How do we do change of basis?

40 Views Asked by At

We consider in $\mathbb{R}^2$ the basis $B=\{ (2,1), (5,3)\}$ and $C=\{ (1,1), (2,3)\}$ and the mapping $f: \mathbb{R}^2 \to \mathbb{R}^2$ such that $f(x,y)=(x+y,-2x+y)$. Let the matrix $Cf_B=\begin{pmatrix} \alpha_{11} & \alpha_{12}\\ \alpha_{21} & \alpha_{22} \end{pmatrix}$. I want to compute the value of $\alpha_{12}+2\alpha_{22}$.

I have found that $f(2,1)=(3,-3)$ and $f(5,3)=(8,-7)$.

So is $f_B=\begin{pmatrix} 3 & -3\\ 8 & -7 \end{pmatrix}$ ? Or is it defined in an other way?

Because I get that $Cf_B=\begin{pmatrix} 11 & -10\\ 30 & -27 \end{pmatrix}$ and this should be wrong. Is the change of basis meant? If so, how can it be done?

1

There are 1 best solutions below

2
On

HINT

  • write down $C_{fS}$ in the standard basis then $w_S=f(v)=C_{fS}v_S$
  • consider the matrix for the change of basis from B to the standard basis $M_{SB}$ such that

$$v_SM_{SB}v_B \iff v_B=M_{SB}^{-1}v_S=M_{BS}v_S$$

  • then

$$w_S=C_{fS}v_S \iff M_{SB}w_B=C_{fS}M_{SB}v_B \iff w_B=M_{SB}^{-1}C_{fS}M_{SB}v_B\\\iff w_b=M_{BS}C_{fS}M_{SB}v_B \iff w_B=C_{fB}v_B \quad C_{fB}=M_{BS}C_{fS}M_{SB}$$