Matrix function decomposition of the form $A(\theta)=B(\theta)^\top Q B(\theta)$ Suppose I have a matrix function $A:\mathbb{R}^n\to\mathbb{R}^{n\times n}$, for which I know the following properties hold:

*

*$A(\theta)$ is real, symmetric and bounded for all $\theta$

*$A(\theta)$ is a $\mathcal{C}_1$ function, i.e., $A(\theta)$ and $\frac{\partial A(\theta)}{\partial \theta}$ are continuous.

*There exist two constants $0<c_1\le c_2$ such that for all $\theta$, $c_1I\preccurlyeq A(\theta)\preccurlyeq c_2 I$. (so it is positive definite for all $\theta$)

Is it true that this matrix can always be represented in the following form?
$$A(\theta)=B(\theta)^\top Q B(\theta),$$
where $0\prec Q\in\mathbb{R}^{n\times n}$, $B:\mathbb{R}^n\to\mathbb{R}^{n\times n}$ and $\det(B(\theta))\neq 0$ for all $\theta$.
My intuition says yes, just take power to a half (so $B(\theta) = A(\theta)^{\frac{1}{2}}$ and $Q=I$. But, I cannot find any reference that 'exploits' this fact. So, my question is, could you give a reference that shows/claims this, or prove this yourself (:P)?
Edit: I was also thinking about eigenfunction of the matrix function. However, I'm not sure how this connects exactly to the decomposition above...
 A: I suppose that you want $B(\theta)$ to be $C^1$. If so, yes, you can take $B(\theta)=A(\theta)^{1/2}$ and $Q=I$.
Denote the set of all real symmetric matrices by $\mathcal S$ and the set of all positive definite matrices by $\mathcal P$. Then $\mathcal P$ is open in the $\frac{n(n+1)}{2}$-dimensional real vector space $\mathcal S$. The positive definite square root function $g(A)= A^{1/2}$ is well-defined on $\mathcal P$. In fact, given $ A$, its positive definite square root uniquely determined by the Lagrange interpolation polynomial that maps each distinct eigenvalue $\lambda$ of $A$ to $\sqrt{\lambda}$.
Now let $ A_0$ be any positive definite matrix and let $B_0=A_0^{1/2}$. Define $f(B)= B^2$ on $S$. Then $f$ is $C^1$ and its Fréchet derivative $Df:X\mapsto BX+XB$ is nonsingular at $B_0$. Therefore, by inverse function theorem, $f$ has a $C^1$ local inverse defined on some neighbourhood $W$ containing $A_0$. Since $\mathcal P$ is open, we may assume that $W\subseteq \mathcal P$. As this local inverse of $f$ gives a positive definite square root function on $W$, it must agree with $g$. Hence $g$ is $C^1$ on $W$. It follows that $g$ is $C^1$ on $\mathcal P$ because $ A_0$ is arbitrary.
Finally, when $A$ is a $C^1$ function of $\theta$, its square root $A(\theta)^{1/2}=g(A(\theta))$ is also $C^1$, by the chain rule.
