Let $M$ be the space of three by three matrices. Let $A = \begin{bmatrix}1&0&-1\\-1&1&0\\0&-1&1\end{bmatrix}$.
Interpreting $A$ as a linear operator from $M$ to $M$ (rather than as a linear operator from $\Bbb R^3$ to $\Bbb R^3$), determine the dimensions of the nullspace and columnspace.
In problem 11.2 c), how did we determine the dimensions of "nullspace" and "column space"?
Why are the values 3, and 6, when they should be 2 (pivot columns), and 1 (free variable)?
Hint: If all that is giving you difficulty in understanding the problem is the fact that the domain and codomain of our linear transformation is not of the form $\Bbb R^n$, but is instead a space of matrices $M$, we can redescribe the problem.
Reinterpreting a matrix $\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$ in $M$ instead as a vector $\begin{bmatrix}a\\b\\c\\d\\e\\f\\g\\h\\i\end{bmatrix}$ in $\Bbb R^9$, we would have our transformation $A$ as:
$$A\left(\begin{bmatrix}a\\b\\c\\d\\e\\f\\g\\h\\i\end{bmatrix}\right) = \begin{bmatrix}a-g\\b-h\\c-i\\d-a\\e-b\\f-c\\g-d\\h-e\\i-f\end{bmatrix}$$
From here you can describe $A$ using a matrix (it won't be $3\times 3$ anymore, do you see what size it will be?) and then use your familiar techniques of row reduction or whatever else.