<s>[INST] Basis of kernel and image of a linear transformation - verification The transformation matrix I found is: $$\begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 0 & 0\end{pmatrix}$$
Is this how a basis for $\ker$ and $\mathrm{im}$ is calculated?
$$\begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 0 & 0\end{pmatrix} \sim \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{pmatrix}$$
Therefore a basis for $\ker$ is $\{0\}$.
For the image:
$$\begin{pmatrix} 1 & 1 & 0 \\ -1 & 1 & 0 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$$
Therefore a basis for $\mathrm{im}$ is $\left\{ \begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix} \right\}$.
 [/INST] This CW answer intends to remove the question from the unanswered queue.

As DonAntonio already noted, your calculation is correct.
</s>