## Extending a Linearly Independent Set of Vectors to a Basis for a Vector Space

Suppose we have a set of two linearly independent vectors
$\left\{ \mathbf{v_1} , \mathbf{v_2} \right\}$
in
$\mathbb{R}^3$
, so that the only solution to
$\alpha \mathbf{v_1} + \beta \mathbf{v_2} =0$
is
$\alpha = \beta =0$
.
Two vectors cannot be a basis for a three dimensional space, We can add more vectors which are not combinations of these vectors to get a spanning space, but if this expanded set is to be a basis, it must also be linearly independent.
Examnple:
$\left\{ \begin{pmatrix}1\\1\\2\end{pmatrix} , \begin{pmatrix}-1\\1\\1\end{pmatrix} , \mathbf{v} \right\}$
is to be a basis for
$\mathbb{R}^3$
.
The first two vectors are linearly independent so thatbthe only solution to
$\alpha \begin{pmatrix}1\\1\\2\end{pmatrix} + \beta \begin{pmatrix}-1\\1\\1\end{pmatrix} =0$
is
$\alpha =\beta =0$
.
We have to solve
$\alpha \begin{pmatrix}1\\1\\2\end{pmatrix} + \beta \begin{pmatrix}-1\\1\\1\end{pmatrix} + \gamma \begin{pmatrix}a\\b\\c\end{pmatrix} =0$
for
$a,b,c$
.such that the only solution is
$\alpha = \beta = \gamma =0$
.
We have
\begin{aligned} \alpha - \beta + \gamma a &= 0 \\ \alpha + \beta + \gamma b &= 0 \\ 2 \alpha + \beta + \gamma c &=0 \end{aligned}
.
We can put
$a=b=0$
in the first two equations then
$\alpha =\beta =0$
as before.
In the thrid equation we can put
$\gamma =0, c=1$
to give the basis
$\left\{ \begin{pmatrix}1\\1\\2\end{pmatrix} , \begin{pmatrix}-1\\1\\1\end{pmatrix} , \begin{pmatrix}0\\0\\1\end{pmatrix} \right\}$
.