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I have encountered a case of a 3x3 symmetric matrix where 2 of the eigenvectors corresponding to 2 different eigenvalues are equal.

In order to find an orthogonal/orthonormal basis for the eigenspace all the vectors in this basis must be mutually orthogonal. However running Gram Schmidt on 2 equal vectors returns zero, which makes perfect sense because the vectors are linearly dependent of each other, well they are identical, in other words there doesn't exist an orthogonal component of a vector to itself.

We also know the basis must consist of 3 vectors, if I understand this correctly? So the question is how can one find a vector that is mutually orthogonal to the 2 others, but is still a "valid" eigenvector, would we pick any vector that is mutually orthogonal to the other 2 vectors? If yes, how can one do that easily?

UPDATE

Conclusion/correction: eigenvectors corresponding to different eigenvalues must be different (assuming we are not dealing with 0).

Different scenario where $x_1=x_2$ and $λ_1=λ_2$:

However let's assume , the basis for the eigenspace must still consist of 3 mutually orthogonal vectors right? So in that case how would one procede to find the 2nd basis-element for the eigenspace? If they are equal they can't be mutually orthogonal, but if they are mutually orthogonal in this case wouldn't they correspond to different eigenvalues? If I remember correctly we would get 2 free variables, thus 2 eigenvectors for the repeated eigenvalue (twice) as a result from the same RREF. In case these are not orthogonal we would use GS on one of them.

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    $\begingroup$ I can't believe that different eigenvalues share the same eigenvector. Since a matrix corresponds to a linear transformation, I can not imagine that the same transformation maps a vector to two different images. So something is wrong in your consideration. $\endgroup$ – Stefan Gyürki Dec 2 '15 at 11:38
  • $\begingroup$ I'm not a 100% about this in the case of a real symmetric matrix but at lest for Hermitian matrices this is true: If $Ax_1=\lambda_1x_1$ and $Ax_2=\lambda_2x_2$ and $\lambda_1\neq\lambda_2$ then we have that $(x_1,x_2)=0$ i.e. they are orthogonal and hence since both $x_1$ and $x_2$ must be different from $0$ since they're eigenvectors, they must be distinct. $\endgroup$ – Erik Olesen Dec 2 '15 at 11:55
  • $\begingroup$ Perhaps you've had a mishap in the calculation of one of them? $\endgroup$ – Erik Olesen Dec 2 '15 at 11:56
  • $\begingroup$ yes thanks just found a missing minus, so you are all right, they are different these eigenvectors. However let's assume $x_1=x_2$ and $λ_1=λ_2$, the basis for the eigenspace must still consist of 3 mutually orthogonal vectors right? So in that case how would one procede to find the 2nd basis-element for the eigenspace? If they are equal they can't be mutually orthogonal, but if they are mutually orthogonal in this case wouldn't they correspond to different eigenvalues? $\endgroup$ – jibo Dec 2 '15 at 12:38
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Can't be true I think. Please share this very special matrix.

$Ax = \lambda x$

You claim $x_1 = x_2$ but $\lambda_1 \neq \lambda_2$? Can only be the case when $x = 0$ for one of them, but that is not allowed for an eigenvector?

A way to determine the eigenvectors is to view them as null-space of $A-\lambda I$

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