TITLE: Is this inequality involving the Frobenius norm right?
QUESTION [4 upvotes]: Let $A$ be a generic (or varying) square, real $ n \times n$ matrix. Let $G$ be a fixed $n \times k$ matrix, $k < n.$ Denote by $||.||_F$ the Frobenius norm.
Is it true that $||AG||_F \geq c(G) ||A||$, where $c(G)$ is a positive constant only depending on the matrix $G?$
P.S. I tend to think this is true when none of the elements of $G$ is zero, so that a minimum modulus of the elements of $G$ exists.
Thanks in advance!

REPLY [5 votes]: As stated$^*$ this problem has nothing to do with the Frobenius norm.
The map $T: A \mapsto AG$ is a linear transformation from a
finite-dimensional vector space, so for any norms we have
a constant $c>0$ such that $\|AG\| \geq c \|A\|$ for all $A$, 
provided $T$ is injective (and if $T$ is not injective then
of course any nonzero $A \in \ker T$ will give a counterexample).
This is thanks to the standard compactness argument for equivalence of
all norms on a finite-dimensional real or complex vector space.
(*) "As stated" $=$ if we care only about the existence of some 
positive $c$, not its size and its dependence on $G$.
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