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My book lists out a bunch of steps for orthogonal diagonalization of a symmetric matrix, and one of them is

For each eigenvalue of multiplicity $>1$, find a set of $k$ linearly independent eigenvectors. If this set is not orthonormal, then apply the Gram-Schmidt orthonormalization process.

Let me start by saying that I don't actually know what the Gram-Schmidt orthonormalization process is, and I don't feel I should have to know it in order to understand my assigned reading since it was never actually taught in my course, but anyway I have a question:

Is this saying that a set of linearly independent eigenvectors corresponding to a certain eigenvalue may not be orthogonal, but that it's possible to make them that way? How could that be possible? The eigenvectors are always going to be at the same angle from each other no matter what scalar you multiply them by, right?

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The Gram-Schmidt process does not really "make the vectors orthonormal"; as you note it, you can't do that if all you do is multiply them by scalars. What it does is producing, by taking linear combinations of the vectors, another set of vectors, that is orthonormal and spans the same subspace. In your case, this means that you obtain an orthonormal basis for your eigenspace (assuming it has dimension $k$), which is exactly what you need to have an orthonormal diagonalization of your matrix.

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For each eigenvalue, there is an infinite number of eigenvectors, since if $Tv=\lambda v$ and $Tw=\lambda w$, then $T(av+bw)=aTv+bTw=a(\lambda v)+b(\lambda w)=\lambda(av+bw)$, so any linear combination of eigenvectors of the same eigenvalue is also an eigenvector of that eigenvalue. In fact, the eigenvectors for any distinct eigengalue form a subspace called an eigenspace.

Now, for a real symmetric matrix, the eigenspaces of different eigenvalues are orthogonal, but a random pair of eigenvectors in the same eigenspace might not be. So, for each distinct eigenvalue, you’ll need to use Gram-Schmidt or some other method to find an orthonormal set of vectors that span its eigenspace.

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