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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 $\left\{ \begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix} \right\}$.

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    $\begingroup$ Looks fine to me.! $\endgroup$
    – DonAntonio
    May 22, 2013 at 22:04

1 Answer 1

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This CW answer intends to remove the question from the unanswered queue.


As DonAntonio already noted, your calculation is correct.

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