$A,B\in \mathbb{R}^{n\times n}$,$A,B\succeq 0$,$B-A\succeq 0$, prove $\Vert A \Vert_F \leq \Vert B \Vert_F $
(explain of sign: If $X \succeq 0$, then there exists
another matrix $Y \succeq 0$ such that $Y^2 = X$. Such $Y$ is not unique in general, but we may denote it as $X^{1/2}$.)
This is my homework. I have no idea at all but suspecting the "$Y^2 = X$" should be "$Y^TY = X$".
I tried to let $A=X^2, B=Y^2, B-A=Z^2$, then $\|B\|_F^2$ can be wrote as the sum of the products of the elements of $X$ and $Z$, but that seems doesn’t work.
$\|A+B\|_F^2=\|A\|_F^2+2\langle A,B\rangle_F+\|B\|_F^2$ may be useful, where $\langle A,B\rangle_{F}=\operatorname{trace}\left(A^{T}B\right)$.
Best Answer
Assume that $A, B$ are both symmetric.
Let $Z := B - A \succeq 0$. We have $$\|B\|_F^2 - \|A\|_F^2 = \mathrm{tr}((A + Z)^2) - \mathrm{tr}(A^2) = \mathrm{tr}(Z^2) + 2\mathrm{tr}(AZ) \ge 0$$ where we use $\mathrm{tr}(AZ) = \mathrm{tr}(A^{1/2}Z A^{1/2}) \ge 0.$
Another way.
Let $A = UVU^\top$ be the EVD. Let $C = U^\top BU$. From $B - A \succeq 0$, we have $C - V \succeq 0$. Thus, $C_{ii} - V_{ii} \ge 0$ for all $i$.
We have $$\|B\|_F^2 - \|A\|_F^2 = \mathrm{tr}(B^2) - \mathrm{tr}(A^2) = \mathrm{tr}(C^2) - \mathrm{tr}(V^2) \ge \sum_{i=1}^n (C_{ii})^2 - \sum_{i=1}^n (V_{ii})^2 \ge 0.$$