# Find the column space of the matrix $\left(\begin{smallmatrix} 1 & 0 & -1 & 0 & 1\\ 0 & 1 & -1 & 2 &0\\ \end{smallmatrix}\right)$

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

• 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. – Bor Kari Apr 10 at 21:09
• 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.) – Théophile Apr 10 at 21:12

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.

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 $$\mathbf R^2$$.

• So is the image space generated by all the linearly independent columns of the matrix representing the transformation? – Mohamad Moustafa Apr 10 at 21:28
• The vector columns, independent or not, are a set of generators of the image space. – Bernard Apr 10 at 21:41
• "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? – Mohamad Moustafa Apr 10 at 21:56
• If you have a basis, you have the whole space. – Bernard Apr 10 at 21:59
• 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? – Mohamad Moustafa Apr 10 at 22:05