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Thread: Prove span

  1. #1
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    Prove span

    It's an easy question...

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

    Show that $\displaystyle [a, b]^T$ is in Span{u,v} for all a and b. In other words $\displaystyle R^2$ = 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 $\displaystyle u = [1, 2]^T$ and $\displaystyle v = [1, -2]^T$

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

    Since vectors u and v are lin. ind., they span $\displaystyle \mathbb{R}^2$; 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
    $\displaystyle x\begin{bmatrix}1 \\ 2\end{bmatrix}+ y\begin{bmatrix}1 \\ -2\end{bmatrix}= \begin{bmatrix}a \\ b\end{bmatrix}$
    for x and y. If you can then $\displaystyle span{u, v}= R^2$.

    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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