The geometric multiplicity of an eigenvalue of a symmetric matrix necessarily equals to its algebric multiplicity.
The Attempt at a Solution
If a matrix is symmetric, then the matrix is diagonalizable. Since the matrix is diagonalizable, there must be eigenvectors correspond to each eigenvalues.
So, I did the proof, but I'm not so sure if it sounds right. I think there could be something more tricky or missing. Would you guys check if this sounds right to you?