# Kernel and Image of a linear transformation T given on a basis

Let V be a 4-d vector space. $T:V \rightarrow V$is a linear operator whose effect on basis {$e_1, e_2,e_3,e_4$} is

$Te_1= 2e_1- e_4$

$Te_2= -2e_1 + e_4$

$Te_3= -2e_1 + e_4$

$Te_4= e_1$

Find a basis for Ker T and Image T. Calculate the rank and nullity of T.

$$A =\begin{bmatrix} 2 & -2 & -2 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ -1 & 1 & 1 & 0\\ \end{bmatrix}$$

RREF: $$A =\begin{bmatrix} 1 & -1 & -1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$$

• I apologize on the lack of format, I am not sure how to get it right Oct 2, 2013 at 19:46
• Read a little about LaTeX, or get into any other question and do "edit" to see how the formating is done. Oct 2, 2013 at 19:47
• fixed the original format Oct 2, 2013 at 19:51
• Nice. Have you already studied about matrix representations of linear transformations? Oct 2, 2013 at 19:52
• to be honest, I'm in a theoretical physics class and have no idea what's going on... Oct 2, 2013 at 19:55

The matrix corresponding to $\;T\;$ and the given basis is in fact

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

From this we can see the matrix rank = the transformation's image's dimension, is one, and thus its kernel (both of the matrix and the transformation) has dimension three.

If you don't understand this I can't see a way to explain you any further without solving completely the question and you don't understanding a thing...

• my orignial parameters were wrong, and I have input in rref Oct 2, 2013 at 20:09

First, notice that if you write the matrix $A$ defined as $Ax = T(x)$

Then if we consider the basis ${\cal B} = \{e_1,e_2,e_3,e_4\}$ we have that

$$A = [T(e_1)\quad T(e_2)\quad T(e_3)\quad T(e_4)]$$

That's because every $x\in V$ can be writed as $x = \alpha_1e_1+\alpha_2e_2+\alpha_3e_3+\alpha_4e_4$, with $\alpha_i\in\mathbb{K}$ (with $\mathbb{K}$ )

Then, as $T$ is linear

Then you must determine the Kernel and Image of $A$ and then write those vectors on basis ${\cal B}$
Remember that $Rank(T) + Kern(T) = Dim(V)$. So if $Dim(V) = 4$ and $Rank(T) = 1$, finding the Kernel should be pretty easy.
You can think of Rank as the dimension of the subspace of $V$ given by $T$.