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**Summary::**Let ##A \in \Bbb R^{n \times n}## be a symmetric matrix and let ##\lambda \in \Bbb R## be an eigenvalue of ##A##. Prove that the geometric multiplicity ##g(\lambda)## of ##A## equals its algebraic multiplicity ##a(\lambda)##.

Let ##A \in \Bbb R^{n \times n}## be a symmetric matrix and let ##\lambda \in \Bbb R## be an eigenvalue of ##A##. Prove that the geometric multiplicity ##g(\lambda)## of ##A## equals its algebraic multiplicity ##a(\lambda)##.

We know that if ##A## is diagonalizable then ##g(\lambda)=a(\lambda)##.

__So all we have to show is that ##A## is diagonalizable.__

I found a proof by contradiction: Assuming ##A## is not diagonalizable we have

$$(A- \lambda_i I)^2 v=0, \ (A- \lambda_i I) v \neq 0$$

Where ##\lambda_i## is some repeated eigenvalue. Then

$$0=v^{\dagger}(A-\lambda_i I)^2v=v^{\dagger}(A-\lambda_i I)(A-\lambda_i I) \neq 0$$

Which is a contradiction (where ##\dagger## stands for conjugate transpose).

But I do not really understand it; what is meant by 'generalized eigenvalues of order ##2## or higher'?

I asked about this proof a while ago but the answer I got did not really convince me...

We could go over an alternative proof of course, the aim is understand how to show that real symmetric matrices are diagonalizable.

Thank you!