Proof that for an $n {\times} n$ matrix $A$ , it has $\text{rank}(A) = n$ if the proposition: $AB = 0 \rightarrow B = 0$ is true

Question

I am struggling to prove that for an $n{\times}n$ matrix A, it has $\operatorname{rank}(A)=n$ if the proposition $AB = 0 \rightarrow B = 0$ is true.

This question belongs to the chapter about ranks and inverses and it should be possible to think of a proof using elementary matrices, etc, without having to use theory from vector spaces. Any ideas on how to proceed?

Thanks in advance!

Answer 1

Since $A$ has full rank, it is invertible. So pre-multiplying both sides of the equation $AB = 0$ by $A^{-1}$ yields $B= 0$.

In other words, by pre-multiplying $A$ by some elementary matrices, $E_1, \ldots, E_n$, say, we obtain the identity matrix $I_n$ or order $n$. So from $AB = 0$ we obtain $$ \left(E_n \ldots E_1 \right) \big( AB \big) = \big( \left(E_n \ldots E_1 \right) A \big) B = 0, $$ which gives $B = 0$.

Answer 2

Let $v_1,\ldots,v_n$ denote the columns of $A$. $A$ has rank $n$ if these are linearly independent. Let $\sum_i c_iv_i=0$. Let $B$ be the matrix with $(c_1,\ldots,c_n)^T$ as first column and 0 else. Then the assumption implies that $c_1=\ldots=c_n=0$. Thus, $A$ has full rank. The other direction is analogous.

Answer 3

Suppose A does not have full rank, then find a basis for its null space, and let B be the matrix whose columns are formed from that basis. Then the image of B is the null space of A, and B is not zero as the null space of A has dimension at least 1, but AB is zero, as the image of AB is zero.