I am studying following theorem from Numerical linear Algebra book by Trefethen and Bau.
Theorem : If $A = A^{*}$, then the singular values of A are the absolute values of the eigenvalues of A.
Proof: A Hermitian matrix has a complete set of eigenvectors, and all the eigenvalues are real.
$$(1)\quad A = X \Lambda X^{-1}$$
An equivalent statement that (1) holds with $X$ equal to some unitary matrix $Q$ and $\Lambda$ a real diagonal matrix. But we can write
$$ (2) \quad A = Q \Lambda Q^{*}= Q|\Lambda|sign(\Lambda)Q^{*}$$
where $|\Lambda|$ and $sign(\Lambda)$ denote the diagonal matrix whose entries are the number $|\lambda_{j}|$ and $sign(\lambda_{j})$, respectively. (We could equally well have put the factor $sign(\Lambda)$ on the left of $|\Lambda|$ instead of right). Since sign($\Lambda)Q^{*}$ is unitary whenever Q is unitary, (2) is an SVD of A, with the singular values equal to the diagonal entries of $|\Lambda|$, $|\lambda_{j}|$. If desired, these numbers can be put into non-increasing order by inserting suitable permutation matrices as factors in the left hand unitary matrix of (2), Q , and the right-hand unitary matrix sign$(\Lambda)Q^{*}$.
I cannot follow the logic in this proof after equation (2). Can any one give an insight on this? Thanks in advance.