Prove span

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January 25th 2010, 05:27 PM
Tander

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}
May 22nd 2010, 09:02 PM
dwsmith

Quote:

Originally Posted by Tander

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.
May 23rd 2010, 03:53 AM
HallsofIvy

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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