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I am studying linear algebra and I have a very basic question.

Why symmetric matrix is always diagonalizable even if it has repeated eigenvalues?

I've seen that sufficient orthonormal eigenvectors can be generated by applying Gram-Schmidt process to the eigenspace of repeated eigenvalue.

I know what this means. I have solved bunch of exercise problems and I've been always able to generate full set of orthonormal basis of eigenspace of repeated eigenvalue.

But what I want to know is this: The dimension of eigenspace of repeated eigenvalue with multiplicity of "k" is always "k"? Is it impossible that eigenspace of repeated eigenvalue of symmetric matrix is a 1-dimensional line?

Thank you.

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If $A$ is symmetric, and zero is an eigenvalue with the dimension of the eigenspace less than the multiplicity of zero, then there will be a vector with $A^2v=0$ but $Av\ne0$. But then $0\ne\|Av\|^2=(Av)^t Av=v^tA^2v=0$, a contradiction.

For general eigenvalues $\lambda$, consider $A-\lambda I$ instead.

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