How do I rewrite vectors in other basis' given change of coordinate matrices? $\displaystyle β= \begin{bmatrix}2\\2\\\end{bmatrix}$,$\displaystyle \begin{bmatrix}4\\-1\\\end{bmatrix}$
$\displaystyle C= \begin{bmatrix}1\\3\\\end{bmatrix}$,$\displaystyle \begin{bmatrix}-1\\-1\\\end{bmatrix}$
From B to C Change of coordinate matrix is:\begin{bmatrix}1 & 3/4 \\ -1 & -13/4\end{bmatrix}
From C to B Change of coordinate matrix is:\begin{bmatrix}13/10 & 3/10 \\ -2/5 & -2/5\end{bmatrix}
Find $\displaystyle\begin{bmatrix}2\\2\\\end{bmatrix}_C$and $\displaystyle  \begin{bmatrix}1\\3\\\end{bmatrix}_β$

Let the basis vectors for $B$ be $\mathbf{b}_1$ and $\mathbf{b}_2$.  Then one can write the $\beta$ vectors as
$$
 \begin{align}
  \boldsymbol{\beta}_1 &= 2\,\mathbf{b}_1 + 2\,\mathbf{b}_2 = \beta_{11}^b\mathbf{b}_1 + \beta_{12}^b\,\mathbf{b}_2 \\
  \boldsymbol{\beta}_2 &= 4\,\mathbf{b}_1 - \,\mathbf{b}_2 = \beta_{21}^b\mathbf{b}_1 + \beta_{22}^b\,\mathbf{b}_2 
  \end{align}
$$
Similarly, if the basis vectors for $C$ are $\mathbf{c}_1$ and $\mathbf{c}_2$, we have
$$
 \begin{align}
  \mathbf{C}_1 &= \,\mathbf{c}_1 + 3\,\mathbf{c}_2 = C_{11}^c\mathbf{c}_1 + C_{12}^c\,\mathbf{c}_2\\
  \mathbf{C}_2 &= -\mathbf{c}_1 -\,\mathbf{c}_2 = C_{21}^c\mathbf{c}_1 +C_{22}^c\,\mathbf{c}_2 \,.
 \end{align}
$$
We can express $\boldsymbol{\beta}_1$ and $\boldsymbol{\beta}_2$ in the $\mathbf{c}$-basis as
$$
  \boldsymbol{\beta}_1 = \beta_{11}^c\,\mathbf{c}_1 + \beta_{12}^c\,\mathbf{c}_2 ~,~~
  \boldsymbol{\beta}_2 = \beta_{21}^c\,\mathbf{c}_1 + \beta_{22}^c \,\mathbf{c}_2 \,.
$$
Similarly, for $\mathbf{C}_1$ and $\mathbf{C}_2$, 
$$
  \mathbf{C}_1 = C_{11}^b\,\mathbf{b}_1 + C_{12}^c\,\mathbf{b}_2 ~,~~
  \mathbf{C}_2 = C_{21}^b\,\mathbf{b}_1 + C_{22}^c \,\mathbf{b}_2 \,.
$$
The coordinate transformation matrices are
$$
  \begin{bmatrix} \beta_{11}^c \\ \beta_{12}^c \end{bmatrix} 
  = \begin{bmatrix}
      1 & 3/4 \\ -1 & -13/4
    \end{bmatrix}
   \begin{bmatrix} \beta_{11}^b \\ \beta_{12}^b\end{bmatrix}
$$
and
$$
  \begin{bmatrix} C_{11}^b \\ C_{12}^b \end{bmatrix} 
  = \begin{bmatrix}
      13/10 & 3/10 \\ -2/5 & -2/5
    \end{bmatrix}
   \begin{bmatrix} C_{11}^c \\ C_{12}^c \end{bmatrix}
$$
The required results follow directly from the transformation rules.
