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.
 A: $\newcommand{\img}{\mathrm{im \,}}$> 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.
A: 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 $0$.
A: 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?
