I'm interested to know if the map: $$S:\mathcal{L}(\mathbb{R}^n)\rightarrow\mathcal{L}(\mathbb{R}^n)$$ $$S(T)=\sqrt{T^*T}$$ Is differentiable. I'm aware $T\mapsto T^* T$ is differentiable, and I know that since $T^*T$ is positive self-adjoint, the square root exists, but not sure where to go from here... My ideia was that, matrix-wise, choosing the correct basis, $T^*T$ is diagonal, then the map would be equivalent to taking the square root component-wise, which should be differentiable, though I don't know if that's a valid way to justify it...

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    $\begingroup$ What is $S$ if $n=1$? $\endgroup$ – user251257 Nov 3 '18 at 13:36
  • $\begingroup$ @user251257 I guess it would be $|x|$, which isn't differentiable at $0$... I was afraid of that. Does that mean the map isn't differentiable at the null matrix? How about for $T\neq 0$? $\endgroup$ – MathNewbie Nov 3 '18 at 13:55
  • $\begingroup$ how do you define square root of a matrix? The question is, when $T^*T$ has a 0 eigenvalue, that is when $T$ is not invertible. $\endgroup$ – user251257 Nov 3 '18 at 13:59
  • $\begingroup$ @user251257 I'm taking the square root of $T^*T$ as the unique self-adjoint positive matrix such that $\sqrt{T^*T}\sqrt{T^*T}=T^*T$ $\endgroup$ – MathNewbie Nov 3 '18 at 14:08

We have that

$\displaystyle dS(0) \cdot V= \lim_{t \to 0} \cfrac{S(0+tV)-S(0)}{t}=\lim_{t \to 0} \cfrac{S(tV)}{t}=\lim_{t \to 0} \cfrac{\sqrt{t^2V^* V}}{t}=\lim_{t \to 0} \cfrac{|t|}{t}\sqrt{V^* V}$

The two one-sided limits have different values. Hence $S$ is not differentiable at $0$.

On the other hand, we can prove that $$f: GL(\mathbb{R}^n) \rightarrow \mathcal{P}(\mathbb{R}^n)$$ $$f(T)=\sqrt{T^*T}$$ is differentiable.

Define the functions $g:GL(\mathbb{R}^n) \rightarrow \mathcal{P}(\mathbb{R}^n)$ given by $g(T)=T^*T$ and $\varphi:\mathcal{P}(\mathbb{R}^n) \rightarrow \mathcal{P}(\mathbb{R}^n)$ given by $\varphi(T)=\sqrt{T}$, where $\mathcal{P}(\mathbb{R}^n)=\{T \in \mathcal{L}(\mathbb{R}^n) : T \textrm{ is positive definite}\}$. We have that $f=\varphi \circ g$.

$g$ is differentiable and its derivative is $dg(T) \cdot V=T^*V+V^*T$.

To check that $\varphi$ is differentiable, define the map $\psi: \mathcal{P}(\mathbb{R}^n) \rightarrow \mathcal{P}(\mathbb{R}^n)$ given by $\psi(T)=T^2$. Let's show that $d\psi(T) \cdot V=TV+VT$ is invertible by showing it is injective. We know that $T \in \mathcal{P}(\mathbb{R}^n)$. Let $H \in \mathcal{L}(\mathbb{R}^n)$ such that $d\psi(T) \cdot H=0$. Then, it follows that $TH=-HT$. So for every eigenvector $v$ of $T$, that is, the vectors that satisfies $Tv=\lambda v$, we have that $THv=-HTv=-\lambda Hv$. Since $T$ is positive definite, we have that $-\lambda$ can't be eigenvalue for $T$, thus $Hv=0$ which implies $H=0$. So $d\psi$ is injective.

Then by the Inverse Function Theorem, it follows that $\psi^{-1}=\varphi$ is differentiable.

Hence $f=\varphi \circ g$ is differentiable.

  • $\begingroup$ How about away from the origin? As in a punctured neighborhood? $\endgroup$ – MathNewbie Nov 4 '18 at 11:57
  • $\begingroup$ I edited the answer defining the function now on $GL(\mathbb{R}^n)$. $\endgroup$ – White Crow Nov 4 '18 at 20:11
  • $\begingroup$ I actually made a mistake. I fixed it now. $\endgroup$ – White Crow Nov 5 '18 at 21:11

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