Basis of image of operator I am trying to solve the following question: 

I have found the Matrix $M(T)$ as:
\begin{bmatrix}
    0&-2&2&0\\
-2&0&0&2\\
2&0&0&-2\\
0&2&-2&0\\
\end{bmatrix}
For $b$, I solve $MX=0$ and get the answer.
Can anyone please help me with the $c$ part?
 A: The "image of T" is the set of all vectors v such that Tu= v for some vector u.  In particular, $\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$ is in the image of T if and only if there exist $\begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix}$ such that $\begin{bmatrix}0 & -2 & 2 & 0 \\ -2 & 0 & 0 & 2 \\ 2 & 0 & 0 & -2 \\ 0 & 2 & -2 & 0 \end{bmatrix}\begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix}= \begin{bmatrix}-2x+ 2y \\ -2w+ 2z \\ 2w- 2z \\ 2x- 2y\end{bmatrix}= \begin{bmatrix}a \\ b \\ c \\ d \end{bmatrix}$.
That is, we must have -2x+ 2y= a, -2w+ 2z= b, 2w- 2z= c, and 2x- 2y= d.  The critical observation now is that a= -2x+ 2y= -(2x- 2y)= -d and b= -2w+ 2z= -(2w- 2z)= -c. 
Any vector in the image of T is of the form $\begin{bmatrix}a \\ b \\ -b \\ -a \end{bmatrix}= \begin{bmatrix}a \\ 0 \\ 0 \\ -a\end{bmatrix}+ \begin{bmatrix}0 \\ b \\ -b \\ 0 \end{bmatrix}= a\begin{bmatrix}1 \\ 0 \\ 0 \\ -1\end{bmatrix}+ b \begin{bmatrix}0 \\ 1 \\ -1 \\ 0 \end{bmatrix}$.
A: The image is the column space.   So $\{\begin{pmatrix}0\\1\\-1\\0\end{pmatrix},\begin{pmatrix}1\\0\\0\\-1\end{pmatrix}\}$ would do.
Or, $\{\begin{pmatrix}0&1\\-1&0\end{pmatrix},\begin{pmatrix}1&0\\0&-1\end{pmatrix}\}$, if you want. 
