For $A(t)$ differentiable, taking positive matrices as values, how show $\sqrt{A(t)}$ is differentiable?

On p. 150 of Lax's Linear Algebra, he mentions that is is not hard to show that if $R(t)$ is a differentiable matrix-valued function of a single variable, whose values are positive matrices, then the square root function $\sqrt{R(t)}$, is differentiable as well. I have been trying to prove this without success.

The definition Lax gives for the derivative of a matrix valued function is that $$\lim_{h\to 0}\left\|\frac{A(t_0+h) - A(t_0)}{h}- \Omega_{t_0}\right\| = 0$$ for some matrix $\Omega_{t_0}$, where the double bars denotes the matrix norm. I believe this is equivalent to the existence of the $n^2$ entry-wise derivatives.

I have been trying to manipulate the limit $$\lim_{h\to 0}\frac{\sqrt{A(t_0+h)}-\sqrt{A(t_0)}}{h}$$

In the scalar-valued case, the proof I have seen for the differentiability of the square root function involves rationalizing the numerator by multiplying top and bottom by $\sqrt{x_0+h} + \sqrt{x_0}$. We might try a similar thing with the matrix-valued function, but we would need $\sqrt{A(t_0 + h)}$ to commute with $\sqrt{A(t_0)}$ in order to clear the numerator, and they don't necessarily commute.

So I'm kind of out of ideas. How might we prove the differentiability of $\sqrt{A(t)}$?

• By positive matrix, you mean positive definite matrix? That is typically when a matrix square root is defined. Commented Jun 24, 2013 at 21:57
• If you just want to prove that it is differentiable, here is an option. The map $A\longmapsto A^2$ is smooth (a polynomial) and invertible from positive definite matrices onto positive definite matrices (=open set in the linear space of symmetric matrices). So its inverse, $A\longmapsto \sqrt{A}$ is smooth as well. Now your function is just the composition of the latter with the differentiable path $t\longmapsto A(t)$, whence differentiable by chain rule. If you want to compute the derivative, there is more work. Commented Jun 24, 2013 at 22:33
• @EricAuld Yes. In the hermitian matrices, though, which is not $\mathbb{C}^{n^2}$ but $\mathbb{R}^n\oplus \mathbb{C}^\frac{n(n-1)}{2}$. And yes, your formulation is correct. Commented Jun 24, 2013 at 22:53
• @julien We know that it is invertible, but how are we sure that the derivative is an invertible linear mapping at an arbitrary point? (We need this for the inverse function theorem I believe) Commented Jun 25, 2013 at 18:01
• You're right. You need to check that the derivative of $\phi(A)=A^2$ is invertible at every point. But this derivative is easy, it is: $D\phi_A(H)=AH+HA$. It takes hermitian matrices to hermitian matrices. And if $H$ is in the kernel, then $AH=-HA$. So for very eigenvector of $A$, $Ax=\lambda x$, we have $AHx=-HAx=-\lambda x$. Since $-\lambda<0$ can't be an eigenvalue of $A$, this forces $Hx=0$. So finally $H=0$ and $D\phi_A$ is injective whence invertible. Commented Jun 25, 2013 at 18:20

Here is a mostly elementary approach. There are certainly some better or fancier proofs, but this is what immediately came to my mind. For convenience, I shall write $A_t$ for $A(t)$. Without loss of generality, assume that $t_0=0$. Let $X_h=A_h^{1/2}-A_0^{1/2}$. Proving that $A_t^{1/2}$ is differentiable at $t=0$ is equivalent to showing that $\color{green}{X_h=hX+o(h)}$ for some constant matrix $X$ (which, if exists, is the derivative $\left.\frac{d}{dt}A_t^{1/2}\right|_{t=0}$).
As eigenvalues of a matrix vary continuously with matrix entries, we know that $X_h=o(1)$ (this is the only non-elementary result that we shall use in this proof). In the below, we will show further that $X_h=O(h)$. This is stronger than continuity but weaker than differentiability at $t=0$. Let $u$ be a unit eigenvector corresponding to an eigenvalue $\lambda$ of $X_h$. Let $Y_h = A_h-A_0$. Then $A_0+Y_h=A_h=(A_0^{1/2}+X_h)^2$ and hence \begin{align*} Y_h &= X_hA_0^{1/2} + A_0^{1/2}X_h + X_h^2,\\ u^\ast Y_hu &= 2\lambda\, u^\ast A_0^{1/2}u + \lambda^2,\\ \lambda &= -u^\ast A_0^{1/2}u \pm \sqrt{(u^\ast A_0^{1/2}u)^2 + u^\ast Y_hu}.\tag{1} \end{align*} Since $A_0\succ0$ and $A_t$ is differentiable, we have $0<\lambda_\min(A_0)^{1/2}\le u^\ast A_0^{1/2}u\le\lambda_\max(A_0)^{1/2}$ and $Y_h=h\dot{A}_0+o(h)$. As $X_h=o(1)$, the plus sign in the "$\pm$" in $(1)$ must be taken when $h$ is small. Apply $\sqrt{a+x}=\sqrt{a}\sqrt{1+\frac xa}=\sqrt{a}\left(1+\frac x{2a}+o(x)\right)$ to $(1)$, we get $\lambda=O(h)$ and in turn $\rho(X_h)=O(h)$. As $X_h$ is Hermitian, we infer that $X_h=O(h)$.
We now show that $A_t^{1/2}$ is differentiable at $0$. If $A_t^{1/2}$ is indeed differentiable, then by differentiating $(A_t^{1/2})^2=A_t$ on both sides, we see that the derivative $X=\left.\frac{d}{dt}A_t^{1/2}\right|_{t=0}$ must satisfy the equation $A_0^{1/2}X + XA_0^{1/2} = \dot{A}_0$, or equivalently, $(I\otimes A_0^{1/2} + A_0^{1/2}\otimes I)\,\mathrm{vec}(X) = \mathrm{vec}(\dot{A}_0)$. So, let $X$ be a solution to this equation. Since $A_0$ is positive definite, the solution exists and is unique. Now we can express $A_h- (A_0^{1/2}+hX)^2$ in two different forms: \begin{align*} A_h- (A_0^{1/2}+hX)^2 &= A_h - (A_0 + hA_0^{1/2}X+hXA_0^{1/2} + h^2X^2)\\ &= A_h - (A_0 + h\dot{A}_0 + h^2X^2) = \color{red}{o(h)},\tag{2}\\ A_h- (A_0^{1/2}+hX)^2 &= (A_0^{1/2}+X_h)^2 - (A_0^{1/2}+hX)^2\\ &= \color{red}{A_0^{1/2}(X_h-hX) + (X_h-hX)A_0^{1/2} + o(h)},\tag{3} \end{align*} where we have used the result $X_h=O(h)$ in deriving $(3)$. Compare $(2)$ and $(3)$, we obtain $A_0^{1/2}(X_h-hX) + (X_h-hX)A_0^{1/2}=o(h)$, i.e. $(I\otimes A_0^{1/2} + A_0^{1/2}\otimes I)\mathrm{vec}(X_h-hX) = o(h)$. Since $I\otimes A_0^{1/2} + A_0^{1/2}\otimes I$ is invertible, it follows that $\color{green}{X_h-hX=o(h)}$. QED