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:
$$A =\begin{bmatrix}
1 & -1 & -1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix} $$
Best Answer
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...