Given that a symmetric matrix with real entries has orthogonal eigenvectors, is the converse true? That is, if a matrix has orthogonal eigenvectors, does it have to be symmetrical and real?
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If a matrix $A$ has orthogonal eigenvectors, then there exists an orthogonal $P$ and a diagonal matrix $\Lambda$ such that $$ A=P\Lambda P^\top $$ It follows that $$ A^\top = (P\Lambda P^\top)^\top = (P^\top)^\top\Lambda^\top P^\top=P\Lambda P^\top=A $$ Hence $A$ is symmetric.
Brian Fitzpatrick
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1So, why the eigenvalues are necessarily real? – Andreas H. Aug 14 '14 at 14:23
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@AndreasH. I thought that was in OP's original assumptions. – Brian Fitzpatrick Aug 14 '14 at 15:28
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1He only mentioned orthogonality of eigenvectors, though I presume he might have also assumed that the eigenvalues are real. – bartgol Aug 15 '14 at 01:19
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1This answer is incorrect according to the OP's requirements – DonAntonio Mar 11 '19 at 13:33
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Answer to the question (v1): No, consider e.g. the anti-Hermitian matrix $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, which has eigenvalues $\pm i$, and which is (complex) orthogonal diagonalizable$^{\dagger}$.
$^{\dagger}$ Note that the underlying inner product is implicitly assumed to be a sesquilinear form rather than a bilinear form.
Qmechanic
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If you allow your matrix to have complex eigenvalues/eigenvectors, the answer is no. However, you can say something about that matrix, namely that is normal.
Definition: a matrix $A$ is said to be normal if $A^HA=AA^H$, where $A^H$ denotes the transpose and conjugate matrix of $A$.
The spectral theorem says that being normal and being unitarily diagonalizable are equivalent statements.
Now, if you restrict your attention to matrices with real eigenvalues and real orthogonal eigenvectors that form a complete basis, then your statement is true, as Brian showed you. This is because $A^{\top}=A^H$ for real matrices.
