my proof correct?

Answer 1

It's correct.

Here's another more elementary proof. First notice that for any $M\in\mathcal{M}_n(\mathbb{C})$, $$\exp(M)^\dagger = \left(\sum_{k=0}^{\infty} \frac{M^k}{k!}\right)^\dagger=\sum_{k=0}^\infty \frac{(M^\dagger)^k}{k!}=\exp(M^\dagger)$$

This is correct because, $M\mapsto M^\dagger$ is ($\mathbb{R}$)-linear (therefore continuous) and because $(AB)^\dagger = B^\dagger A^\dagger$.

Now, in the case where $M=itA$ with $A$ hermitian, this implies $$\exp(itA)^\dagger=\exp((iAt)^\dagger)=\exp(-iAt)$$