How to find matrix $K$ such that $KM=0$ with $M$ full-rank? I have a full-rank matrix $M$ of size $(2n,n)$ and I try to find a non-zero matrix $K$ of size $(n,2n)$ such that the product of $K$ and $M$ results in the zero matrix, i.e., $KM=0$.
The null-space of $M$ is trivial since it's a full-rank matrix, so I currently see no way in using the null space of $M$ to find $K$. Are there suggestions to my problem?
 A: It's perhaps easier if you consider a matrix $N$ of shape $n\times 2n$ with rank $n$. Its null space has dimension $n$, and so you can surely find a vector $v\ne0$ such that $Nv=0$.
Now take $L=[v\ v\ v\ \dots\ v]$ with $n$ columns and $NL=0$.
OK, now consider $N=M^T$ and you see that $K=L^T$ fits the bill.
A: I assume we are working over the real numbers $\mathbb{R}$.
Let $m_1, m_2, \dots, m_n$ be the columns of $M$ and let $e_1,e_2,\dots, e_{2n}$ be the standard basis of the column space.
Apply the Gram-Schmidt process to the sequence $m_1, m_2, \dots, m_n,e_1,e_2,\dots, e_{2n}$ to extract an orthonormal basis $f_1,\dots,f_{2n}$.
For $K$ take the matrix whose rows are $f_{n+1}^T,\dots,f_{2n}^T$.
[I actually think this is rather horrid!]
A: $T : F^{n} \to F^{2n}$  be a linear map with  $dim(R(T)) =n$
Define a linear map $S:F^{2n} \to F^{n}$ such that $R(T) \subseteq Ker(S) $
Assume that such linear map $S$ exists.
Then
$S\circ T: F^{n} \to F^{n}$
Claim: $S\circ T=0 $
Choose, $v\in F^{n}$
\begin{align} S\circ T(v) &=S( T(v))=S(0)=0 \end{align}
Since, $T(v) \in Ker(S) \implies S(T(v)) =0$
Now, we prove the existence of such linear map $S$
Define,  $ S_1 : F^{2n}\to F^{2n}/{R(T)} $ by $$S_1(v) =v+R(T) $$
Then, $Ker(S_1) =R(T) $
Now, $F^{2n}/{R(T)}\cong F^n$
Let, $S_2 : F^{2n}/{R(T)}\to F^n$ be an isomorphism, then
$S_2\circ S_1=S$ be the required linear map satisfying   $ST=0$.
