Let $A$ be a $2 \times 4$ matrix and $B$ be a $4 \times 4$ matrix, prove that if $rank(A) = 2$ and $rank(B)=3$ then $AB \neq 0$.
I got stuck at $rank(AB) \leq 2 $
How do I continue from here?
linear algebra
Let $A$ be a $2 \times 4$ matrix and $B$ be a $4 \times 4$ matrix, prove that if $rank(A) = 2$ and $rank(B)=3$ then $AB \neq 0$.
I got stuck at $rank(AB) \leq 2 $
How do I continue from here?
Best Answer
As pointed out in the comments your matrix $A$ is the wrong way round... As, for our purposes, is the inequality! (It's right, just not helpful since what we want is $\mathrm{rank}(AB)\ge1$.)
Here's an equivalent way of saying $AB=0$.
Can you see why this can't happen?