## 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

\[\begin{equation} \begin{aligned} \alpha - \beta + \gamma a &= 0 \\ \alpha + \beta + \gamma b &= 0 \\ 2 \alpha + \beta + \gamma c &=0 \end{aligned} \end{equation}\]

.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 \[\left\{ \begin{pmatrix}1\\1\\2\end{pmatrix} , \begin{pmatrix}-1\\1\\1\end{pmatrix} , \begin{pmatrix}0\\0\\1\end{pmatrix} \right\}\]

.