If $A$ is real and nonsymmetric with Schur decomposition $UTU^H$, then what types of matrices are $U$ and $T$? How are the eigenvalues of $A$ related to $U$ and $T$?
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See here. – S.B. Jul 05 '13 at 11:45
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You have two possibilities:
- (Complex) Schur decomposition: $U$ and $T$ are complex, with $U$ being unitary ($U^*U = {\rm I}$) and $T$ being (usually upper) triangular.
- Real Schur decomposition: $U$ and $T$ are real, with $U$ being orthogonal ($U^TU = {\rm I}$) and $T$ being (usually upper) quasitriangular (it has diagonal blocks of order $1$ and $2$).
In both cases, since the Schur decomposition is a (unitary/orthogonal) similarity of $A$ and $T$, these two have the same eigenvalues.
In case 1 the diagonal elements of $T$ are the eigenvalues of $T$ (and, consequently, of $A$). In case 2, the eigenvalues are the elements of the diagonal blocks of $T$ of order $1$ and the eigenvalues of the diagonal blocks of $T$ of order $2$ (these eigenvalues are complex conjugate pairs and can be computed directly).
Given a normal matrix $A$, columns of $U$ in case 1 are its eigenvectors.
If $A$ is symmetric, then it has only real eigenvalues, so cases 1 and 2 are the same: $U$ is real unitary (meaning orthogonal), $T$ is real diagonal (with the eigenvalues on the diagonal), and the columns of $U$ are (orthonormal) eigenvectors of $A$.
Vedran Šego
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You said that $A$ is real, so Hermitian = symmetric. If $A$ is complex (not necessarily real) Hermitian, then you can only use case 1, in which you get a unitary $U$ and a real diagonal $T$. – Vedran Šego Jul 05 '13 at 20:13
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yeah yeah I know I said A real, I just wanted to ask what would happend if A is Hermitian (with complex). Thanks – piere Jul 06 '13 at 20:41