Diagonalisation spectral theorem

Question

For the proof of the spectral theorem for complex numbers I know that the proof follows that, as T is normal then the algebraic and geometric multiplicities coincide. This means that there will be n eigenvectors generated by n distinct or non-distinct eigenvalues. But then if they aren't distinct how is a basis generated? As doesn't linear independence follow from distinct eigenvalues only?

Answer 1

No, you do not need distinct eigenvalues to get linearly independent eigenvectors. For instance, the $n \times n$ identity matrix only has one eigenvalue, yet all vectors in the standard basis are eigenvectors. In general, linear independence of eigenvectors has nothing to do with distinctness of eigenvalues.