Is it true that $\|A\|_1\le \sqrt{rank(A)}\|A\|_F$?
I met a derivation says: $\|uu'-vv'\|_1\le\sqrt{2}\|uu'-vv'\|_F$, where $u,v$ are unit vectors and $u'$ is transpose. I can't derive this claim.
Thanks!
2026-04-17 08:22:37.1776414157
Matrix norm inequality Frobenius and l1: $\|A\|_1\le \sqrt{rank(A)}\|A\|_F$ and $\|uu'-vv'\|_1\le\sqrt{2}\|uu'-vv'\|_F$
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No.
Let $A = \left[1,1,1,1\right]^T$. Then $\|A|_1 = 4$, $\|A\|_F = 2$, and $rank(A) = 1$. For this matrix $\|A\|_1 > \sqrt{rank(A)}\|A\|_F$
However for any $m\times n$ matrix $A$: $$\|A\|_1 \leq \sqrt{m}\|A\|_F$$