Let $A$ be a matrix of size $5\times 5$ with rank(A)=3 .Prove that there exists a matrix $B$ such that $AB=0$

Question

I have to prove that the rank of the matrix $B=2$ and given that $B$ is a $5\times 5$ non zero matrix.

My attempt : Let the columns of $B$ be $X_1,X_2,...,X_5$.Then $AX_1=0$,$AX_2=0$,...,$AX_5=0$.Now since the rank of the matrix $A$ is given as $3$ .So there are $2$ free variables and the possible dimensions of $X_1,..,X_5$ are $2$.

Hence combining we can find the possible rank of $B$ is 2 .

Where am I going wrong in the proof?Also I have not been taught linear transformation so I cannot use it here.

Answer 1

$B$ can be any $5 \times k$ matrix whose columns are in the null space of $A$ which is 2 dimensional. So $B$ has rank at most $2$.