For the vector space $\complex{4}$ we have the two bases, \begin{align*} B&=\set{ \colvector{1 \\ -2 \\ 1 \\ -2},\, \colvector{-1 \\ 3 \\ 1 \\ 1},\, \colvector{2 \\ -3 \\ 3 \\ -4},\, \colvector{-1 \\ 3 \\ 3 \\ 0} } & C&=\set{ \colvector{1 \\ -6 \\ -4 \\ -1},\, \colvector{-4 \\ 8 \\ -5 \\ 8},\, \colvector{-5 \\ 13 \\ -2 \\ 9},\, \colvector{3 \\ -7 \\ 3 \\ -6} } \end{align*}
The change-of-basis matrix from $B$ to $C$ requires writing each vector of $B$ as a linear combination the vectors in $C$, \begin{align*} \vectrep{C}{\colvector{1 \\ -2 \\ 1 \\ -2}} &=\vectrep{C}{ (1)\colvector{1 \\ -6 \\ -4 \\ -1}+ (-2)\colvector{-4 \\ 8 \\ -5 \\ 8}+ (1)\colvector{-5 \\ 13 \\ -2 \\ 9}+ (-1)\colvector{3 \\ -7 \\ 3 \\ -6} } =\colvector{1\\-2\\1\\-1}\\ \vectrep{C}{\colvector{-1 \\ 3 \\ 1 \\ 1}} &=\vectrep{C}{ (2)\colvector{1 \\ -6 \\ -4 \\ -1}+ (-3)\colvector{-4 \\ 8 \\ -5 \\ 8}+ (3)\colvector{-5 \\ 13 \\ -2 \\ 9}+ (0)\colvector{3 \\ -7 \\ 3 \\ -6} } =\colvector{2\\-3\\3\\0}\\ \vectrep{C}{\colvector{2 \\ -3 \\ 3 \\ -4}} &=\vectrep{C}{ (1)\colvector{1 \\ -6 \\ -4 \\ -1}+ (-3)\colvector{-4 \\ 8 \\ -5 \\ 8}+ (1)\colvector{-5 \\ 13 \\ -2 \\ 9}+ (-2)\colvector{3 \\ -7 \\ 3 \\ -6} } =\colvector{1\\-3\\1\\-2}\\ \vectrep{C}{\colvector{-1 \\ 3 \\ 3 \\ 0}} &=\vectrep{C}{ (2)\colvector{1 \\ -6 \\ -4 \\ -1}+ (-2)\colvector{-4 \\ 8 \\ -5 \\ 8}+ (4)\colvector{-5 \\ 13 \\ -2 \\ 9}+ (3)\colvector{3 \\ -7 \\ 3 \\ -6} } =\colvector{2\\-2\\4\\3}\\ \end{align*}
Then we package these vectors up as the change-of-basis matrix, \begin{equation*} \cbm{B}{C}= \begin{bmatrix} 1 & 2 & 1 & 2 \\ -2 & -3 & -3 & -2 \\ 1 & 3 & 1 & 4 \\ -1 & 0 & -2 & 3 \end{bmatrix} \end{equation*}
Now consider a single (arbitrary) vector $\vect{y}=\colvector{2\\6\\-3\\4}$. First, build the vector representation of $\vect{y}$ relative to $B$. This will require writing $\vect{y}$ as a linear combination of the vectors in $B$, \begin{align*} \vectrep{B}{\vect{y}} &=\vectrep{B}{\colvector{2\\6\\-3\\4}}\\ &=\vectrep{B}{ (-21)\colvector{1 \\ -2 \\ 1 \\ -2}+ (6)\colvector{-1 \\ 3 \\ 1 \\ 1}+ (11)\colvector{2 \\ -3 \\ 3 \\ -4}+ (-7)\colvector{-1 \\ 3 \\ 3 \\ 0} } &=\colvector{-21\\6\\11\\-7} \end{align*}
Now, applying Theorem CB we can convert the representation of $\vect{y}$ relative to $B$ into a representation relative to $C$, \begin{align*} \vectrep{C}{\vect{y}} &=\cbm{B}{C}\vectrep{B}{\vect{y}}&&\knowl{./knowls/theorem.CB.knowl}{\text{Theorem CB}}\\ &= \begin{bmatrix} 1 & 2 & 1 & 2 \\ -2 & -3 & -3 & -2 \\ 1 & 3 & 1 & 4 \\ -1 & 0 & -2 & 3 \end{bmatrix} \colvector{-21\\6\\11\\-7}\\ &=\colvector{-12\\5\\-20\\-22}&&\knowl{./knowls/definition.MVP.knowl}{\text{Definition MVP}} \end{align*}
We could continue further with this example, perhaps by computing the representation of $\vect{y}$ relative to the basis $C$ directly as a check on our work (Exercise CB.C20). Or we could choose another vector to play the role of $\vect{y}$ and compute two different representations of this vector relative to the two bases $B$ and $C$.