0

I was reading this question here.

The answer says the change of basis from the basis $E$ into the basis $B$ is the matrix $P^{-1}$ (with the columns of $B^{-1}$).

But shouldn't it be $P$ instead?

For consider this: Let $e_i \in E$ denote the standard basis of $\mathbb R^3$. If we let $P=B$ be the matrix containing the vectors $b_i$ of the new basis then

$$ Pe_i = b_i$$

that is, $P$ transforms $e_i$ into the new basis $b_i$. Doesn't this make $P$ the basis transformation matrix from $E$ into $B$?

Community
  • 1
learner
  • 2,047
  • 21
  • 33

1 Answers1

2

Call the vectors of $B$ $b_1$, $b_2$, $b_3$. When you think of say $b_1$ as being the vector $(-1,1,0)$, you are thinking of it being written in terms of the basis $E$. When $b_1$ is written in terms of $B$, it is simply $(1,0,0)$. So, when we want to map the basis $B$ to the basis $E$, that is when we take the vector $b_1$ and map it to $(-1,1,0)$, because that is $b_1$ in terms of $E$. Thus, $P$ transforms $B$ to $E$. And $P^{-1}$ transforms $E$ to $B$.

J126
  • 17,707