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We know that a normal matrix will have, or can be made to have (by orthogonalization, if not all its eigenvalues are distinct), orthogonal eigenvectors. This means that they are also linearly independent (i.e. they form a basis for the vector space concerned). However, in the case when these eigenvectors are not all orthogonal (i.e. when some share the same, degenerate, eigenvalues and weren't orthogonal and we did not orthogonalize them), will they still be linearly independent?

In other words, are the eigenvectors of a normal matrix always linearly independent, regardless of their mutual orthogonality?

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  • $\begingroup$ you can obtain a basis of eigenvectors. a set of distinct eigenvectors need not be linearly independent $\endgroup$ – user251257 Aug 6 '15 at 23:23
  • $\begingroup$ So, Hermitian matrices with a nondegenerate eigenspectrum has an orthonormal eigenbasis. If the spectrum is degenerate it is not the case that there will be an eigenbasis for the column space (here I speak about square matrices). If there is no basis then the $n$ eigenvectors are not linearly independent. $\endgroup$ – Tucker Aug 6 '15 at 23:26
  • $\begingroup$ Technically if $v$ is an eigenvector, then $\alpha v$ is also an eigenvector, for all scalars $\alpha \neq 0$, but $v$ and $\alpha v$ are not linearly independent. @Tucker, for Hermitian matrices there will always exist an orthonormal basis of eigenvectors. $\endgroup$ – Calle Aug 6 '15 at 23:34

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