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I apologize for asking a question though there are pretty much questions on math.stackexchange with the same title, but the answers on them are still not clear for me.

I have this linear operator:

$$ Ax = (2x_1-x_2-x_3, x_1-2x_2+x_3, x_1+x_2-2x_3); $$

And I need to find the basis of the kernel and the basis of the image of this transformation.

First, I wrote the matrix of this transformation, which is:

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

I found the basis of the kernel by solving a system of 3 linear equations:

$$ \begin{pmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix} $$

It is

$$ kerA = (1,1,1) $$

But how can I find the basis of the image? What I have found so far is that I need to complement a basis of a kernel up to a basis of an original space. But I do not have an idea of how to do this correctly. I thought that I can use any two linear independent vectors for this purpose, like

$$ imA = \{(1,0,0), (0,1,0)\} $$

because the image here is $\mathbb{R}^2$

But the correct answer from my textbook is:

$$ imA = \{(2,1,1), (-1,2,1)\} $$

And by the way I cannot be sure that there is no error in the textbook's answer.

So could anyone help me with this. I will be very grateful, thank you in advance.

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    $\begingroup$ There are many bases for the image. To get one such, find what $A$ does to the standard basis, and throw away the linearly dependent one. $\endgroup$ – André Nicolas Aug 12 '14 at 18:51
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    $\begingroup$ A basis of the image is the columns in the original matrix which correspond to the pivot columns in the row reduced matrix. So presumably the first and second columns of your row reduced matrix are pivot columns, so the first two columns of your original matrix are a basis. There may be a sign error in the answer of your book for the $2$ in the second basis vector. That's just my suspicion, I haven't actually worked it out. $\endgroup$ – Ben West Aug 12 '14 at 18:53
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    $\begingroup$ @BenWest thanks to you and André Nicolas, now it seems clear to me. $\endgroup$ – user907860 Aug 12 '14 at 19:03
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Reducing your matrix $$ \begin{pmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{pmatrix} $$ to row-echelon form gives $$\begin{pmatrix} 1 & -2 & 1\\0 & 1 & -1\\0 & 0 & 0\end{pmatrix},$$

and a basis for the image of $A$ is given by a basis for the column space of your matrix, which we can get by taking the columns of the matrix corresponding to the leading 1's in any row-echelon form.

This gives the basis $\{(2,1,1), (-1,-2,1)\}$ for the image of $A$.

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  • $\begingroup$ Actually does this work for any A or can it happen that the column we pick are linearly dependent? $\endgroup$ – user519338 Mar 29 '18 at 23:50
  • $\begingroup$ @user519338 This method should work for any matrix A. $\endgroup$ – user84413 Jul 7 '18 at 20:45
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The image of a basis of the domain gives a spanning set for the image. You may have to reduce this spanning set to get a basis for the image.

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  • $\begingroup$ could you please provide an example? $\endgroup$ – user907860 Aug 12 '14 at 18:38
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    $\begingroup$ Let's say you had a linear transformation $T:\mathbb{R}^2\rightarrow\mathbb{R}^3$ Suppose also that $T(1,0)=(3,7,1)$ and $T(0,1)=(-6,-14,-2)$. Since any vector in $\mathbb{R}^2$ can be written as a linear combination of $(1,0)$ and $(0,1)$, using the linear transformation properties, we have that the image of any $\mathbb{R}^2$ vector is a linear combination of $(3,7,1)$ and $(-6,-14,-2)$. Thus these two vectors span the image of $T$. In this case, these image vectors are linearly dependent, so you would have to get rid of one of them to get a basis. $\endgroup$ – paw88789 Aug 12 '14 at 18:51

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