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}

Printable View

- Jan 25th 2010, 05:27 PMTanderProve 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} - May 22nd 2010, 09:02 PMdwsmith
$\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. - May 23rd 2010, 03:53 AMHallsofIvy
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.