Change of basis.

Example 9.1.1   Let $\overrightarrow{e_1} =\begin{pmatrix}1 \\ 0 \end{pmatrix}, \overrightarrow{e_2} =\begin{pmatrix}0 \\ 1 \end{pmatrix}$ be the standard basis of $\mathbb{R} ^2$.

Define a new basis $\overrightarrow{f_1} =\begin{pmatrix}1 \\ 1 \end{pmatrix}, \; \overrightarrow{f_2} =\begin{pmatrix}-1 \\ 1 \end{pmatrix}$. Let $\overrightarrow{x} =\begin{pmatrix}1 \\ 2 \end{pmatrix}_e = \begin{pmatrix}\alpha_1 \\ \alpha_2 \end{pmatrix}_f$.

We define a matrix
$$\begin{align*}T_{ef} = & \begin{pmatrix}1 & -1 \\ 1 & 1 \end{pmatrix} \\ \quad ... ...parrow \\ \overrightarrow{f_1} & \overrightarrow{f_2} \end{matrix}\end{align*}$$

We have

• $T_{ef}(\overrightarrow{e_i} )=\overrightarrow{f_i} , \; i=1,2$.
• $\vert T_{ef}\vert=2 \neq 0$ and $T_{ef}^{-1}=\begin{pmatrix}\frac12 & \frac12 \\ -\frac12 & \frac12 \end{pmatrix}$

Question:

How to calculate the coordinates $\alpha_{1,2}$ of a given vector $\overrightarrow{x} =\begin{pmatrix}1\\ 2 \end{pmatrix}$ in the new basis $\{ \overrightarrow{f_1} ,\overrightarrow{f_2}\}$?

$\overrightarrow{x} = \alpha_1 \overrightarrow{f_1} +\alpha_2 \overrightarrow{f_2}$

$\hskip .5cm = \alpha_1 \begin{pmatrix}1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmat... ...begin{pmatrix}1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix}0 \\ 1 \end{pmatrix}$

$\hskip .5cm = \begin{pmatrix}1 & -1 \\ 1 & 1 \end{pmatrix} \biggl[ \alpha_1 \... ...x}1 \\ 0 \end{pmatrix} + \alpha_2 \begin{pmatrix}0 \\ 1 \end{pmatrix} \biggr]$

$\hskip .5cm = \begin{pmatrix}1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix}\alpha_1 \\ \alpha_2 \end{pmatrix}_e =\begin{pmatrix}1 \\ 2 \end{pmatrix}_e$

Therefore:

$$\begin{pmatrix}\alpha_1 \\ \alpha_2 \end{pmatrix}= \begin{pm... ...nd{pmatrix}= \begin{pmatrix}\frac32 \\ \frac12 \end{pmatrix}.$$

Let's check the result:

$$\alpha_1 \overrightarrow{f_1} +\alpha_2 \overrightarrow{f_2} ... ...\frac12 \end{pmatrix} = \begin{pmatrix}1 \\ 2 \end{pmatrix}_e.$$