# Can matrix inequality induces matrix norm inequality for positive semidefinite matrix?

I am wondering whether matrix inequality induces matrix norm inequality for positive semidefinite matrix.

For example, considering two positive semidefinite matrices $$A$$ and $$B$$, when $$A \succeq B$$, does it imply $$||A||\geq ||B||$$($$||\cdot||$$ is any matrix norm)?

I think perhaps it is true for 2-norm. Since $$A \succeq B$$ implies $$\lambda_k(A)> \lambda_k(B),\ \text{for}\ k=1,2...n$$, where $$\lambda_k$$ is the k-largest eigenvalues. So, $$\lambda_\max(A)> \lambda_\max(B)$$.

Therefore, \begin{align*} \|A\|_{2} &= \sqrt{\lambda_{\max }\left(A^{*} A\right)} \\ &= \sqrt{\lambda_{\max }\left(A^{T} A\right)} \\ &= \sqrt{\lambda^2_{\max }\left(A\right)} \\ &\geq \sqrt{\lambda^2_{\max }\left(B\right)} \\ &= \sqrt{\lambda_{\max }\left(B^{T} B\right)} \\ &= \|B\|_{2}. \end{align*}

So, $$A \succeq B \rightarrow \|A\|_{2}\geq ||B||_2$$.

May I ask whether this reduction is correct? and does this hold for any other matrix norm?

• I believe that if $\|\cdot\|$ is any unitarily invariant norm, then $A \succeq B$ implies that $\|A\| \geq \|B\|$. At least, this implication definitely holds for the spectral, Frobenius, and nuclear norms Jul 6, 2021 at 17:52