13.2 verify that ....... is a basis for R^2 find [v]_beta

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The discussion confirms that the set $\beta=\left\{\begin{bmatrix} 0\\2 \end{bmatrix},\begin{bmatrix} 3\\1 \end{bmatrix}\right\}$ is indeed a basis for $\mathbb{R}^2$. The user demonstrates how to express the vector $v=\left[\begin{array}{c}6\\8\end{array}\right]$ as a linear combination of the basis vectors, resulting in the coordinates $[v]_{\beta}=\left[\begin{array}{c}3\\2\end{array}\right]$. The augmented matrix method is utilized to solve for the coefficients, confirming the calculations align with standard linear algebra techniques.

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karush
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Verify that
$\beta=\left\{\begin{bmatrix}
0\\2
\end{bmatrix}
,\begin{bmatrix}
3\\1
\end{bmatrix}\right\}$
is a basis for $\Bbb{R}^2$
Then for $v=\left[
\begin{array}{c}6\\8\end{array}
\right]$, find $[v]_\beta$
ok, I presume next is
$c_1\begin{bmatrix}
0\\2
\end{bmatrix}
+c_2\begin{bmatrix}
3\\1
\end{bmatrix}= \left[
\begin{array}{c}6\\8\end{array}
\right]$
by augmented matrix we get (the book did this?)
$\left[ \begin{array}{cc|c} 0 & 3 & 6 \\ 2 & 1 & 8 \end{array} \right]
=\left[ \begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & 2 \end{array} \right]$
hence
$[v]_{\beta}=\left[
\begin{array}{c}3\\2\end{array}
\right]$
following an example I don't think I understand the notation of $[v]_{\beta}$
 
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$[v]_\beta$ denotes the coordinates of vector $v$ in basis $\beta$.

Could you say from which textbook this notation is taken?
 
https://www.physicsforums.com/attachments/9053
this one
 

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