# Differentiability of matrix square root operator

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...

• What is $S$ if $n=1$? – user251257 Nov 3 '18 at 13:36
• @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 $T\neq 0$? – MathNewbie Nov 3 '18 at 13:55
• 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. – user251257 Nov 3 '18 at 13:59
• @user251257 I'm taking the square root of $T^*T$ as the unique self-adjoint positive matrix such that $\sqrt{T^*T}\sqrt{T^*T}=T^*T$ – 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.

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