Show that there exists a matrix that when pre-multiplied by the design matrix yields the identity matrix. I'm given a matrix $C=(\underline{0}_{k}, I_{k})$, where $\underline{0_{k}}$ is the  $k \times 1$ zero vector, and $I_{k}$ is the $k \times k$ identity matrix. I'm asked to show that there exists some matrix $M$ such that $C=MX$ and $M\underline{1}_{n}=\underline{0}_{k}$.
Where $X$ is a full column rank $n \times (k+1)$ design matrix with the first column being a vector of one's (for the intercept).
I can notice that each row in $M$ has to sum to zero. I have tried approaching this via pre/post multiplying $C=MX$ by a number of choices, and by trying to find values for the rows of $M$ that would satisfy:
$\sum_{j}M_{ij}=0$
and
$\sum_{j}M_{ij}X_{ji}=1$ $\quad$and$\quad$ $\sum_{j}M_{ij}X_{ji'}=0$
with little success. Can anyone offer me a hint or a pointer on how to approach this problem?

Hint: Let $N$ be any $k \times n$ matrix for which and $N \underline{1}_n = 0$.  Because $X$ has $\underline{1}_n$ as its first column, the nullspace of $N X$ automatically contains the nullspace as $C$. 
If two matrices $A,B$ have the same nullspace, then they are row-equivalent. That is, there exists and invertible matrix $P$ such that $A = PB$.
Note: Because $C$ has rank $k$, we must have $n \geq k$. Because $X$ has full column rank, we must have $n \leq k+1$.
