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$A$ is the linear mapping $f(x)= Ax,\mathbb{R} \rightarrow \mathbb{R}$

$$f\left( \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix}\right)= \begin{pmatrix} x_{2}-x_{3}\\ x_{1}+3x_{2}-2x_{3}\\ x_{1}-4x_{2}+5x_{3} \end{pmatrix}$$

What's the image $\text{Im}(f)$? What's the basis of the image?

So far I have only calculated the image / basis of matrix but I hope / think it's actually the same for these mappings.

So I form $f$ to a matrix $\begin{pmatrix} 0 & 1 & -1\\ 1 & 3 & -2\\ 1 & -4 & 5 \end{pmatrix}$

Then transpose this matrix: $\begin{pmatrix} 0 & 1 & 1\\ 1 & 3 & -4\\ -1 & -2 & 5 \end{pmatrix}$

Now get as many zero lines as possible using Gauss:

Take third line and add it to second line:

$\begin{pmatrix} 0 & 1 & 1\\ 0 & 1 & 1\\ -1 & -2 & 5 \end{pmatrix}$

Now make second line negative by $\cdot(-1)$ and add it to first line:

$\begin{pmatrix} 0 & 0 & 0\\ 0 & -1 & -1\\ -1 & -2 & 5 \end{pmatrix}$

Transpose back we have $\begin{pmatrix} 0 & 0 & -1\\ 0 & -1 & -2\\ 0 & -1 & 5 \end{pmatrix}$

Thus $\text{Im}(f)= \text{span} \left(\left\{ \begin{pmatrix} 0\\ -1\\ -1 \end{pmatrix},\begin{pmatrix} -1\\ -2\\ 5 \end{pmatrix} \right\}\right )$ and basis $B= \left\{ \begin{pmatrix} 0\\ -1\\ -1 \end{pmatrix},\begin{pmatrix} -1\\ -2\\ 5 \end{pmatrix} \right\}$


Can you please tell me if I did it correctly and if not how to do it correctly?

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    $\begingroup$ Keep in mind that there’s no such thing as the basis of a vector space. For instance, $\{(0,1,1)^T,(1,3,-4)^T\}$ is another easily-found basis for $\operatorname{im}(f)$. $\endgroup$ – amd Mar 13 '17 at 17:44
  • $\begingroup$ @amd Oh okay good to know :) What I did above is correct by the way? $\endgroup$ – cnmesr Mar 13 '17 at 17:49
  • $\begingroup$ Looks OK to me. You can check your work by prepending the vectors you got to $A$ and row-reducing. $\endgroup$ – amd Mar 13 '17 at 17:52
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You start with the matrix $$ f = \left[ \begin{array}{rrr} 0 & 1 & -1 \\ 1 & 3 & -2 \\ 1 & -4 & 5 \\ \end{array} \right]= \left[ \begin{array}{rrr} f_{1} & f_{2} & f_{3} \end{array} \right]. $$

This looks like an academic exercise, so we look for trivial combinations of rows or columns. We see the column property $$ f_{1} - f_{2} = f_{3}. $$

The image of $f$ is the combination of all independent column vectors: $$ \text{Im} \left( f \right) = a_{1} f_{1} + a_{2} f_{2} = a_{1} \left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right] + a_{2} \left[ \begin{array}{r} 1 \\ 3 \\ -4 \end{array} \right] % = % \left[ \begin{array}{r} a_{2} \\ a_{1} + 3 a_{2} \\ a_{1} - 4 a_{2} \end{array} \right]. $$ The column combination formula provides the kernel $$ \text{Ker} \left( f \right) = % \left[ \begin{array}{r} 1 \\ -1 \\ -1 \end{array} \right]. $$

A minimal basis is any two linearly independent vectors from $$ \text{span} \left\{ \, \left[ \begin{array}{r} 0 \\ 1 \\ -1 \end{array} \right] , \, \left[ \begin{array}{r} 1 \\ 3 \\ -4 \end{array} \right] \, \right\} $$ For example $v_{1} = \left[ \begin{array}{r} 0 \\ 1 \\ -1 \end{array} \right]$, and $v_{2} = \left[ \begin{array}{r} 1 \\ 3 \\ -4 \end{array} \right].$

You could provide an orthogonal basis using the process of Gram and Schmidt: $u_{1} = \left[ \begin{array}{r} 0 \\ 1 \\ -1 \end{array} \right]$, and $u_{2} = \left[ \begin{array}{r} 1 \\ 3 \\ -4 \end{array} \right].$

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  • $\begingroup$ Thank you very much for this detailled answer :) I was just going to calculate the kernel next as well :P Can you please tell me if my solution is fine, too? Thank you again! $\endgroup$ – cnmesr Mar 13 '17 at 19:03
  • $\begingroup$ Yes. You have another basis. $\left[ \begin{array}{r} 0 \\ -1 \\ -1 \end{array} \right] = -\left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right];$ and $\left[ \begin{array}{r} -1 \\ -2 \\ 5 \end{array} \right] = \left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right] - \left[ \begin{array}{r} 1 \\ 3 \\ -4 \end{array} \right].$ $\endgroup$ – dantopa Mar 13 '17 at 19:30

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