Smoothness of (sum of squares) of singular values?

Question

Lately, I found out that the follwoing function is smooth: $h:GL_n^+ \to \mathbb{R} \, , \, h(A)=\sum_{j=1}^n [s_j(A)-1]^2$ where $s_j(A)$ are the singular values of $A$.

I came to this conclusion in a rather twisted way (explained below) and I would like to find a more straightforward way.

Proof of smoothnes of $h$:

Look at $dist^2(A,O(n)) =\underset{X \in O(n)}{\text{min}} \|A - X\|^2$

[where $\|\cdot \|$ is the standard Frobenius (Euclidean) norm]

It turns out* that the closest orthongonal matrix to $A$ is $Q(A)=A(\sqrt{A^t A})^{-1}$, and that $$\|A - Q(A)\|^2 = dist^2(A,O(n)) = \sum_{j=1}^n [s_j(A)-1]^2$$

*For details see these two answers: (1),(2).

It is a fact that the positive square root of a matrix is smooth when considerd as a function $S^+ \to S^+$ where $S^+$ is the manifold of symmetric positive definite matrices.

It follows that the function $Q:GL_n^+ \to O(n) \, , \, Q(A)=A(\sqrt{A^t A})^{-1}$ is smooth.

Hence $h:GL_n^+ \to \mathbb{R} \, , \, h(A)=\sum_{j=1}^n [s_j(A)-1]^2=\|A - Q(A)\|^2$ is smooth.

Question: Is there a simple way to show $h$ is smooth? (without going through all the detour of characterizing it as the distance from $O(n)$ and proving that the closest matrix is smooth)

Remark:

As far as I understand, the singular values themselves are not always smooth functions of the matrix: They are eigenvalues of some matrix ($\sqrt{A^tA}$) and hence roots of some characteristic polynomial, and roots of a polynomial are not smooth if there are multiple roots.

Answer 1

Given a matrix $A \in \mathrm{GL}_n(\mathbb{R})$, consider the coefficients of the characteristic polynomial of $\sqrt{A^tA}$

$$ \chi_{\sqrt{A^tA}} \left( x \right) = \sum_{k=0}^n (-1)^{n+k} e_{n-k}(A) x^k $$

as functions of $A$. Since $A \mapsto \sqrt{A^tA}$ is smooth as a composition of smooth functions and the coefficients of the characteristic polynomial of a matrix are polynomials in the entries of the matrix, you see that $e_i \colon \mathrm{GL}_n(\mathbb{R}) \rightarrow \mathbb{R}$ are smooth functions. Let $p_1(A)$ be the sum of roots of the polynomial $\chi_{\sqrt{A^tA}}$ and $p_2(A)$ be the sum of square of roots of $\chi_{\sqrt{A^tA}}$. You are interested in the smoothness of the function $p_2(A) - 2p_1(A) + n$. Newton's identities show that

$$ p_2(A) - 2p_1(A) + n = e_1(A)^2 - 2e_2(A) - 2e_1(A) + n $$

is smooth. More generally, this argument (together with the fundamental theorem of symmetric polynomials) shows that any symmetric polynomial function of the roots of a polynomial whose coefficients depends smoothly on some set of parameters will also be smooth.