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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.

Rodrigo de Azevedo
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user123
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    For a Hermitian matrix the Schur decomposition is, in fact, diagonalization, i.e. $T$ is diagonal, so $\exp(\lambda T)=\exp(\operatorname{diag}{\lambda t_k})=\operatorname{diag}{\exp(\lambda t_k)}$. – A.Γ. Jun 06 '17 at 09:48

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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.

Yiorgos S. Smyrlis
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  • Do you happen to have a reference for this? I encountered a reference book for matrix computational algorithms about nine months ago, but misplaced it, and cannot recall the authors. Thank you in advance --- And a minute later I actually found (although I have been searching for days): "Functions of Matrices: Theory and Computation" by Nicholas J. Higham, 2008. – ChrisR Sep 16 '17 at 07:10