# Find bases for $\ker T,\text{im}\,T$ with respect to the members of $B$

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.

• Do you know how to solve system of linear equations by Gaussian elimination? Dec 18, 2016 at 12:35
• 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. Dec 18, 2016 at 12:35
• @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 :) Dec 18, 2016 at 12:37
• @ZoranLoncarevic yes but how is that useful ? Dec 18, 2016 at 12:37
• 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. Dec 18, 2016 at 16:34

$$\newcommand{\img}{\mathrm{im \,}}$$> 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.

• 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 Dec 18, 2016 at 12:55

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?

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 $0$.