I am studying for my reexamination. I failed the first try and I have a really hard time understanding most of linear algebra. But I am working as hard as my head can.
I have the vectors shown as
$$
\begin{pmatrix}
1 \\
1
\end{pmatrix}
,
\begin{pmatrix}
-1 \\
-2
\end{pmatrix},
$$
\begin{pmatrix}
2 \\
0
\end{pmatrix}
I need to show that these three vectors expand the vector space $\Bbb R^2$ and I don't know how to show that the right way.
I hope someone can help me.
Let $v_1 = (1,1) , v_2 = (-1, -2), v_3 = (2, 0) $
And let $P = (x, y) $ be any point in $\mathbb{R}^2$, then if the three given vectors span $\mathbb{R}^2$ we can always find at least one triple $(c_1, c_2, c_3)$ such that
$ c_1 v_1 + c_2 v_2 + c_3 v_3 = P $
In matrix-vector form, this translated into the system
$\begin{bmatrix} 1 && -1 && 2 \\ 1 && -2 && 0 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} $
always have a solution $(c_1, c_2, c_3) $. The augmented matrix of the above system is
$\begin{bmatrix} 1 && -1 && 2 && x \\ 1 && -2 && 0 && y \end{bmatrix} $
Upon Gauss-Jordan elimination to transform this into reduced-row echelon form it becomes
$\begin{bmatrix} 1 && 0 && 4 && 2x-y \\ 0 && 1 && 2 && x-y \end{bmatrix} $
So by selecting $c_3 = t $ for an arbitrary $ t \in \mathbb{R} $ we can solve for $c_1 $ and $c_2 $ as follows:
$c_1 = (2x - y) - 4 t $
$c_2 = (x-y) - 2 t $
And this means there is an infinite number of ways we can express $P$ as a linear combination of $v_1, v_2, v_3$. Hence, these vectors span $\mathbb{R}^2$.