How to prove tr($Z^TZ$) is the sum of singular values

Question

Ask a maybe trivious question:
We know the product of singular values is |determinant of tha matrix| as in:
Singular value proofs

Then how to prove:

$tr(Z^TZ)^{1/2}$ $=\sigma_1(Z) +...+ \sigma_r(Z) $? where $r$ is the rank of $Z$

thanks

Answer 1

That should be interpreted as $$\text{tr}\left((Z^T Z)^{1/2}\right) = \ldots$$ not $$\left(\text{tr}(Z^T Z)\right)^{1/2} = \ldots$$ The singular values are the nonzero eigenvalues of $(Z^T Z)^{1/2}$, i.e. the positive semidefinite square root of $Z^T Z$. The trace of a matrix is the sum of its eigenvalues.