# Find the coordinate vector

• Feb 5th 2010, 09:11 AM
wopashui
Find the coordinate vector
Let $\displaystyle \beta$ = {$\displaystyle v_1, v_2, v_3$} be the basis for $\displaystyle R^3$ where the v's are
$\displaystyle [1, 2, -2]^T$, $\displaystyle [-1, -1, 3]^T$ $\displaystyle [2, 6, -1]^T$ respectively.

Find the coordinate vector with respect to $\displaystyle \beta$ for the vector v =$\displaystyle [3, 5, -6]^T$
• Feb 5th 2010, 09:16 AM
felper
You have two options: build the matrix that changes the bases (i don't know his english name :( ), or do a linnear equation system. do you need more help?
• Feb 5th 2010, 09:40 AM
tonio
Quote:

Originally Posted by wopashui
Let $\displaystyle \beta$ = {$\displaystyle v_1, v_2, v_3$} be the basis for $\displaystyle R^3$ where the v's are
$\displaystyle [1, 2, -2]^T$, $\displaystyle [-1, -1, 3]^T$ $\displaystyle [2, 6, -1]^T$ respectively.

Find the coordinate vector with respect to $\displaystyle \beta$ for the vector v =$\displaystyle [3, 5, -6]^T$

It must be $\displaystyle \begin{pmatrix}\;3\\\;5\\\!\!-6\end{pmatrix}=a\begin{pmatrix}\;1\\\;2\\\!\!-2\end{pmatrix}$ $\displaystyle +b\begin{pmatrix}\!\!-1\\\!\!-1\\\;3\end{pmatrix}+c\begin{pmatrix}2\\6\\\!\!\!-1\end{pmatrix}$ , with $\displaystyle a,b,c,\in\mathbb{R}\Longleftrightarrow\begin{array }{l}\;\;a-b+2c=\;\;3\\2a-b+6c=\;\;5\\\!\!\!\!-2a+3b-c=\!-6\end{array}$ . Well, solve this system now.

Tonio