We are given the vector space $V=F^3$ where $F$ is a field and we are given $B=\{v_1,v_2,v_3\}$ is a basis in $V$, and $T: V \rightarrow V$ is a linear transformation. We are also given:
$$[T]_B=\begin{bmatrix} -1 & -1 & -3 \\ -5 & -2 & -6 \\ 2 &1 &3 \end{bmatrix}$$Find $\operatorname{Ker}(T)$.
Here is what I tried: for $v \in V$ with $[v]_B=(a,b,c)$ we get that $v$ is in the kernel if and only if $[T]_B \cdot [v]_B = 0$. A simple calculation shows that the requirment is $a=0, b=-3\alpha,c=\alpha$ for some $\alpha \in F$. So we get $\operatorname{Ker}(T)=\{-3\alpha \cdot v_2+ \alpha \cdot v_3 : \alpha \in F\}$.
This is the most simplified answer I got, but the answer should be $\operatorname{Ker}(T)=\operatorname{Span}\left\{(0,-3,1)\right\}$.
Can someone help me get that answer? I'm not sure how to continue.
You got it right. Note that\begin{align}-3\alpha v_2+\alpha v_3&=\alpha(-3v_2+v_3)\\&=\alpha(0,-3,1).\end{align}So,\begin{align}\{-3\alpha v_2+\alpha v_3\,|\,\alpha\in\mathbb{R}\}&=\bigl\{\alpha(0,-3,1)\,|\,\alpha\in\mathbb R\bigr\}\\&=\operatorname{span}\bigl\{(0,-3,1)\bigr\}.\end{align}