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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

No One
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2 Answers2

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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.)

user1551
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  • "$0\preceq A\preceq B$, we have $0\le\lambda_i(A)\le\lambda_i(B)$ " Why is this true? – Rioghasarig Dec 25 '21 at 16:29
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    @Rioghasarig This is due to the variational characterisation of eigenvalues, a.k.a. Courant-Fischer minimax principle. – user1551 Dec 25 '21 at 20:56
  • @user1551 I'm struggling to see why this follows. The eigenvectors of the two matrices aren't generally the same. – idl Dec 26 '21 at 04:11
  • @idl $\lambda_k^\downarrow(A)= \max\limits_{\dim V=k}\min\limits_{x\in V,|x|=1}x^\ast Ax \le\max\limits_{\dim V=k}\min\limits_{x\in V,|x|=1}x^\ast Bx =\lambda_k^\downarrow(B).$ – user1551 Dec 26 '21 at 05:35
  • I see that we have a maxmin condition btw the eigenvalues of $B - A$. But I don't see how you can split open the maxmin, I'd expect it's not linear in $A$ and $B$ unless they are mutually diagonalizable. – idl Dec 28 '21 at 04:36
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    @idl For any subspace $V$, define $m_B(V)=\min\limits_{x\in V,|x|=1}x^\ast Bx$ and define $m_A(V)$ analogously. Now let $V_A=\arg\max\limits_{\dim V=k}m_A(V)$ and $x_B=\arg\min\limits_{x\in V_A,|x|=1}x^\ast Bx$. Then $$ \max_{\dim V=k}m_B(V)\ge m_B(V_A) =x_B^\ast Bx_B\ge x_B^\ast Ax_B \ge m_A(V_A)=\max_{\dim V=k}m_A(V). $$ – user1551 Dec 28 '21 at 06:50
  • Thanks @user1551! That's very nice. – idl Dec 28 '21 at 07:02
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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.

Chee Han
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    +1, but in the same vein, I think it's clearer to write $$\operatorname{tr}(B^2-A^2)=\operatorname{tr}\bigl(B(B+A)-(B+A)A\bigr)=\operatorname{tr}\bigl((B+A)^{1/2}(B-A)(B+A)^{1/2}\bigr)\ge0.$$ – user1551 Dec 25 '21 at 08:04
  • @user1551 Oh absolutely, but I thought I wrote down my original line of thought instead (: – Chee Han Dec 25 '21 at 08:09