Question

# The vector x is in H = Span {v_{1}, v_{2}} and find the beta-coordinate vector [x]_{beta}

Alternate coordinate systems
The vector x is in $$H = Span \ {v_{1}, v_{2}}$$
and find the beta-coordinate vector $$[x]_{\beta}$$

2021-01-26

Let vector x is in a vector space V and $$\beta = {b_{1}, b_{2}, \cdots, b_{n}}$$ is a basis for V,
then the beta- coordinates of x are the weights $$c_{1}, c_{2}, ..., c_{n}\ or\ [x]_{\beta} = \beta\begin{bmatrix}c_{1}\\c_{2}\\\cdots\\c_{n} \end{bmatrix}\ such\ that\ x = c_{1} b_{1} + c_{2} b_{2} + \cdots + c_{n} b_{n}.$$
Given:
The vectors $$v_{1} = \begin{bmatrix}11\\-5\\10\\7\end{bmatrix},v_{2} \begin{bmatrix}14\\-8\\13\\10\end{bmatrix} , and \ x=\begin{bmatrix}19\\-13\\18\\15\end{bmatrix}$$
Calculation:
Here $$\beta = {v_{1}, v_{2}}\ and\ H = Span {v_{1}, v_{2}}.$$
Therefore, for showing x is in H, show that the vector equation $$x = x_{1} v_{1} + x_{2} v_{2}$$ has a solution.
Write the given vectors as an augmented matrix $$[v_{1} v_{2} x]$$ and find the row reduced form of the matrix.
$$\begin{bmatrix}11 & 14 & 19\\ -5 & -8 &-13\\10 & 13& 18\\ 7 &10 & 15\end{bmatrix} (R_{1}\rightarrow R_{1} + R_{2})\overrightarrow{R_{3}\rightarrow R_{3} + 2R_{2} } \begin{bmatrix}6 & 6 & 6\\ -5 & -8 &-13\\0 & 3 & 8\\ 7 &10 & 15\end{bmatrix}$$
$$\begin{bmatrix}6 & 6 & 6\\ -5 & -8 &-13\\0 & 3 & 8\\ 7 &10 & 15\end{bmatrix} R_{1} \rightarrow \frac{1}{6} R_{1} \begin{bmatrix}1 & 1 & 1\\ -5 & -8 &-13\\0 & 3 & 8\\ 7 &10 & 15\end{bmatrix} \begin{bmatrix}1 & 1 & 1\\ -5 & -8 &-13\\0 & 3 & 8\\ 7 &10 & 15\end{bmatrix} (R_{2}\rightarrow R_{2}-5R_{1}) \overrightarrow{R_{4}\rightarrow R_{4}-7R_{1}} \begin{bmatrix}1 & 1 & 1\\ 0 & -3 &-8\\0 & 3 & 8\\ 0 & 3 & 8\end{bmatrix} \begin{bmatrix}1 & 1 & 1\\ 0 & -3 &-8\\0 & 3 & 8\\ 0 & 3 & 8\end{bmatrix} (R_{3}\rightarrow R_{3} +R_{2})\overrightarrow{R_{4}\rightarrow R_{4}+R_{2}} \begin{bmatrix}1 & 1 & 1\\ 0 & -3 &-8\\0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$$
On further simplification the matrix becomes,
$$\begin{bmatrix}1 & 1 & 1\\ 0 & -3 &-8\\0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix} R_{2} \rightarrow - \frac{1}{3} R_{2} \begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 8/3\\0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$$
$$\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 8/3\\0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix} R_{1} \rightarrow R_{1} \rightarrow R_{2} \begin{bmatrix}1 & 0 & -5/3\\ 0 & 1 & 8/3\\0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$$
Thus, the vector x is in $$H = Span {v_{1}, v_{2}}$$. Since, the system has a solution.
The vector equation gives,
$$x = x_{1} v_{1} + x_{2} v_{2}$$
$$= - \frac{5}{3} v_{1} + \frac{8}{3} v_{2}$$
By the above definftion the beta-coordinate vector for x is, $$[x]_{\beta} =\begin{bmatrix}-5/3 \\8/3 \end{bmatrix}.$$
Therefore, the beta-coordinate vector of x is $$[x]_{\beta} = \begin{bmatrix}-5/3 \\8/3\end{bmatrix}$$