Show that a vector $\langle h,k\rangle$ is in span${u, v}$ for all $h$ and $k$

Question

I have a question with two vectors and it asks to prove that a third vector is in the span of the two vectors.

Let $u = \left[ {\begin{array}{c} 2 \\ -1 \\ \end{array} } \right]$ and $v = \left[ {\begin{array}{c} 2\\ 1\\ \end{array} } \right]$. Show that $\left[ {\begin{array}{cc} h \\ k \\ \end{array} } \right]$ is in Span$\{u, v\}$ for all $h,k$.

This was my answer but I'm not sure if it's correct.

All $h, k$ are in Span$\{u, v\}$ because $u$ and $v$ are not multiples of each other, therefore they produce a plane in $\mathbb R^2$ which means any vector $\langle h, k\rangle$ is on the plane.

Is that correct reasoning? I'm not sure how to show it mathematically.

Answer 1

Your reasoning is good. Since the vectors are linearly independent they span $\mathbb{R}^2$ and since any vector $(h,k)$ is in $\mathbb{R}^2$ it must be in Span$\{\vec{u},\vec{v}\}$.

If you want to find the coefficients needed for $$ c_1 \vec u + c_2 \vec v = \begin{bmatrix} h\\ k \end{bmatrix} $$ then you would proceed as others have suggested by row reducing the matrix $$ \begin{bmatrix} 2 && 2 && h\\ -1 && 1 && k \end{bmatrix}. $$

Answer 2

Note that $$ \frac{h-2k}{4}\,\vec u+\frac{h+2k}{4}\,\vec v=\begin{bmatrix}h\\ k\end{bmatrix} $$

Answer 3

Hint: you show that: $\exists x,y$ such that: $x\begin{pmatrix} 2 \\ -1 \end{pmatrix}+y\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} h \\ k \end{pmatrix}$. This means you solve for $x,y$ in terms of $h,k$.