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I am confused by Nagabhushan S N's answer in Do Real Symmetric Matrices have 'n' linearly independent eigenvectors?.

It seems his first sentence matches his last, but they contradict the two points enumerated in the middle of his answer? I may be misunderstanding this, because I feel like I've seen this in other sources as well where it is stated that "$n\times n$ Symmetric matrices have $n$ orthogonal (hence linearly independent) eigenvectors."

But isn't it that a symmetric matrix has $m$ orthogonal eigenectors, where $m$ is the number of distinct eigenvalues? If you have an eigenvalue with algebraic multiplicity > 1, then there's no guarantee of orthogonality between the eigenvectors corresponding to that eigenvalue? To get orthogonal eigenvectors, I believe you have to perform Gram-Schmidt orthogonalization as Nagabhushan alluded to in their answer.

In this case, where the symmetric matrix has $m$ distinct eigenvalues, does it still have $n$ linearly independent eigenvectors?

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Yes. The spectral theorem guarantees that the matrix is diagonalizable, and the eigenspaces are orthogonal. Now if you take in each eigenspace, an orthogonal basis (for example you can use indeed Gram-Schmidt to construct it), you end with $n$ orthogonal eigenvectors that form a basis of your space.

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  • $\begingroup$ Yeah, I think that's my general understanding. Do you have to do any special to ensure that the eigenvectors in each eigenspace are linearly independent, or do they just come linearly independent by default? $\endgroup$
    – 24n8
    Commented Aug 16, 2020 at 17:16
  • $\begingroup$ Two eigenvectors in two different eigenspaces are always linearly independant. However, in a given eigenspace, you have to be careful and take eigenvectors that form a basis, unless you cannot be sure that they are independant. $\endgroup$ Commented Aug 16, 2020 at 17:18

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