Prove that a vector $v \in \ker T$ if and only if $[v]_{B_V} \in \operatorname{null} A$

Let $V$ and $W$ be vector spaces over a field $F$ with $\dim V = n,$ $\dim W = m$ with bases $B_V = (v_1,\ldots,v_n)$ and $B_W = (w_1,\ldots,w_m),$ respectively. Assume $T:V \to W$ is a linear transformation and $A = \mathcal{M}_T(B_V,B_W) = (x_1,\ldots,x_n).$ Prove that a vector $v \in \ker T$ if and only if $[v]_{B_V} \in \operatorname{null} A = \{v \in F^n: Av = 0 \in F^m \}.$

Assume that $v \in \ker T \Rightarrow T(v) = 0.$ Let $v = c_1v_1+\cdots+c_nv_n$ so that $[v]_{B_V} = \begin{pmatrix} c_1 \\ \vdots \\ c_n \end{pmatrix}.$ We have that $T(v) = T(c_1v_1+\cdots+c_nv_n) = c_1T(v_1)+\cdots+c_nT(v_n) = 0.$ Since the $T(v_i)$ can be uniquely written as a combination of the $w_j$ from the basis $B_v,$ we obtain $$0 = c_1T(v_1)+\cdots+c_nT(v_n) = c_1 (a_{11}w_1+\cdots+a_{m1}w_m)+\cdots+c_n(a_{1n}w_1+\cdots+a_{mn}w_m)$$ $$\hspace{1.95cm} = (c_1a_{11}+c_2a_{12} + \cdots + c_na_{1n})w_1 + \cdots + (c_1a_{m1}+c_2a_{m2}+\cdots+c_na_{mn})w_m,$$ implying that $c_1a_{11}+ \cdots + c_na_{1n} = \cdots = c_1 a_{m1} + \cdots + c_n a_{mn} = 0$ since the $w_j$ are linearly independent. \ Thus $$A[v]_{B_V} = c_1x_1 + \cdots + c_n x_n = c_1 \begin{pmatrix}a_{11} \\ \vdots \\ a_{m1} \end{pmatrix} + c_2 \begin{pmatrix}a_{12} \\ \vdots \\ a_{m2} \end{pmatrix} + \cdots + c_n \begin{pmatrix} a_{1n} \\ \vdots \\ a_{mn} \end{pmatrix} = \begin{pmatrix} c_1a_{11} + \cdots + c_n a_{1n} \\ \vdots \\ c_1a_{m1}+\cdots + c_na_{mn} \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix},$$ so that $[v]_{B_V} \in \operatorname{null} A.$

Now assume that $[v]_{B_V} \in \operatorname{null} A.$ Then $A[v]_{B_V} = c_1x_1+\cdots+c_nx_n = 0,$ implying that $c_1a_{11} + \cdots + c_na_{1n} = \cdots = c_1a_{m1} + \cdots + c_n a_{mn} = 0.$ But $$T(v) = T(c_1v_1+\cdots+c_nv_n) = c_1T(v_1)+\cdots+c_nT(v_n) = (c_1a_{11}+\cdots+c_na_{1n})w_1+\cdots + (c_1a_{m1}+\cdots+c_na_{mn})w_m =$$ $$\hspace{2.6cm} = 0w_1+ \cdots + 0w_m = 0.$$ Hence $T(v) = 0,$ proving that $v \in \ker T. \Box$

Is my proof correct?