Show that two given vectors span the real plane

by RoRi

April 25, 2016

Given vectors $A = (2,1)$ and $B = (1,3)$ show that every vector $C = (c_1, c_2)$ in $\mathbb{R}^2$ is of the form $C = xA + yB$ for some $x,y \in \mathbb{R}$ . Express $x$ and $y$ in terms of $c_1$ and $c_2$ .

Proof. (Note: I’m doing this without any real linear algebra since we don’t get to that until Volume 2. If you do know linear algebra then you would show the two vectors $A$ and $B$ are linearly independent and you’d be done.) Since

$C = xA+ yB + (2x+y, x+ 3y) = (c_1, c_2)$

we have

$\begin{align*} c_1 &= 2x+y \\ c_2 &= x + 3y. \end{align*}$

This implies

$x = c_2 - 3y \quad \implies \quad c_1 = 2c_2 - 6y + y.$

Therefore,

$\begin{align*} y &= \frac{1}{5} \left( 2c_2 - c_1 \right) \\ x &= \frac{1}{5} \left( 3c_1 - c_2 \right). \end{align*}$

This also shows that any vector in $\mathbb{R}^2$ can be obtained as $xA + yB$ since given $C = (c_1, c_2)$ we compute $x$ and $y$ by the formulas above $. \qquad \blacksquare$

One comment

February 18, 2017 at 8:28 am

Alan Gonzaga says:

There is an error on the first line
Should be an “=” in the place of one of the “+”

There is C = xA + yB “+”
Should be C = xA + yB “=”

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Stumbling Robot

Show that two given vectors span the real plane

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