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If it given that the eigenvectors corresponding to distinct eigenvalues of a matrix are orthogonal, and the eigenvalues are real, can I get the conclusion that the matrix is symmetric/self-adjoint?

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No, consider $\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}.$

If you add the condition that the eigenspaces sum to the entire space, the answer is...

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    @Scorpio19891119: Did you read the last sentence in my answer? If the matrix has a full set of eigenvectors, which your matrix does, then the answer is yes.2012-11-03