Matrix exponential using the Schur decomposition

Question

I have a Hermitian $m\times m$ matrix, say $A$. I can use Schur decomposition and transform the matrix in to $A=QTQ^{\dagger}$. Is it then possible to calculate straightforward the matrix exponential using $\exp[A]=Q\cdot\exp[-a T]\cdot Q^{\dagger}$, where $a>1$ is a scalar and $\dagger$ denotes the conjugate transpose of $Q$. Thanks for any suggestion.

Answer 1

If $A=Q^*DQ$, where $Q^*Q=I$, then $$ \mathrm{e}^{tA}=Q^*\mathrm{e}^{tD}Q, $$ for all $t\in\mathbb R$ (even $t\in\mathbb C$.)

In particular, if $A$ is symmetric (in general, hermitian), the tridiagonal matrix provided by the Schur decomposition is diagonal.