User:IssaRice/Linear algebra/Change of basis example in two dimensions

This example comes from this video. To make it easier to go back and forth between this page and the video, the notation on this page tries to follow that of the video (where the discussion overlaps), but we distinguish between matrices and vectors.

We are working in $R^{2}$ , the plane. To be slightly pedantic, we will distinguish between matrices and vectors: $[\begin{matrix} 1 \\ 0 \end{matrix}] \in R^{2, 1}$ and $(1, 0) \in R^{2}$ .

Jennifer's basis vectors: ${\vec{b}}_{1} : = (2, 1)$ and ${\vec{b}}_{2} : = (- 1, 1)$ .

To Jennifer, ${\vec{b}}_{1}$ looks like $(1, 0)$ and ${\vec{b}}_{2}$ looks like $(0, 1)$ .

If Jennifer says " $(- 1, 2)$ ", to us (in the standard basis) this is the vector $- 1 {\vec{b}}_{1} + 2 {\vec{b}}_{2} = - 1 (2, 1) + 2 (- 1, 1) = (- 4, 1)$ .

We can also write the above calculation as $[\begin{matrix} {\vec{b}}_{1} & {\vec{b}}_{2} \end{matrix}] [\begin{matrix} - 1 \\ 2 \end{matrix}] = [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] [\begin{matrix} - 1 \\ 2 \end{matrix}] = [\begin{matrix} - 4 \\ 1 \end{matrix}]$ .

Notice that $[\begin{matrix} {\vec{b}}_{1} & {\vec{b}}_{2} \end{matrix}] {\vec{e}}_{1} = {\vec{b}}_{1}$ and $[\begin{matrix} {\vec{b}}_{1} & {\vec{b}}_{2} \end{matrix}] {\vec{e}}_{2} = {\vec{b}}_{2}$ , i.e., this matrix transforms our (standard) basis vectors into Jennifer's basis vectors.

How can we write this using change of basis notation? When Jennifer says " $(- 1, 2)$ ", this is the vector $v \in R^{2}$ such that, when written in Jennifer's coordinate system, it has coordinates $(- 1, 2)$ . In other words, it is the vector $v$ such that $[v]^{({\vec{b}}_{1}, {\vec{b}}_{2})} = [\begin{matrix} - 1 \\ 2 \end{matrix}]$ . To find out what this vector means in our coordinate system, we must compute $[v]^{({\vec{e}}_{1}, {\vec{e}}_{2})}$ .

We can write $[I]_{({\vec{b}}_{1}, {\vec{b}}_{2})}^{({\vec{e}}_{1}, {\vec{e}}_{2})} [v]^{({\vec{b}}_{1}, {\vec{b}}_{2})} = [v]^{({\vec{e}}_{1}, {\vec{e}}_{2})}$ .

Now consider the linear transformation $T : R^{2} \to R^{2}$ defined by $T {\vec{e}}_{1} : = {\vec{b}}_{1}$ and $T {\vec{e}}_{2} : = {\vec{b}}_{2}$ . Since $({\vec{e}}_{1}, {\vec{e}}_{2})$ is a basis of $R^{2}$ , there is exactly one such linear transformation, i.e., our specification is well-defined. What is the matrix of $T$ ? We can look at where it takes the standard basis vectors to see that the first column is ${\vec{b}}_{1}$ and the second column is ${\vec{b}}_{1}$ , i.e., we have $[T]_{({\vec{e}}_{1}, {\vec{e}}_{2})}^{({\vec{e}}_{1}, {\vec{e}}_{2})} = [\begin{matrix} {\vec{b}}_{1} & {\vec{b}}_{2} \end{matrix}] = [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] = [I]_{({\vec{b}}_{1}, {\vec{b}}_{2})}^{({\vec{e}}_{1}, {\vec{e}}_{2})}$ .

To summarize, we can write the same equation in multiple ways:

Equation	Description
$- 1 {\vec{b}}_{1} + 2 {\vec{b}}_{2} = - 1 [\begin{matrix} 2 \\ 1 \end{matrix}] + 2 [\begin{matrix} - 1 \\ 1 \end{matrix}] = [\begin{matrix} - 4 \\ 1 \end{matrix}]$	Linear combination of Jennifer's basis vectors
$[\begin{matrix} {\vec{b}}_{1} & {\vec{b}}_{2} \end{matrix}] [\begin{matrix} - 1 \\ 2 \end{matrix}] = [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] [\begin{matrix} - 1 \\ 2 \end{matrix}] = [\begin{matrix} - 4 \\ 1 \end{matrix}]$	Matrix multiplication
$[I]_{({\vec{b}}_{1}, {\vec{b}}_{2})}^{({\vec{e}}_{1}, {\vec{e}}_{2})} [v]^{({\vec{b}}_{1}, {\vec{b}}_{2})} = [v]^{({\vec{e}}_{1}, {\vec{e}}_{2})}$	Change of coordinate equation
$T (- 1 {\vec{e}}_{1} + 2 {\vec{e}}_{2}) = - 1 {\vec{b}}_{1} + 2 {\vec{b}}_{2}$
$[T]_{({\vec{e}}_{1}, {\vec{e}}_{2})}^{({\vec{e}}_{1}, {\vec{e}}_{2})} [\begin{matrix} - 1 \\ 2 \end{matrix}] = [T]_{({\vec{e}}_{1}, {\vec{e}}_{2})}^{({\vec{e}}_{1}, {\vec{e}}_{2})} [(- 1, 2)]^{({\vec{e}}_{1}, {\vec{e}}_{2})} = [T (- 1, 2)]^{({\vec{e}}_{1}, {\vec{e}}_{2})} = [(- 4, 1)]^{({\vec{e}}_{1}, {\vec{e}}_{2})} = (- 4, 1)$