Find Beta Matrix

Question

Find the matrix $B$ for the transformation $A = \begin{bmatrix} 3 & 1 \\ -1 & 4 \end{bmatrix}$ in the basis $\left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\}$.

Am I overcomplicating this? Do I just put the basis into its own $2 \times 2$ matrix and then use the formula $S^{-1}AS$? I feel like I know how to do this, but want to make sure I'm going in the right direction.

Answer 1

Note that the transformation from basis

$$\mathcal{B}=\left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\}$$

to the standard basis is

$$v_S=Sv_B=\begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}v_B$$

thus given in the standard basis

$$w_S=Av_S$$

in the new basis we have

$$Sw_B=ASv_B\implies w_B=S^{-1}ASv_B$$

Can you follow these steps? What is not clear to you?

Answer 2

Let $\beta = \left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\}$ and $\gamma$ be the standard basis. Then \begin{align} B &= [\mathsf{L}_A]_\beta \\ &= [\mathsf{Id}]_\gamma^\beta A [\mathsf{Id}]_\beta^\gamma \\ &= ([\mathsf{Id}]_\beta^\gamma)^{-1} A [\mathsf{Id}]_\beta^\gamma \\ &= \left( \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \right)^{-1} \begin{bmatrix} 3 & 1 \\ -1 & 4 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 4 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} -1 & 4 \\ 5 & -7 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} 2 & -1 \\ 3 & 5 \end{bmatrix}. \end{align}