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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

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2 Answers 2

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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.

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    $\begingroup$ +1, but in the same vein, I think it's clearer to write $$\operatorname{tr}(B^2-A^2)=\operatorname{tr}\bigl(B(B+A)-(B+A)A\bigr)=\operatorname{tr}\bigl((B+A)^{1/2}(B-A)(B+A)^{1/2}\bigr)\ge0.$$ $\endgroup$
    – user1551
    Dec 25, 2021 at 8:04
  • $\begingroup$ @user1551 Oh absolutely, but I thought I wrote down my original line of thought instead (: $\endgroup$
    – Chee Han
    Dec 25, 2021 at 8:09
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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.)

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  • $\begingroup$ "$0\preceq A\preceq B$, we have $0\le\lambda_i(A)\le\lambda_i(B)$ " Why is this true? $\endgroup$
    – Rioghasarig
    Dec 25, 2021 at 16:29
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    $\begingroup$ @Rioghasarig This is due to the variational characterisation of eigenvalues, a.k.a. Courant-Fischer minimax principle. $\endgroup$
    – user1551
    Dec 25, 2021 at 20:56
  • $\begingroup$ @user1551 I'm struggling to see why this follows. The eigenvectors of the two matrices aren't generally the same. $\endgroup$
    – idl
    Dec 26, 2021 at 4:11
  • $\begingroup$ @idl $\lambda_k^\downarrow(A)= \max\limits_{\dim V=k}\min\limits_{x\in V,\|x\|=1}x^\ast Ax \le\max\limits_{\dim V=k}\min\limits_{x\in V,\|x\|=1}x^\ast Bx =\lambda_k^\downarrow(B).$ $\endgroup$
    – user1551
    Dec 26, 2021 at 5:35
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    $\begingroup$ @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 $$ \max_{\dim V=k}m_B(V)\ge m_B(V_A) =x_B^\ast Bx_B\ge x_B^\ast Ax_B \ge m_A(V_A)=\max_{\dim V=k}m_A(V). $$ $\endgroup$
    – user1551
    Dec 28, 2021 at 6:50

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