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Math Help - Prove span

  1. #1
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    Prove span

    It's an easy question...

    Let u = [1, 2]^T and v = [1, -2]^T

    Show that [a, b]^T is in Span{u,v} for all a and b. In other words R^2 = Span{u,v}
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  2. #2
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    Quote Originally Posted by Tander View Post
    It's an easy question...

    Let u = [1, 2]^T and v = [1, -2]^T

    Show that [a, b]^T is in Span{u,v} for all a and b. In other words R^2 = Span{u,v}
    \begin{bmatrix}<br />
1 & 1\\ <br />
2 & -2<br />
\end{bmatrix}\to \begin{bmatrix}<br />
1 & 1\\ <br />
0 & -4<br />
\end{bmatrix}\to \begin{bmatrix}<br />
1 & 1\\ <br />
0 & 1<br />
\end{bmatrix}\to \begin{bmatrix}<br />
1 & 0\\ <br />
0 & 1<br />
\end{bmatrix}

    Since vectors u and v are lin. ind., they span \mathbb{R}^2; therefore, any other real vector will be a linear combination of u and v.
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  3. #3
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    Directly from the definition: solve
    x\begin{bmatrix}1 \\ 2\end{bmatrix}+ y\begin{bmatrix}1 \\ -2\end{bmatrix}= \begin{bmatrix}a \\ b\end{bmatrix}
    for x and y. If you can then span{u, v}= R^2.

    That is the same as solving x+ y= a, 2x- 2y= b. Multiply the first equation by 2 and add: 4x= 2a+ b so x= (2a+ b)/4. Multiply the first equation by 2 and subtract: 4y= 2a- b so y= (2a- b)/4. Those exist for all a and b.
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