Let $$\ v_1 = \begin{pmatrix} 2 \\ 0 \end{pmatrix} \text{ and } \ v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ We define $S = \{e_1, e_2\}$ where $$ e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \text{ and }\ e_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ is the canonical basis.
Let $\alpha$ be a linear map from $\mathbb{R^2}$ to $\mathbb{R^2}$ such that $$\alpha(v_1) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \text{ and } \alpha(v_2) = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$$
1) Write the matrix $M_{T,T}(\alpha)$ without computations
2) Write the base change matrix $C_{S,T}(\alpha)$ without computations
3) Compute the matrix $C_{T,S}$ and $M_{S,S}(\alpha)$
Could you please give me a hint? For 1) and 2) should I write $M \times S \times T$ and $C\times S\times T$, respectively?
Too long for a comment ... we have $$\begin{align} M_{T,T}&=\begin{pmatrix}1&5\\0&2\end{pmatrix},\\ C_{S,T}&=\begin{pmatrix}2&1\\0&1\end{pmatrix},\\ C_{T,S}&=C_{S,T}^{-1},\\ \text{and finally}\quad M_{S,S}&=C_{S,T}M_{T,T}C_{T,S}. \end{align}$$
As a reminder have a look at the indices: $(S,S)=(S,T),(T,T),(T,S).$ So $C_{T,S}$ transfers an $S$-based vector the corresponding $T$-based vector, $M_{T,T}$ carries out the linear map in $T$-base and finally $C_{S,T}$ changes the coordinates of that mapped vector in base $S$.