Question

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

Alternate coordinate systems
ANSWERED
asked 2021-01-25
The vector x is in \(H = Span \ {v_{1}, v_{2}}\)
and find the beta-coordinate vector \([x]_{\beta}\)

Expert Answers (1)

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 \([x]_{\beta} = \begin{bmatrix}-5/3 \\8/3\end{bmatrix}\)

27
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...