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Let $U$ be the vector $ \begin{bmatrix} 2\\ -1 \end{bmatrix} $ and let $V=\begin{bmatrix} 2\\ 1 \end{bmatrix}. $

Show that the \begin{bmatrix} h\\ k \end{bmatrix}

is in the $\text{Span}\{U,V\}$ for all $h$ and $k$.

I am not sure how to solve this question. I do not seem to grasp what is being asked.

I do think you have to make a system of linear equation. $$ \left\{ \begin{array}{l} 2x+2y=h, \\ -1x+1y=k. \end{array} \right. $$

But not sure how to keep going.

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Notice that $U$ and $W$ are linear independent, so $span \{U,W\}=\mathbb{R}^2$and $\begin{bmatrix} H\\ K \end{bmatrix} \in \mathbb{R}^2$ for all $H,K \in\mathbb{R}$.

$W$ and $U$ are linear independent, because $W \neq \alpha U$ and $U \neq \alpha W$ for all $\alpha \in\mathbb{R}$.

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  • $\begingroup$ I think I see what you are saying because as point (2,1) and (2,-1) are two opposities and so they cover the entire plane. $\endgroup$
    – Fernando Martinez
    Sep 1, 2014 at 20:10
  • $\begingroup$ Yes, in $\mathbb{R}^2$ it's the same. $\endgroup$
    – agha
    Sep 1, 2014 at 20:12
  • $\begingroup$ @FernandoMartinez almost. They're not really opposites, but they are linearly independent which is enough from the general theory to conclude you get everything. This is the usual way to show existence for systems where you don't want to work out the specifics, since the notion of spanning is so strong. $\endgroup$
    – Adam Hughes
    Sep 1, 2014 at 20:31
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To see that these vectors are linearly independent, write the system of equations

$$\begin{bmatrix}2&2\\-1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\vec{0}$$

By elimination, it's easy to see that this has no non-trivial solutions. From here, it's simple to see that $\{U,V\}$ spans all of $\mathbb R^2$.

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A more elementary solution : Show there is a solution to the system of equations you wrote (solve for x and y and treat H and K as constants)

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