# For $4\times 3$ matrix $M$, for any $3\times4$ matrix $N$, $\exists 0 \neq v \in \mathbb{C}^4$ such that $MNv = 0$.

Define $$M := \left(\begin{matrix}1&-1&2\\2&-1&1\\-4&1&0\\3&-2&3\end{matrix}\right)$$. Prove or disprove: For every $$3 \times 4$$ complex matrix $$N$$, there is a non-zero vector $$v \in \mathbb{C}^4$$ such that $$MNv = 0$$.

I do not quite sure how to start this problem. Writing $$u := Nv := (x, y , z,w)$$. If $$Mu = 0$$, by doing some elementary row operations on the augmented matrix, then I should have that

$$\left(\begin{matrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{matrix}\right).$$ That is, $$u = (0,0,0,w)$$ for any $$w$$ would be fine. But I don't know how to proceed, could anyone give me a hint?

I made some mistakes above, and I think the correct way to think about this problem is given in the answer.

• The vector $u=Nv$ is to be of dimension $3$. Why $u=(x,y,z,w)$? A simple argument: any matrix $3\times 4$ has a non-trivial kernel, i.e. there exists a non-zero $v$ such that $Nv=0$. – A.Γ. Aug 17 at 21:05
• @A.Γ. Yes you are right. I made a mistake. – mathdoge Aug 17 at 21:11

The rank of a product of matrices cannot be higher than the rank of either factor. Here, without looking at the entries, you know that $$\operatorname{rk} M\leq 3$$, therefore $$\operatorname{rk} MN\leq 3$$ as well, and by the rank-nullity theorem $$MN$$ must have a non-trivial kernel no matter what $$N$$ is.