Class of matrices such that every element of their kernel contains a zero coordinate Consider the following class of matrices:
$$S=\{A\in M_n(\mathbb{R}): Ax=0\Rightarrow \exists i: x_i =0\}$$
I have two questions:

*

*Is this set known by any name?

*What are some equivalent ways of characterizing this behavior for a given $A$? E.g. would it be correct to state that $A\in S$ if and only if $A$ has a column $a_i$ which is not contained in the linear span of its remaining columns: $\exists i: a_i \not \in \langle a_j\rangle_{j\neq i}$?

 A: The set $S$ has not been well-defined (what is $x$?). Do you mean
$$S = \{A \in \mathcal{M}_n(\mathbb{R}) : \forall x \in \mathrm{Ker}(A), ~\exists i \in [1,n], ~x_i=0\} ?$$
If yes, the set $S$ is large since it contains all invertible matrices.
Moreover, $S = S_1 \cup \ldots \cup S_n$, where $S_i$ denotes the set of all matrices whose kernel is entirely contained
in the hyperplane $\{x \in \mathbb{R}^n : x_i=0\}$.
Indeed, if no such hyperplane contains $\mathrm{Ker}(A)$, then for each $i$ one can find a vector in $\mathrm{Ker}(A)$ whose $i$-th coordinate is not zero. By taking suitable linear combinations of such vectors, we get a vector in $\mathrm{Ker}(A)$ without any null coordinate.
Last, $S_i$ is exactly the set of all matrices $A \in \mathcal{M}_n(\mathbb{R})$ whose $i$-th column is not a linear combination of the other ones, so your guess is correct. Indeed, saying that the $i$-th column is a linear combination of the other ones is equivalent to say that $\mathrm{Ker}(A)$ contains some vector whose $i$-th coordinate is $1$ (equivalently, is not $0$).
A: An alternate definition are matrices that consist of columns for which some of them are orthogonal to all the others.
Consider $x$ such that $Ax=\sum_i x_ia_i=0$ and assume that the first column is orthogonal to all the others. Then, we have that $a_1^TAx=||a_1||^2_2x_1=0$, which implies that we necessarily have $x_1=0$.
The general case follows from the same arguments repeated multiple times.
