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- Thread starter AxiomOfChoice
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- #2

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In the case of repeated eigenvalues, it may or may not be possible to find independent eigenvectors. For example, the identity matrix has only one eigenvalue, 1, repeated n times. In this case, EVERY nonzero vector is an eigenvector, and if you simply choose n of them at random, most likely they won't be linearly independent. (However, in this case at least it's POSSIBLE to choose n independent eigenvectors.)

For some matrices with repeated eigenvectors, it's not even possible to choose n independent eigenvectors: for example, the matrix

[0 1]

[0 0]

has only one (repeated) eigenvalue, 0. Any eigenvector [x y]^T must satisfy y = 0, so this restricts the eigenvectors to a one-dimensional subspace, and therefore you can't have two independent eigenvectors.

On the other hand, for a positive definite matrix, you are guaranteed to be able to find a linearly independent (even orthonormal) set of eigenvectors. This is even true for any Hermitian matrix by the spectral theorem (see, e.g., Horn and Johnson, "Matrix Analysis," theorem 2.5.6).

Since a positive-definite or positive-semidefinite matrix is by definition Hermitian, and so therefore is their product, then the answer to your second question is yes, it's true, but the proof is not so elementary.

- #3

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Is it possible for a Hermitian matrix to have a repeated eigenvalue, or several repeated eigenvalues? Or do we know that [itex]n \times n[/itex] Hermitian matrices will have [itex]n[/itex] distinct eigenvalues?

- #4

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Is it possible for a Hermitian matrix to have a repeated eigenvalue, or several repeated eigenvalues? Or do we know that [itex]n \times n[/itex] Hermitian matrices will have [itex]n[/itex] distinct eigenvalues?

Sure, the identity matrix is Hermitian and all of its eigenvalues are 1.

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