Suppose $A$ is an $m\times n$ matrix. Solve the homogeneous system of linear equations $Ax=0$. We know that,
- $x$ has nonzero solution, if and only if $\text{rank}(A)<n$.
- $\#$ free parameters of $x$ = $\text{dim}(\text{null}(A))$ = $n-\text{rank}(A)$.
Note that here "nonzero" means "not all entries of $x$ are zero". But for a nonzero $x$, there is a case where some entries of $x$ must always be zero. For example, $$ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix},$$then the first entry of $x$ must always be zero. In other words, the null space of $A$, which should be a subspace of $\mathbb{R}^3$, is actually a subspace of $\mathbb{R}^2$. Seeing from the basis of $\text{null}(A)$, then some dimension of these vector(s) is always zero.
I call this the "degenerated null space". I was wondering if there is any particular properties about it, e.g. what is the iff condition for a matrix $A$ s.t. $\text{null}(A)$ is degenerated? What is the relationship between $A$ and $A'$, the submatrix of $A$ which deletes columns w.r.t. always-zero dimension?
Also looking for related materials. Would really appreciate it if you could provide some links, materials about it :)
Your kernel is "degenerated" iff there is some $e_i$ so that $e_i$ is orthogonal to the kernel. But as the kernel is defined to be everything that is orthogonal to the rows of $A$ we get that this is the case iff $e_i$ is in the space spanned by the rows of $A$. For example in your case $$ e_1 = 2(1,1,1) - (1,2,2)$$ So the kernel is degenerated in the $i$-th coordinate iff $$ xA = e_i $$ has a solution.
Then $A$ and $A'$ are basically the same matrix, just with some $0$-columns removes. Your question here is quite broad, so I’m not sure what relationship you’re interested in. But note that the kernel of $A$ is isometrically isomorphic to some $\mathbb R^k$ where $k$ is the dimension of the kernel anyway.