How does rank of PSD symmetric matrix being equal to number of nonzero eigenvalues follow from SVD?

Question

I understand that the rank of a positive semi-definite matrix is equal to the number of non-zero singular values of the matrix.

$$\operatorname{rank}(M) = \{ \sigma \mid \sigma \ne 0 \}$$

This is somehow related to the spectral decomposition (or singular value decomposition, as some call it), but I cannot figure out how.

This question touches on that, but I cannot figure out the relationship:

Relation between rank of a symmetric positive semi-definite matrix and its number of non-zero eigen values (or singular values)

How does the rank of a PSD matrix being equal to number of nonzero eigenvalues, follow from the spectral decomposition?

Answer 1

For any matrix, PSD or not, the rank of the matrix equals the number of nonzero singular values, and hence equals the number of nonzero eigenvalues.