Verification of change of basis calculation

Question

A short question just to check regarding a change of basis: Let $$A = \left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \end{bmatrix} \right\}$$ be a basis and let $A'$ be the standard basis of $\mathbb{R}^2$, would I be right in stating that the change of basis matrix is $P = \begin{bmatrix} 2 & 2 \\ 1 & 5 \end{bmatrix}$ then $P^{-1} = \begin{bmatrix} \frac{5}{8} & -\frac{1}{4} \\ -\frac{1}{8} & \frac{1}{4} \end{bmatrix}$ and that $C =\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ written in the standard basis is $$P^{-1}C = \begin{bmatrix} -\frac{9}{8} \\ \frac{5}{8} \end{bmatrix} .$$ Is this the correct working, for some reason it's not agreeing with the answer in the literature which gives \begin{bmatrix} 2 \\ 9 \end{bmatrix} Thanks.

Answer 1

You should be careful with "the" matrix for a change of basis, since the order matters: the matrix converting coordinates from basis $A$ to basis $A'$ is the inverse of the matrix converting coordinates in the other direction.

You filled a matrix with as columns the base vectors of $A$ expressed w.r.t. the base vectors of $A'$, the standard basis. This gives you a change of basis matrix $P_{A'A}$ that converts coordinates in the following direction: $$[\vec x]_{A'} = P_{A'A}[\vec x]_{A}$$ where $[\vec x]_B$ denotes the coordinate vector of $\vec x$ w.r.t. a basis $B$.

You use it the other way around, there was no need to find the inverse matrix for the conversion in the direction asked.

If a vector $\vec c$ has coordinate vector: $$[\vec c]_{A} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$$ with respect to $A$, then the coordinate vector with respect to $A'$ is given by: $$[\vec c]_{A'} = P_{A'A}[\vec c]_{A} = \begin{bmatrix} 2 & 2 \\ 1 & 5 \end{bmatrix}\begin{bmatrix} -1 \\ 2 \end{bmatrix}= \begin{bmatrix} 2 \\ 9 \end{bmatrix} $$

If you are given a coordinate vector w.r.t. the standard basis and you would want to know its coordinate vector w.r.t. the intially given basis $A$, then you would need $P_{AA'} = P_{A'A}^{-1}$.