Let $V$ be a finite-dimensional vector space with (ordered) basis $\beta=(b_1,...,b_n)$, and let $T:V\rightarrow V$ be a linear transformation. Let $B=[T]_\beta$ by the $\beta$-matrix of $T$
Prove that for all $\vec{v}\in V, v\in$ ker$(T)$ iff $[\vec{v}]_\beta \in$ ker $(B)$.
The proof I came up with includes a step that I'm not sure I'm allowed to do
$T\vec{v}=\vec{0}$
$[T\vec{v}]_\beta=[\vec{0}]_\beta$
$[T]_\beta\ [\vec{v}]_\beta=\vec{0}$
I'm unsure if I can distribute the $[\ ]_\beta$ between the two. I feel like I can since it would perform the same operation, but is there a proof that this distribution property applies?
Yes $[Tv]_\beta = [T]_\beta[v]_\beta$. Note that for $v = a_1b_1 + \cdots + a_nb_n \in V$ we have: $$ \begin{align} [Tv]_\beta &= [a_1T(b_1) + \cdots + a_nT(b_n)]_\beta \\ &= a_1[T(b_1)]_\beta + \cdots + a_n[T(b_n)]_\beta \end{align} $$
The last equality follows because choosing a basis defines an isomorphism. In particular, the map $[\cdot]_\beta: V \to \mathbb{R}^n$ which sends $v = a_1b_1 + \cdots + a_nb_n$ to $[v]_\beta = (a_1, \ldots , a_n)^T$ is $\mathbb{R}$-linear.
On the other hand, we have:
$$ \begin{align} [T]_\beta[v]_\beta &= ([T(b_1)]_\beta, \ldots , [T(b_n)]_\beta)\cdot (a_1, \ldots , a_n)^T \\ & =a_1[T(b_1)]_\beta + \cdots + a_n[T(b_n)]_\beta \end{align} $$
As you can see, both give the same expression.