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Change of basis: general formulae.

Given an n-dimensional vector space V and two bases $e=\{ \overrightarrow{e_1} , \dots,\overrightarrow{e_n}\}$ and $f=\{ \overrightarrow{f_1} , \dots,\overrightarrow{f_n}\}$ for V.

Let $\overrightarrow{x}\in V$; denote $\overrightarrow{x} =
\begin{cases}
x_1\overrightarrow{e_1} + \cdots + x_n \over...
...} =
\begin{pmatrix}\alpha_1 \\ \dots \\ \alpha_n \end{pmatrix}_f
\end{cases}.$

The coefficients $(x_1, \dots , x_n)$ are given. What are $(\alpha_1, \dots, \alpha_n)$?

Suppose $\overrightarrow{f_i} =\sum_{j=1}^{j=n}t_{ji}\overrightarrow{e_j} =\begin{pmatrix}t_{1i} \\ \dots \\ t_{ni} \end{pmatrix}_e$.


\begin{displaymath}i \rightarrow
\underbrace{
\begin{pmatrix}
\overbrace{t_{11}}...
...ots \\ \vdots \\ t_{ni} \end{pmatrix}}_{\overrightarrow{f_i} }
\end{displaymath}

What is the matrix Af?

Tef:
$\overrightarrow{f_1} =\cos 3t \cos \frac{\pi}{3} - \sin t \sin \frac{\pi}{3}
=\...
...t 3}{2} \sin t
= \begin{pmatrix}\frac 12 \\ -\frac{\sqrt 3}{2} \end{pmatrix}_e$
$\overrightarrow{f_2} =\cos t \cos \frac{\pi}{6} - \sin t \sin \frac{\pi}{6}
=\f...
...- \frac 12 \sin t
= \begin{pmatrix}\frac{\sqrt 3}{2} -\frac 12 \end{pmatrix}_e$
$\Longrightarrow
T_{ef}=
\begin{pmatrix}\frac 12 & \frac{\sqrt 3}{2} \\ -\frac{\sqrt 3}{2} & -\frac 12
\end{pmatrix}$

$T_{ef}^{-1}=\begin{pmatrix}-1 & -\sqrt 3 \\ \sqrt 3 & 1 \end{pmatrix}
\begin{C...
...rt 3}{2} & -\frac 12
\end{pmatrix}= \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$ O.K.


\begin{align*}A_f
&= \begin{pmatrix}-1 & -\sqrt 3 \\ \sqrt 3 & 1 \end{pmatrix}\b...
...& -\sqrt 3 \end{pmatrix}\renewcommand{\arraystretch}{1.0}\\
&=A_f.
\end{align*}



Noah Dana-Picard
2001-02-26