Faster way to check whether a set of $3 \times 3$ matrices is the subspace of some vector space?

Question

Is the following subset of the vector space $V=Mat_{3,3}(\mathbb{R})$ are subspaces of $V?$ $\\$

The $3\times 3$ matrices such that the vector $A$ \begin{align} A\begin{bmatrix} 7 \\ 6 \\ 8 \\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix} \end{align}

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I notice that this is essentially a homogenous system of linear equations $A\underline{v}=\underline{0}$, but am not sure how I can use this to do something. Instead, I have just been trying to think of counterexamples that don't satisfy the subspace criterion: \begin{align} \text{If } \emptyset \neq X \subseteq V,\text{ then } X\leq V \iff X \text{ closed under vector addition and scalar multiplication (subspace criterion)}. \end{align}

but it will take so long to consider individual examples of A that satisfy the condition $Av=0$ and then trying to find a counterexample that fails the subspace criterion. Furthermore, how will I even know when I've exhausted enough options?

Answer 1

Let $v = \pmatrix{7\cr 6\cr 8}$. You want to show that $$ V = \{A \in M_3({\bf R}) \mid Av =0\} $$ is a vector subspace of $M_3({\bf R})$. This is proven by showing that for all $A,B\in V$, and all $\lambda \in {\bf R}$, $\lambda A + B \in V$. And indeed if $Av=0$, $Bv=0$, then by definition of the sum of matrices, $$(\lambda A+B) v = \lambda A v + B v = 0.$$

Answer 2

Let $v\in \mathbb{R}^3$ be a fixed vector.

Then it is easy to check that $A\mapsto Av$ is a linear map from $M_{3\times 3}(\mathbb{R})$ to $\mathbb{R}^3$. The set you describe is the kernel (or null space) of this map, so from general principles is a subspace.