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Assume that $V$ is a $4$-dimension vector space and $B=\{e_1,e_2,e_3,e_4\}$ is a basis of $V$. If $T : V \to V$ is a linear transformation such that the matrix representation of $T$ with respect to $B$ is $$ \begin{bmatrix} 1 & 0 & 2 & 1 \\ -1 & 2 & 1 & 3 \\ 1 & 2 & 5 & 5 \\ 2 & -2 & 1 & -2 \\ \end{bmatrix} $$
Find a basis for $\ker T$, $\text{im}\,T$ with respect to the members of $B$.


Note 1 : I'm sorry but I have no idea about this problem... I have never found the basis of kernel and image with the use of matrix representation.

Note 2 : By $e_1$ I mean $(1,0,0,0)$ . You can guess what $e_2,e_3,e_4$ are.

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  • $\begingroup$ Do you know how to solve system of linear equations by Gaussian elimination? $\endgroup$ Dec 18, 2016 at 12:35
  • $\begingroup$ Do you know the definiton of ker $(T)$ and im$(T)$? I am sure there are lot of similar kind of questions on MSE already. $\endgroup$ Dec 18, 2016 at 12:35
  • $\begingroup$ @ArpitKansal Yes i know the definitions ... no, this kind of question ( finding the basis of kernel and image when matrix representation is given ) is not between the similar questions :) anyway, please help me :) $\endgroup$ Dec 18, 2016 at 12:37
  • $\begingroup$ @ZoranLoncarevic yes but how is that useful ? $\endgroup$ Dec 18, 2016 at 12:37
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    $\begingroup$ Dear @ArmanMalekzade: Did you really try to find similar questions? ofc by similar i dint mean the exact question.For instance you can see here and here and plenty more. $\endgroup$ Dec 18, 2016 at 16:34

3 Answers 3

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$\newcommand{\img}{\mathrm{im \,}}$> I have no idea about this problem

First of all, you might be interested in Gowers's article about "fake difficult" and also the thread here.


Exercises:

  • Write down what is $\img(T)$ by definition. (This is the starting point of the problem. One has to know exactly what this is in order to go on.)
  • $\img(T)$ consists of all the linear combinations of the columns of $T$.

  • Find a basis for $im(T)$. (This is a very standard exercise in linear algebra. Every standard textbook should at least have a related example.)
  • Essentially, you are asked to find a set of linearly independent vectors among the columns of $T$. A systematic way to do it is by Gaussian elimination.

  • Write down what is $\ker(T)$.
  • Find a basis for $\ker(T)$.

I have never found the basis of kernel and image with the use of matrix representation.

You are making things complicated. Since $V=\mathbb{R}^4$ and $T$ is represented with respect to the standard basis $\{e_1,\cdots,e_4\}$ , one can view $T$ as its representing matrix.

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  • $\begingroup$ The problem is about $im(T)$ . I can't find it's basis ... even a hint is more useful than the humiliation you made me feel ... what you wrote is not the answer ... its not a good behavior in this website sir $\endgroup$ Dec 18, 2016 at 12:55
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Well, you say you don't know how to find kernel and image from matrix representation, but I assume that you could if I gave you explicit formula for $T$?

Then, we have

$$\begin{pmatrix} 1 & 0 & 2 & 1 \\ -1 & 2 & 1 & 3 \\ 1 & 2 & 5 & 5 \\ 2 & -2 & 1 & -2 \\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{pmatrix} = \begin{pmatrix} x_1 +2x_3 + x_4\\ -x_1+2x_2+x_3+3x_4 \\ x_1+2x_2+5x_3+5x_4 \\ 2x_1-2x_2+x_3-2x_4 \\ \end{pmatrix}$$ or $$T(x_1,x_2,x_3,x_4) = (x_1 +2x_3 + x_4, -x_1+2x_2+x_3+3x_4, x_1+2x_2+5x_3+5x_4, 2x_1-2x_2+x_3-2x_4)$$

Now, you can easily find all $x$ such that $Tx = 0$ and all $y$ that are of the form $Tx = y$, right?

...

Probably not.

To find the kernel you need to solve linear system $Tx = 0$, and how do you solve linear systems? Well, you write them as matrices. I will leave to you to solve system with extended matrix given as $$\left(\begin{array}{cccc|c} 1 & 0 & 2 & 1 & 0\\ -1 & 2 & 1 & 3 & 0\\ 1 & 2 & 5 & 5 & 0\\ 2 & -2 & 1 & -2 & 0\\ \end{array}\right)$$

And what about the image? Well take a look again at what typical element of $\operatorname{im} T$ looks like:

$$\begin{pmatrix} x_1 +2x_3 + x_4\\ -x_1+2x_2+x_3+3x_4 \\ x_1+2x_2+5x_3+5x_4 \\ 2x_1-2x_2+x_3-2x_4 \\ \end{pmatrix}= x_1\begin{pmatrix} 1 \\ -1 \\ 1 \\ 2 \\ \end{pmatrix} + x_2 \begin{pmatrix} 0\\ 2 \\ 2\\ -2\\ \end{pmatrix} + x_3\begin{pmatrix} 2\\ 1\\ 5\\ 1\\ \end{pmatrix} + x_4\begin{pmatrix} 1 \\ 3 \\ 5 \\ -2 \\ \end{pmatrix}$$

Thus, $\operatorname{im} T = \operatorname{span}\{(1,-1,1,2),(0,2,2,-2),(2,1,5,1),(1,3,5,-2)\}$. Now, this doesn't need to be base, but it is generating set so you can reduce it to base. Can you finish the task now?

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Hint:

For the image of $T$, sort out all linearly independant columns of $T$. For the kernel of $T$, check which vectors in $V$ map to $0$.

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