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I got a math problem from my linear algebra textbook that comes with no answer. I just got my head on tilt on this one.

If $B_1$, $B_2$, and $B_3$ are bases for $\mathbb{R}^2$, and if:

$$P_{B_1 \to B_2}=\begin{bmatrix} 4 & 2 \\ 3 & 2 \end{bmatrix} ~~~\text{and}~~~ P_{B_2 \to B_3}=\begin{bmatrix} 6 & -1 \\ 2 & 3 \end{bmatrix}$$

Then what is $P_{B_3 \to B_1}$?

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    The two key points of this question are the fact that linear transformations over $\Bbb R^2$ are transitive, the linear transformation is matrix multiplication and sometimes have inverse (inverse concept in general and how that applies to change of basis). – 2017-01-04

2 Answers 2

2

Hint: it may help to consider what these different matrices (should) do:

  • $\color{blue}{[x]_{B_2}}=P_{B_1 \to B_2}\color{red}{[x]_{B_1}}$
  • $\color{purple}{[x]_{B_3}}=P_{B_2 \to B_3}\color{blue}{[x]_{B_2}}$
  • $\color{red}{[x]_{B_1}}=P_{B_3 \to B_1}\color{purple}{[x]_{B_3}}$

Combining the first two:

$$\color{purple}{[x]_{B_3}}=P_{B_2 \to B_3}\color{blue}{[x]_{B_2}} = P_{B_2 \to B_3}\underbrace{P_{B_1 \to B_2}\color{red}{[x]_{B_1}}}_{\color{blue}{[x]_{B_2}}}$$ So: $$\color{purple}{[x]_{B_3}}= \underbrace{P_{B_2 \to B_3}P_{B_1 \to B_2}}_{P_{B_1 \to B_3}}\color{red}{[x]_{B_1}}$$ And recall that $P_{B_3 \to B_1} = P_{B_1 \to B_3}^{-1}$.

Hover for answer:

Comparing shows that: $P_{B_1 \to B_3}=P_{B_2 \to B_3}P_{B_1 \to B_2}$ so $P_{B_3 \to B_1}=\left( P_{B_2 \to B_3}P_{B_1 \to B_2} \right)^{-1}$.

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$B_3 \rightarrow B_1= ([B_2 \rightarrow B_3] * [B_1 \rightarrow B_2])^{-1}$

or $ [B_2 \rightarrow B_3]^{-1} * [B_1 \rightarrow B_2]^{-1}$

Finally,

$$B_1 \rightarrow B_3 =\begin{bmatrix} 6 & -1 \\ 2 & 3 \\ \end{bmatrix} \begin{bmatrix} 4 & 2 \\ 3 & 2 \\ \end{bmatrix} = \begin{bmatrix} 21 & 10 \\ 17 & 10 \\ \end{bmatrix}$$

$$[B_1 \rightarrow B_3]^{-1}=[B_3 \rightarrow B_1]=\begin{bmatrix} 21 & 10 \\ 17 & 10 \\ \end{bmatrix}^{-1}=\begin{bmatrix} \frac{1}{4} & \frac{-1}{4} \\ \frac{-17}{40} & \frac{21}{40} \\ \end{bmatrix}$$

so $P_{B_3 \rightarrow B_1}=\begin{bmatrix} \frac{1}{4} & \frac{-1}{4} \\ \frac{-17}{40} & \frac{21}{40} \\ \end{bmatrix}$