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

This example comes from this video.

We are working in $R^{2}$ , the plane.

Jennifer's basis vectors: ${\vec{b}}_{1} : = [\begin{matrix} 2 \\ 1 \end{matrix}]$ and ${\vec{b}}_{2} : = [\begin{matrix} - 1 \\ 1 \end{matrix}]$

To Jennifer, ${\vec{b}}_{1}$ looks like $[\begin{matrix} 1 \\ 0 \end{matrix}]$ and ${\vec{b}}_{2}$ looks like $[\begin{matrix} 0 \\ 1 \end{matrix}]$ .

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

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.