Why two vectors cannot span ${\bf R}^3$?
By John Thompson
Consider two vectors $$ u_{1} = \begin{bmatrix} -1 \\ 3 \\ 2 \\ \end{bmatrix},\quad u_{2} = \begin{bmatrix} 6 \\ 1 \\ 1 \\ \end{bmatrix} $$
I can tell they don't span $R^3$ because $R^3$ requires three vectors to span it.
But is there another way I can see that it does not span $R^3$, for example, in terms of overdetermined system, or not a square matrix? Just trying to develop a mature understanding of this topic, that's all.
$\endgroup$ 43 Answers
$\begingroup$You want to find $(a,b,c)$ such that $(a,b,c)$ cannot be written as $x(-1,3,2)+y(6,1,1)$. But this is easy -- in fact, since most $(a,b,c)$ cannot be written in that way, you can just pick one at random and start solving!
For example, pick $(a,b,c)=(0,0,1)$.
Since $-x+6y=0$ we must have $x=6y$. Now plug that into $3x+y=0$, giving $18y+y=0$ which is the same as $19y=0$, and so $y=0$.
But then $x=y=0$ and $x(-1,3,3)+y(6,1,1)=(0,0,0)$, which is not what we wanted. So $(0,0,1)$ is impossible to make, and thus we have the counterexample we need.
$\endgroup$ 1 $\begingroup$Let $u=\begin{bmatrix}a\\b\\c\end{bmatrix}$, with $a,b,c\in\mathbb R$. Let us see if $u\in\langle u_1,u_2\rangle$. It is true if and only if there are $x,y\in\mathbb R$ such that $u=xu_1+yu_2$, which is equivalent to$$\left\{\begin{array}{l}-x+6y=a\\3x+y=b\\2x+y=c\end{array}\right.$$If you try to solve the system, you will see that, for some values of $a$, $b$, and $c$; it has no solution. Therefore, $\langle u_1,u_2\rangle\neq\mathbb{R}^3$.
$\endgroup$ 4 $\begingroup$$\dim\operatorname{span}(\{v_1,v_2\}) \le 2$ while $\dim(\mathbb{R}^n) = n$ for all $n$.
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