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?

Thanks in advance!

  • $\begingroup$ 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 $\endgroup$ Jul 6, 2021 at 17:52
  • $\begingroup$ And yes, the steps you have taken are correct. $\endgroup$ Jul 6, 2021 at 17:52
  • $\begingroup$ It also works for the entrywise infinity norm. (!) By Cauchy-Schwarz, the largest-modulus entry of A and B is on the diagonal, and the diagonals of A, B, and B-A are nonnegative. $\endgroup$ Jul 6, 2021 at 19:39
  • $\begingroup$ Related : math.stackexchange.com/questions/374500/monotone-matrix-norms $\endgroup$
    – Desura
    Jul 6, 2021 at 19:55
  • $\begingroup$ Thanks so much for your helps! $\endgroup$
    – YRS
    Jul 7, 2021 at 10:19


You must log in to answer this question.

Browse other questions tagged .