# Inequality between Frobenius and nuclear norm

Let $M$ be a square matrix, $\|\cdot\|_*$ be the nuclear (trace) norm, and $\|\cdot\|_F$ be the Frobenius norm. The following inequality holds between the norms:

$$\|M\|^2_* \leq \text{rank}(M) \|M\|^2_F.$$

This is pretty easy to show by using the definitions of the norms in terms of the singular values $\sigma_i$, since $\|M\|_* = \sum_i \sigma_i$ and $\|M\|_F = \sqrt{\sum_i \sigma_i^2}$ and the result follows by Cauchy-Schwarz.

However, out of curiosity I have been trying to prove this using the definitions of the norms as $\|M\|_* = \text{trace} (\sqrt{M^* M})$ and $\|M\|_F = \sqrt{\text{trace}(M^* M)}$. Can the above inequality be shown using these definitions and without invoking the singular values explicitly?

For convenience, let $|M| = \sqrt{M^*M}$.
It suffices to note that $$\langle A,B \rangle = \operatorname{Tr}(AB^*) = \operatorname{Tr}(B^*A)$$ is an inner product on the space of $m \times n$ matrices. From there, the properties of an inner product are enough to prove the Cauchy-Schwarz inequality: $$|\operatorname{Tr}(AB^*)| = |\langle A,B \rangle| \leq \|A\| \cdot \|B\| \ = \sqrt{\operatorname{Tr}(A^*A)\operatorname{Tr}(B^*B)} = \sqrt{\operatorname{Tr}(|A|^2) \operatorname{Tr}(|B|^2)}$$ Now, if $M$ has rank $r$, then the polar decomposition tells us that there exists a partial isometry $U$ (i.e. $U^*U$ is an orthogonal projection) with $\operatorname{rank}(U) = M$ such that $M = U|M|$ and $|M| = U^*M$. From there, $$\operatorname{Tr}(|M|)^2 = \langle M,U \rangle^2 \leq \langle M, M \rangle \langle U,U \rangle = \operatorname{Tr}(M^*M) \cdot \operatorname{rank}(M)$$
• Thanks, this makes sense, although I was under the impression that the polar decomposition (for matrices) is always defined with $U$ being unitary. Is this a common generalization?