Find $x_{b2}$ if $x_{b1} = \begin{pmatrix}2 \\ 5 \end{pmatrix}_{b1}$
Let $$ B_{1} = \begin{pmatrix} 3 & -1 \\ -5 & 4 \\ 5 & -3 \end{pmatrix}\, \ \;\;\;\;\;\;\;\;\;\; B_{2}= \begin{pmatrix} 2 & 1 \\ -1 & 3 \\ 2 & -1 \end {pmatrix} $$
I know I have to use the equation $x_{b2} = V^{-1}Ux_{b1}$, where $V =B_{2}$ by V is not invertible since it is not a square matrix. The textbook provides a solution in the back of the book, but no examples on how to deal with non-square matrices. How do I solve these type of questions.
The coordinates of your vector, $x$, relative to the basis $B_1$ are $\begin{bmatrix} 2 \\ 5 \end{bmatrix}$. This means that $$ x = B_1 \begin{bmatrix} 2 \\ 5 \end{bmatrix} = 2\begin{bmatrix} 3 \\ -5 \\ 5 \end{bmatrix} + 5\begin{bmatrix} -1 \\ 4 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 \\ 10 \\ -5 \end{bmatrix}$$
To find $x$'s coordinates relative to the other basis, you need to solve the system: $$ B_2 \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -1 & 3 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 1 \\ 10 \\ -5 \end{bmatrix}$$