We define a matrix $Z$ to be a null-space matrix for $A$ if any vector in $N(A)$ (the null space of $A$) can be expressed as a linear combination of the columns of $Z$.
Let $Z$ be an $n\times r$ null-space matrix for the matrix $A$. If $Y$ is any invertible $r\times r$ matrix, prove that $\hat{Z}=ZY$ is also a null-space matrix for $A$.
First note that $Z$ is a null-space matrix for $A$ if and only if $$ N(A)\subseteq \mbox{Ran} Z $$ since the linear combinations of the columns of $Z$ are precisely the vectors in the range of $Z$.
Now if $Y$ is invertible, $Y(\mathbb{R}^r)=\mathbb{R}^r$, so $$ \mbox{Ran}ZY=\mbox{Ran}Z. $$
The result follows.
Let $v$ be any vector in $N(A)$.
Then by the definition of nullspace matrix, there is a $w\in\mathbb{R}^r$ s.t. $Zw = v$. Because $Y$ is invertible, there is a $w'\in\mathbb{R}^r$ such that $w = Yw'$.
Therefore since there is a $w'$ such that $ZYw' = v$ for any $v\in N(A)$, $ZY$ is a nullspace matrix for $A$.
