\[\left\{ \mathbf{v_1} , \mathbf{v_2} \right\}\]
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
\[\begin{equation} \begin{aligned} \alpha - \beta + \gamma a &= 0 \\ \alpha + \beta + \gamma b &= 0 \\ 2 \alpha + \beta + \gamma c &=0 \end{aligned} \end{equation}\]
.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 \[\left\{ \begin{pmatrix}1\\1\\2\end{pmatrix} , \begin{pmatrix}-1\\1\\1\end{pmatrix} , \begin{pmatrix}0\\0\\1\end{pmatrix} \right\}\]
.