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I'm asked to find a basis of the image of the linear transformation $f:\mathbb{R^4} \rightarrow \mathbb{R^3}$ defined as $f(v) = (v_1-v_3+v_4,2v_1+v_2+2v_3+v_4,3v_1-v_2+v_4)$.

I found the matrix of the linear transformation $Af = \begin{pmatrix} 1 & 0 & -1 & 1\\ 2 & 1 & 2 & 1\\ 3 & -1 & 0 & 1 \\ \end{pmatrix} $ with rank equal to 3.

I tried to find a basis this way : $Im f=\{w\in\mathbb{R^3} | \exists v\in\mathbb{R^4}s.t f(v)=w\}$ i.e $$ \left\{\begin{pmatrix}v_1-v_3+v_4\\2v_1+v_2+2v_3+v_4\\3v_1-v_2+v_4\end{pmatrix}\right\}=\left\{v_1\begin{pmatrix}1\\2\\3\end{pmatrix} + v_2\begin{pmatrix}-1\\2\\0\end{pmatrix} + v_3\begin{pmatrix}0\\1\\-1\end{pmatrix} + v_4\begin{pmatrix}1\\1\\1\end{pmatrix}\right\} $$ The problem is that I have 4 vectors here, and I know the image should be of the dimension 3 since the rank of the matrix is 3.But I don't know how to compute the basis formed from 3 vectors ,I think one of them is a linear combination of another , but I don't know wich one .Any help please ? Thank you !

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    $\begingroup$ choose a random vector on the image and find two linearly independent ones. Moreover: if the rank of the matrix is $3$ then we knows that the image is $\Bbb R^3$, then any basis of $\Bbb R^3$ will work! $\endgroup$ – Masacroso Jun 13 '17 at 20:58
  • $\begingroup$ Try column-reducing (like row-reducing, but with the columns). $\endgroup$ – user357980 Jun 13 '17 at 20:59
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    $\begingroup$ @Masacroso that makes sense, so I can chose the standard basis and It will be fine , right ? $\endgroup$ – Eduard Valentin Jun 13 '17 at 21:01
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    $\begingroup$ Yes. The unique vector subspace of dimension $3$ in $\Bbb R^3$ is $\Bbb R^3$ itself. $\endgroup$ – Masacroso Jun 13 '17 at 21:03
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You have noted that the span of the four columns of $A$ is the image.

It remains to find a subset of these columns that are linearly independent. Just try them in order. You can grab the first two columns, since they are clearly linearly independent. Then you can check if the third column can be written as a linear combination of the first two; this is not the case (you can check this directly, or show that the determinant of the first three columns is nonzero, etc.), so the first three columns form a basis.

More generally for larger problems where such a direct approach is hard, column reduction (as user357980 suggested) will always work.

As you noted in the comments, the standard basis works as well. To arrive at this, you would need to justify that the rank is $3$, as Masacroso noted. This prevents cases where image is lower-dimensional, e.g. if the columns are all scalar multiples of the first column.

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  • $\begingroup$ I did that also, I checked the determinant formed by the first 3 vectors , I saw that is not zero , so then I could chose those 3 first vectors as a base for my image and lefting the 4 th vector behind ? $\endgroup$ – Eduard Valentin Jun 13 '17 at 21:05
  • $\begingroup$ @EduardValentin Yes, exactly. $\endgroup$ – angryavian Jun 13 '17 at 21:06
  • $\begingroup$ But if I chose those 3 vectors, how am I sure that the 4 th vector can be formed as a linear combination of the first 3 ,since is part of the image he should come from somewere , right? $\endgroup$ – Eduard Valentin Jun 13 '17 at 21:09
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    $\begingroup$ If you have three independent vectors in $\mathbb{R}^3$, you can span everything in $\mathbb{R}^3$. $\endgroup$ – Siong Thye Goh Jun 13 '17 at 21:14
  • $\begingroup$ Ok, I got it , thank you ! $\endgroup$ – Eduard Valentin Jun 13 '17 at 21:15
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I think any one R^3 vector above out of the 4 can be written as combination of other three.

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  • $\begingroup$ I can't find any vector that can be written as a combination of other three.. $\endgroup$ – Eduard Valentin Jun 13 '17 at 21:02
  • $\begingroup$ Try solving [1;1;1] = c1[1;2;3]+c2[-1;2;0]+c3[0;1;-1], it has a solution. Make a matrix of above system of equations, then we can see it is invertible, so solution exists. $\endgroup$ – ab123 Jun 13 '17 at 21:10
  • $\begingroup$ Oh , you are right .. solution is $(\frac{5}{7},\frac{3}{7},\frac{4}{7})$ $\endgroup$ – Eduard Valentin Jun 13 '17 at 21:14
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If it's not completely obvious which column in a linear combination, then just row reduce?

$\begin{pmatrix} 1 & 0 & -1 & 1\\ 2 & 1 & 2 & 1\\ 3 & -1 & 0 & 1 \\ \end{pmatrix} \to \begin{pmatrix} \color{red}{1} & 0 & -1 & 1\\ 0 & \color{red}{1} & 4 & -1\\ 0 & 0 & \color{red}{1} & \frac{-3}{7} \\ \end{pmatrix} $

Notice that this is row echeolen and not reduced, since we only want to find out with columns have pivots.

From this you see that the basis is: $$b=\{[1,2,3],[0,1,-1],[-1,2,0]\}$$

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I like $0$'s in matrices.

Let $$\begin{bmatrix} 1 \\ 2\\ 3\end{bmatrix} = a \begin{bmatrix} 0 \\ 1\\ -1\end{bmatrix} + b \begin{bmatrix} -1 \\ 2\\ 0\end{bmatrix}$$

Immediately, from the first row, we can see that $b=-1$.

From the third row, we can see that $a=-3$.

Let's verify using the second row, $$a(1)+2b=-3-2=-5$$

Hence the first column is not in the span of the second and third column.

Hence the first three columns form a basis.

Remark: If you know that the dimension is $3$, you can use the standard basis directly.

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