# 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

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.

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

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.)

• "$0\preceq A\preceq B$, we have $0\le\lambda_i(A)\le\lambda_i(B)$ " Why is this true? Dec 25, 2021 at 16:29
• @Rioghasarig This is due to the variational characterisation of eigenvalues, a.k.a. Courant-Fischer minimax principle. Dec 25, 2021 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, 2021 at 4:11
• @idl $\lambda_k^\downarrow(A)= \max\limits_{\dim V=k}\min\limits_{x\in V,\|x\|=1}x^\ast Ax \le\max\limits_{\dim V=k}\min\limits_{x\in V,\|x\|=1}x^\ast Bx =\lambda_k^\downarrow(B).$ Dec 26, 2021 at 5:35
• @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 $$\max_{\dim V=k}m_B(V)\ge m_B(V_A) =x_B^\ast Bx_B\ge x_B^\ast Ax_B \ge m_A(V_A)=\max_{\dim V=k}m_A(V).$$ Dec 28, 2021 at 6:50