Proving a Vector in $\Bbb R^2$ is of the Form $au+bv$

[Niamh1]

Let $a, b \in \Bbb R$ and $u, v \in \Bbb R^2$, with $u = (0, 1)$ and $v = (1, 0)$. Show that every vector in $\Bbb R^2$ is of the form $au + bv$. Under what conditions is this true for general vectors $u = (u_1, u_2)$ and $v = (v_1, v_2)$?

No idea where to begin. Would appreciate any help.

 

[Greg]

Any point in $\mathbb{R}^2$ can be described as the coordinates of the tip of a vector. As any vector can be the hypotenuse of the triangle formed by such a vector and two perpendicular vectors (with either $a$ or $b$ being $0$ if our vector is vertical or horizontal), we can describe any vector with the vector sum given.

What if $<u_1,u_2>$ and $<v_1,v_2>$ are orthogonal? What if they are not orthogonal?

Does that help?

 

[johng1]

Let $w=(w_1,w_2)$ be an arbitrary vector. With your notation, when can you find $a,\,b$ with $w=au+bv$. That is the system in unknowns a and b has a (unique) solution:
$$u_1a+v_1b=w_1$$
$$u_2a+v_2b=w_2$$

Since this is linear algebra you should know that the answer is that the determinant $u_1v_2-v_1u_2\neq0$.