Square root of a matrix not differentiable if $A$ is singular 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 :)
Let $f(A)= \sqrt{A^TA}$. When $A$ is singular, let $A=U\pmatrix{S\\ &amp;0}V^T$ be a singular value decomposition of $A$ and let $H=U\pmatrix{0\\ &amp;I}V^T$. Then the directional derivative of $f$ along the direction $H$, i.e.
$$
\lim_{t\to0}\frac{f(A+tH)-f(A)}{t}=\lim_{t\to0}\frac{|t|}{t}V\pmatrix{0\\ &amp;I}V^T,
$$
does not exist. Hence $f$ is not differentiable at $A$.
In the easiest case $n=1$ we have $ \phi(A)=\sqrt{A^2}=|A|.$
$\phi$ is not differentiable at $A=0.$