Linear transformation in matrix space

Question

Say that $T:S\to S$ is defined by $$T(A)=\frac{(MA+AM)}2$$ where $A\in S$, and S is defined by $$S=\left\{\begin{bmatrix}a&b\\-b&c\end{bmatrix}: a,b,c\in\mathbb R\right\}$$ and $$M=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ Show that T is a linear transformation and build its matrix representation according to the following basis of S: $$\beta_S=\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\-1&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$$ I think I've been able to prove that $T$ is a linear transformation, since $T(B)+T(C)=T(B+C)$ and $cT(B)=T(cB)$, but I'm really lost about how to find the matrix representation for $T$, I haven't been able to find an exercise quite like this in SE.

Answer 1

For convenience, denote the elements of our basis by $\beta_S = \{v_1,v_2,v_3\}$. We find that $$ T(v_1) = \frac 12 \left( \pmatrix{0&0\\1&0} + \pmatrix{0&-1\\0&0}\right) = \pmatrix{0&-1/2\\1/2&0} \\ = (0)v_1 + (-1/2)v_2 + (0)v_3 $$ It follows that the first column of our matrix $[T]_{\beta_S}$ is given by $(0,-1/2,0)$. Repeat this for the remaining two columns. You should find $$ [T]_{\beta_S} = \pmatrix{0&1&0\\-1/2&0&-1/2\\0&1&0} $$

---

An example in which we use the matrix: we find $$ T \pmatrix{1&2\\-2&3} = T (v_1 + 2v_2 + 3v_3) = T \pmatrix{1\\2\\3}_{\beta_S} = [T]_{\beta_S} \pmatrix{1\\2\\3}_{\beta_S} =\\ \left[\pmatrix{0&1&0\\-1/2&0&-1/2\\0&1&0} \pmatrix{1\\2\\3} \right]_{\beta_{S}} = \pmatrix{2\\-2\\2}_{\beta_S} = \\ 2v_1 - 2v_2 + 2v_3 = \pmatrix{2&-2\\2&2} $$