Regarding the matrix form of a linear transformation This is my first question, so I will try and make it brief.
If $V= \mathbb R^2$, let $\mathcal{B}=\{ e_1, e_2 \}$ be any orthonormal basis of $V$.
In class we were asked to find the matrix form, in the base $\mathcal{B}$, of the linear map $T:V \to V$ such that $T(e_j)=-Xe_j$ and $T(Xe_j)=-e_j$, for $j=1,2$. Here $X$ is the matrix $$ X=\begin{bmatrix} 0 & -1\\
1 & 0 \end{bmatrix}.$$
I would be able to find the matrix representation of $T$ if the the matrix $X$ wasn’t present, and I don’t know what to do? I thing that the matrix representation of $T$ is a permutation matrix, but I’m not sure.
Any suggestions? Thanks!!
 A: For the question as given, there is no such matrix.
Multiplying the first condition by $i$, we see that
$$i \cdot T(e_j) = (i)(-i)e_j = e_j.$$
So, by linearity we see that $T(ie_j) = e_j$, which contradicts the other condition.
Edit: Unless I'm missing something, it is still inconsistent. The first condition says that
$$T \left[\begin{array}{c} 1 \\ 0 \end{array}\right] = \left[ \begin{array}{cc} 0&1 \\ -1&0\end{array} \right]\left[\begin{array}{c} 1 \\ 0 \end{array} \right] = \left[\begin{array}{c}0 \\ -1 \end{array} \right] = -e_2, \text{ and}$$
$$T \left[\begin{array}{c} 0 \\ 1 \end{array}\right] = \left[ \begin{array}{cc} 0&1 \\ -1&0\end{array} \right]\left[\begin{array}{c}0 \\ 1 \end{array} \right] = \left[\begin{array}{c} 1 \\ 0 \end{array} \right] = e_1.$$
That is, $T(e_1) = -e_2$ and $T(e_2) = e_1$. Considering the other condition (I'll spare you the working out), we get
$$T(Xe_1) = T(e_2) = -e_1, \text{ and }$$
$$T(Xe_2) = T(-e_1) = -e_2 \implies T(e_1) = e_2.$$
A: The columns of the matrix are the results of applying the transformation to the basis vectors, expressed in that basis.  We get $$\begin{pmatrix}0\quad1\\-1\quad 0\end{pmatrix}$$.
