Square root of a matrix not differentiable if $A$ is singular

Question

Consider the map

$$\phi:M_n(\mathbb{R}) \to M_n(\mathbb{R}), \qquad A \mapsto \sqrt{A^T A}$$

I claim that this map is differentiable if and only if $A$ is invertible.

Just from the inverse function theorem, it can be seen that the map $\phi$ is differentiable at any invertible matrix. Now I need to show that this map is not differentiable at singular matrices. I tried to do the singular value decomposition but it did not work. Any help or reference will be appreciated.

Thanks :)

Answer 1

Let $f(A)= \sqrt{A^TA}$. When $A$ is singular, let $A=U\pmatrix{S\\ &0}V^T$ be a singular value decomposition of $A$ and let $H=U\pmatrix{0\\ &I}V^T$. Then the directional derivative of $f$ along the direction $H$, i.e. $$ \lim_{t\to0}\frac{f(A+tH)-f(A)}{t}=\lim_{t\to0}\frac{|t|}{t}V\pmatrix{0\\ &I}V^T, $$ does not exist. Hence $f$ is not differentiable at $A$.

Answer 2

In the easiest case $n=1$ we have $ \phi(A)=\sqrt{A^2}=|A|.$

$\phi$ is not differentiable at $A=0.$

Answer 3

At singular $A$, choose $x$ of length $1$ such that $Ax=0$. Put $\gamma(t)= txx^T + A$. Then $\gamma$ is smooth, and I claim $\sqrt{\gamma(t)\gamma(t)^T}$ is not differentiable at $0$. For $\sqrt{\gamma(t)^T\gamma(t)}x = |t|x$.

However, I'm now doubting wether $A\mapsto \sqrt{A^TA}$ will be smooth in regions where the multiplicity of an eigenvalue changes. See the answer to this post: Do eigenvalues depend smoothly on the matrix elements of a diagonalizable matrix?