Does Frobenius norm (not operator 2 norm) preserve the positive semidefinite order of matrices? Let $A, B$ be two real symmetric positive semidefinite $n\times n$ matrices (these conditions might be unnecessary). We say $A\le B$ if $B-A$ is positive semidefinite. If $A\le B$, do we necessarily have $\|A\|_F \le \|B\|_F$, where the norm is Frobenius norm?
Note that for a general matrix $A$, $\|A\|_F^2=\text{Trace} (AA^T)$. So we are really asking whether $\text{Trace} (AA^T) \le \text{Trace} (BB^T)$. I have tried a couple of examples and believe this is true but can't find a proof.
the proof posted below is about the operator 2 norm I believe
Symmetric positive semi-definite matrices and norm inequalities
 A: Yes. Since $0\preceq A\preceq B$, we have $0\le\lambda_i(A)\le\lambda_i(B)$ when the eigenvalues of $A$ and $B$ are arranged in the same (ascending or descending) order. It follows that $\operatorname{tr}(A^2)=\sum_i\lambda_i(A)^2\le\sum_i\lambda_i(B)^2=\operatorname{tr}(B^2)$.
(However, despite $\operatorname{tr}(B^2)\ge\operatorname{tr}(A^2)$, the difference $B^2-A^2$ is not necessarily PSD.)
A: Here is an alternative proof that is more direct (I think). Since $B - A$ is symmetric positive semidefinite, we know that eigenvalues of $(B - A)$ are nonnegative. In other words, $\mathrm{Tr}((B - A)^2)\ge 0$. Expanding this using the fact that the trace is linear and $\mathrm{Tr}(CD) = \mathrm{Tr}(DC)$, we get
\begin{align*} 
\mathrm{Tr}(B^2)\ge 2\mathrm{Tr}(AB) - \mathrm{Tr}(A^2) & = \mathrm{Tr}(A^2) + 2\Big[\mathrm{Tr}(AB) - \mathrm{Tr}(A^2)\Big] \\ 
& = \mathrm{Tr}(A^2) + 2\mathrm{Tr}(A(B - A)). 
\end{align*}
The second term is nonnegative since $A$ and $B - A$ are both symmetric positive semidefinite, see here for the proof of this fact.
