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

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.

• Why don't you take $B=0$? Commented Dec 2, 2020 at 18:18
• If you want to restrict what cannot be used, it would likely be more helpful to explain what can be used. IE Are we allowed to use matrix theory about dimensions? Commented Dec 2, 2020 at 18:18
• Yes @Calvin we can use that but am I going wrong? Commented Dec 2, 2020 at 18:20
• If $AB = 0$, what can you say about the dimensions of their kernels? Commented Dec 2, 2020 at 18:21
• Yes, but looking at the matrix $B$ in this way where am I going wrong? Commented Dec 2, 2020 at 18:24

$$B$$ 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 $$2$$.