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I have the following matrix

$$ \begin{pmatrix} 1 & 0 & -1 & 0 & 1\\ 0 & 1 & -1 & 2 &0\\ \end{pmatrix} $$

And I am unsure as to how to write the column space/image for the transformation it represents.

I know that, for a Mapping $M:S\to R$, $\operatorname{Im}(M)$ is the subspace of elements $x$ of $R$ for which there is some element $y$ of $S$ such that $M(y)=x$.

Should I write the Image as $$\operatorname{Im}(M)= \left(\begin{matrix} x_{1}-x_3+x_5\\ x_{2}-x_3+2x_4\\ \end{matrix}\right) $$

for $x_1,x_2,x_3,x_4,x_5 \in\mathbb R$?????

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  • $\begingroup$ If you find the reduced row echelon form, you can write everything in terms of the free variables then write it in terms of two vectors that span a plane. $\endgroup$ – Bor Kari Apr 10 at 21:09
  • $\begingroup$ To put @Omega's answer another way: notice that the first two columns generate $\Bbb R^2$. With only two rows, you can't get any bigger than this, so the other columns are redundant. Therefore the column space is $\Bbb R^2$. (You could make exactly the same argument using, say, columns $3$ and $5$; it's just columns $1$ and $2$ are the obvious choice.) $\endgroup$ – Théophile Apr 10 at 21:12
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That is not the image space, that is just another way to write the transformation. The image space is the entire $R^2$, since for any vector $(x,y)$, at least the vector $(x,y,0,0,0)$ is mapped onto it.

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The image space is generated by the images of the vectors in the basis of $\mathbf R^5$ chosen to represent the transformation. Now, these images are precisely the $5$ column vectors. As they contain the vectors $\;\begin{pmatrix} 1\\0\end{pmatrix}$ and $\;\begin{pmatrix}0\\1\end{pmatrix}$ , which are a basis of $\mathbf R^2$, this means the image space is equal to $\mathbf R^2$.

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  • $\begingroup$ So is the image space generated by all the linearly independent columns of the matrix representing the transformation? $\endgroup$ – Mohamad Moustafa Apr 10 at 21:28
  • $\begingroup$ The vector columns, independent or not, are a set of generators of the image space. $\endgroup$ – Bernard Apr 10 at 21:41
  • $\begingroup$ "As they contain the vectors (1,0) and (0,1) , which are a basis of R2, this means the image space is equal to R2." Why does it mean that? $\endgroup$ – Mohamad Moustafa Apr 10 at 21:56
  • $\begingroup$ If you have a basis, you have the whole space. $\endgroup$ – Bernard Apr 10 at 21:59
  • $\begingroup$ So is the Image space is the space spanned by the 5 column vectors. But in this case since the other three are just a combination of the first two, and the first two span the entire R2 set, the image space is R2? $\endgroup$ – Mohamad Moustafa Apr 10 at 22:05

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