Determinant of basis of nullspace of matrix

Question

Assume we have a matrix $A \in \mathbb{R}^{n \times p}$ that has rank $n$ (or perhaps $k$ more generally). We can define a matrix $N \in \mathbb{R}^{p \times (p-n)}$ that is a basis for the nullspace of $A$, i.e., $N$ is full column rank and $A N = 0$.

We can also define a basis $C \in \mathbb{R}^{p \times n}$ that is a basis for the column space of $A$, where $C$ is also full column rank.

I am interested in the determinant of $N^TN$ and how it can be expressed as the determinant of some function of $A$ and $C$, i.e. not involving $N$.

Thanks for your help!

Answer 1

If $B = \{v_1,\dots, v_n\}$ is a basis for the null space of $A$, then so is $B' = \{c_1v_1,\dots,c_nv_n\}$, where $c_1,\dots,c_n$ are ANY non-zero scalars. Therefore, $N$ is not unique, and you can make $\det(N^TN)$ be any non-zero scalar.

Take a very simple example, $A = \begin{bmatrix}1&0\end{bmatrix}$. The null space of $A$ is spanned by $ N =\begin{bmatrix}0\\1\end{bmatrix}$, but also by $N' = \begin{bmatrix}0\\2\end{bmatrix}$. Clearly, $\det(N^TN) = 1$, and $\det(N'^TN') = 4$.

Answer 2

In the same spirit as Carl Chaanin's answer: We typically choose $N$ as the matrix that contains as columns the normalized basis vectors of the null space. These vectors are the same whether you consider $A$, or $2A$, or any $\alpha A$, $\alpha\neq 0$. The same thing can be said about the matrix $C$.

As a consequence, if you want to express $$\det(N^T N)=\det f(A,C)$$ then this will at the very least not be a simple function of $A$ because you will have to have that $f(A,C)=f(\alpha A, C)$.