Prove that AB = 0 for two matrices if rank(A) < n

Question

Problem

Suppose I have a matrix $A \in \mathbb{K}^{m \times n}$ with $rank(A) < n$.

I need to prove: $\forall k \in \mathbb{N} \; \exists B \neq 0 \in \mathbb{K}^{n \times k} : AB = 0$

Approach

Since $rank(A) < n$ at least one column of $A$ is linearly dependent, which means we can bring the matrix into the row echelon form and get one zero row/column. That implies that for the dot product of the zero row and some arbitrary column of B, some entries in B do not matter, which means we can fill the matrix with zeroes everywhere else and choose these special coefficients such that the matrix multiplication yields the zero matrix $0 \in \mathbb{K}^{m \times k}$

I'm pretty sure my approach is wrong. Also, I can't seem to find out how to prove this in a rigorous mathematical/symbolical way.

Any hints are highly appreciated. I'm new in university and still need to get used to this kind of abstract problem solving thinking.

Answer 1

View $A$ as a linear map from $\mathbb{K}^n$ t0 $\mathbb{K}^m$. Then by the fundamental theorem of linear maps, we have $$\text{dim }\mathbb{K}^n = \text{dim null }A + \text{dim range }A.$$ Since $\text{dim }\mathbb{K}^n = n$ and $\text{dim range }A = \text{rank }A$ we get $$\text{dim null }A = n - \text{rank }A.$$ If $\text{rank }A < n$, then $$\text{dim null }A \geq 1,$$ which means that $A$ has a nonzero nullspace. If you fill the columns of $B$ with vectors in the null space of $A$ then $AB = 0$.

Answer 2

Hint: If $B$ has columns $b_1,\dots,b_k$, then $$ AB = A \pmatrix{b_1 & \cdots & b_k} = \pmatrix{Ab_1 & \cdots & Ab_k}. $$