Find bases for ker(t) and range(t) for a given matrix

Question

Find bases for ker(T) and range(T), and determine whether T is one-to-one or onto.

$T \left({\begin{bmatrix}a_{11} & a_{12} &a_{13} \\a_{21}& a_{22}& a_{23} \end{bmatrix}} \right)$ = $\begin{bmatrix}2a_{11}-2a_{12} & a_{13}+2a_{12} \\0& 0\end{bmatrix}$

I normally can do these type of questions but this seems trivial. So for kernel $2a_{11}-2a_{12} = 0$ and $ a_{13}+2a_{12} = 0$ what's the next step?

Answer 1

Let $a_{11} = t$ then $a_{12} = t$ and $a_{13} = -2t$. The other three entries can be anything. So the kernel is

$$ \left\{ \begin{bmatrix} t & t & -2t \\ x & y & z \end{bmatrix} : t, x, y, z \in \mathbf{F} \right\}. $$

For the range, let $a_{12} = 0$ then you have

$$ \begin{bmatrix} 2a_{11} & a_{13} \\ 0 & 0 \end{bmatrix} $$

so it is clear that the range of $T$ is

$$ \left\{ \begin{bmatrix} a & b \\ 0 & 0 \end{bmatrix} : a, b \in \mathbf{F} \right\}. $$