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.

  • $\begingroup$ 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 $Nv=0$. $\endgroup$ – A.Γ. Aug 17 at 21:05
  • $\begingroup$ @A.Γ. Yes you are right. I made a mistake. $\endgroup$ – 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.


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