True or false? Prove it.

Question

If $A$ is an $n\times n$ invertible matrix and $B$ is an $n\times m$ matrix, then $\operatorname{rank}(AB) = \operatorname{rank}(B)$.

Is this true or false?

I've tried proven that if $B=0$, then $\operatorname{rank}(AB)=\operatorname{rank}(B)$, but I don't know how to do it when $\operatorname{rank}(B)$ isn't zero.

Answer 1

Hint
The statement is true, because $\mathrm{rank}(A) = n$, i.e. all columns of $A$ are linearly independent (wich implies...)

Answer 2

True. Because the dimension of the kernal of $AB$ is the same as the dim of the kernel of $B$.

Answer 3

You surely know that, in general, $$ \operatorname{rank}(AB)\le\operatorname{rank}(B) $$

By the same reason, $$ \operatorname{rank}(A^{-1}AB)\le\operatorname{rank}(AB) $$

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How to see $\operatorname{rank}(AB)\le\operatorname{rank}(B)$? This requires the rank-nullity theorem (but there are other proofs). Suppose $A$ is $m\times n$ and $B$ is $n\times p$. Then we can define linear maps \begin{align} f_A&\colon\mathbb{R}^n\to\mathbb{R}^m, && f_A(v)=Av\\ f_B&\colon\mathbb{R}^p\to\mathbb{R}^n, && f_B(w)=Bw\\ f_{AB}&\colon\mathbb{R}^p\to\mathbb{R}^n && f_{AB}(w)=ABw \end{align} It is obvious that $f_{AB}=f_A\circ f_B$ and that $\ker f_{B}\subseteq\ker f_{AB}$. By the rank-nullity theorem \begin{align} p&=\dim\ker f_B+\operatorname{rank}(B)\\ &=\dim\ker f_{AB}+\operatorname{rank}(AB) \end{align} and, since $\dim\ker f_B\le\dim\ker f_{AB}$, we deduce that $$ \operatorname{rank}(AB)\le\operatorname{rank}(B) $$

Answer 4

Consider the columns of $B$ to form a spanning set for a subspace. Suppose you know the dimension of that subspace and it is $d$. Now, for the columns of $AB$, can you find linear combinations that span that subspace also giving dimension $d$? And if so, is it ever possible that rank$(AB)$ > rank$(B)$?

Answer 5

rank(AB)≤ min{rank(A), rank(B)} implies rank(AB) ≤ rank(B)

Given A is invertible.

rank(B)=rank(A−1AB)≤ min{rank(A−1), rank(AB)},A−1 means the inverse of A. i.e, rank(B)≤rank(AB)

Thus, rank (AB)=rank(B).