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

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

What is the matrix A_f?

T_ef:

$\overrightarrow{f_1} =\cos 3t \cos \frac{\pi}{3} - \sin t \sin \frac{\pi}{3} =\... ...t 3}{2} \sin t = \begin{pmatrix}\frac 12 \\ -\frac{\sqrt 3}{2} \end{pmatrix}_e$

$\overrightarrow{f_2} =\cos t \cos \frac{\pi}{6} - \sin t \sin \frac{\pi}{6} =\f... ...- \frac 12 \sin t = \begin{pmatrix}\frac{\sqrt 3}{2} -\frac 12 \end{pmatrix}_e$

$\Longrightarrow T_{ef}= \begin{pmatrix}\frac 12 & \frac{\sqrt 3}{2} \\ -\frac{\sqrt 3}{2} & -\frac 12 \end{pmatrix}$

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

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