How to find matrix $K$ such that $KM=0$ with $M$ full-rank? [closed]

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?

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.

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!]