Consider the vector \(\vect{y}\in\complex{4}\)
\begin{equation*}
\vect{y}=\colvector{6\\14\\6\\7}\text{.}
\end{equation*}
We will find several vector representations of \(\vect{y}\) in this example. Notice that \(\vect{y}\) never changes, but the representations of \(\vect{y}\) do change. One basis for \(\complex{4}\) is
\begin{equation*}
B=\set{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3,\,\vect{u}_4}=
\set{
\colvector{-2\\1\\2\\-3},\,
\colvector{3\\-6\\2\\-4},\,
\colvector{1\\2\\0\\5},\,
\colvector{4\\3\\1\\6}
}
\end{equation*}
as can be seen by making these vectors the columns of a matrix, checking that the matrix is nonsingular and applying
Theorem CNMB. To find
\(\vectrep{B}{\vect{y}}\text{,}\) we need to find scalars,
\(a_1,\,a_2,\,a_3,\,a_4\) such that
\begin{equation*}
\vect{y}=a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3+a_4\vect{u}_4\text{.}
\end{equation*}
By
Theorem SLSLC the desired scalars are a solution to the linear system of equations with a coefficient matrix whose columns are the vectors in
\(B\) and with a vector of constants
\(\vect{y}\text{.}\) With a nonsingular coefficient matrix, the solution is unique, but this is no surprise as this is the content of
Theorem VRRB. This unique solution is
\begin{align*}
a_1&=2&a_2&=-1&a_3&=-3&a_4&=4\text{.}
\end{align*}
\begin{equation*}
\vectrep{B}{\vect{y}}=\colvector{2\\-1\\-3\\4}\text{.}
\end{equation*}
Suppose now that we construct a representation of \(\vect{y}\) relative to another basis of \(\complex{4}\text{,}\)
\begin{equation*}
C=\set{
\colvector{-15\\9\\-4\\-2},\,
\colvector{16\\-14\\5\\2},\,
\colvector{-26\\14\\-6\\-3},\,
\colvector{14\\-13\\4\\6}
}\text{.}
\end{equation*}
As with \(B\text{,}\) it is easy to check that \(C\) is a basis. Writing \(\vect{y}\) as a linear combination of the vectors in \(C\) leads to solving a system of four equations in the four unknown scalars with a nonsingular coefficient matrix. The unique solution can be expressed as
\begin{equation*}
\vect{y}=\colvector{6\\14\\6\\7}=
(-28)\colvector{-15\\9\\-4\\-2}+
(-8)\colvector{16\\-14\\5\\2}+
11\colvector{-26\\14\\-6\\-3}+
0\colvector{14\\-13\\4\\6}
\end{equation*}
\begin{equation*}
\vectrep{C}{\vect{y}}=\colvector{-28\\-8\\11\\0}\text{.}
\end{equation*}
We often perform representations relative to standard bases, but for vectors in
\(\complex{m}\) this is a little silly. Let us find the vector representation of
\(\vect{y}\) relative to the standard basis (
Theorem SUVB),
\begin{equation*}
D=\set{\vect{e}_1,\,\vect{e}_2,\,\vect{e}_3,\,\vect{e}_4}\text{.}
\end{equation*}
Then, without any computation, we can check that
\begin{equation*}
\vect{y}=\colvector{6\\14\\6\\7}=6\vect{e}_1+14\vect{e}_2+6\vect{e}_3+7\vect{e}_4
\end{equation*}
\begin{equation*}
\vectrep{D}{\vect{y}}=\colvector{6\\14\\6\\7}
\end{equation*}
which is not very exciting. Notice however that the order in which we place the vectors in the basis is critical to the representation. Let us keep the standard unit vectors as our basis, but rearrange the order we place them in the basis. So a fourth basis is
\begin{equation*}
E=\set{\vect{e}_3,\,\vect{e}_4,\,\vect{e}_2,\,\vect{e}_1}\text{.}
\end{equation*}
Then,
\begin{equation*}
\vect{y}=\colvector{6\\14\\6\\7}=6\vect{e}_3+7\vect{e}_4+14\vect{e}_2+6\vect{e}_1
\end{equation*}
\begin{equation*}
\vectrep{E}{\vect{y}}=\colvector{6\\7\\14\\6}\text{.}
\end{equation*}
So for every possible basis of \(\complex{4}\) we could construct a different representation of \(\vect{y}\text{.}\)
