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Let $ \begin{pmatrix} 0&1&0&0&0\\ 0 & 0&0&0&0\\ 0 & 0&0&0&0\\ 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 &1 \end{pmatrix} $ $\in \mathbb R^{5,5}$

I think a basis of the image of this matrix is given by $\cal B_I=\{ e_2,e_4,e_5\}$ and a basis of the kernel by $\cal B_k=\{ e_1,e_3\}$. In the solution it is given by $B_I\{ e_1,e_4,e_5\}$ and $B_k=\{e_1,e_3\}$. Actually the image is simply the linearly independent colums in the reduced matrix form. I just want to make sure that the solution is wrong, if so, a comment if I am right would be enough, thanks!

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Note that the second column $C_2$ is $e_1$, $C_4=e_4$ and $C_5-C_4=e_5$ thus $\cal B_I=\{ e_1,e_4,e_5\}$ for the image is correct.

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    $\begingroup$ ok yes, I see. lol $\endgroup$ – user519338 Mar 29 '18 at 20:27
  • $\begingroup$ @J.D Well done! $\endgroup$ – user Mar 29 '18 at 20:30

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