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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

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

Let's check the result:

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



Noah Dana-Picard
2001-02-26