# Find a basis for the space spanned by the given vectors, v_1...,v_5PS

Find a basis for the space spanned by the given vectors, ${v}_{1}\dots ,{v}_{5}$
$\left[\begin{array}{ccccc}1& -2& 6& 5& 0\\ 0& 1& -1& -3& 3\\ 0& -1& 2& 3& -1\\ 1& 1& -1& -4& 1\end{array}\right]\sim \left[\begin{array}{ccccc}1& 0& 0& -1& -2\\ 0& 1& 0& -3& 5\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 0& 0\end{array}\right]$

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Arnold Odonnell

This problem is equivalent to finding a basis for Col A, where
$\left[\begin{array}{ccccc}1& -2& 6& 5& 0\\ 0& 1& -1& -3& 3\\ 0& -1& 2& 3& -1\\ 1& 1& -1& -4& 1\end{array}\right]\sim \left[\begin{array}{ccccc}1& 0& 0& -1& -2\\ 0& 1& 0& -3& 5\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 0& 0\end{array}\right]$
Since the reduced echelon form A is
$\left[\begin{array}{ccccc}1& -2& 6& 5& 0\\ 0& 1& -1& -3& 3\\ 0& -1& 2& 3& -1\\ 1& 1& -1& -4& 1\end{array}\right]\sim \left[\begin{array}{ccccc}1& 0& 0& -1& -2\\ 0& 1& 0& -3& 5\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 0& 0\end{array}\right]$
we see that the first, second and third columns of A are its pivot columns. Thus a basis for the space spanned by the given vectors is
$\left[\begin{array}{l}1\\ 0\\ 0\\ 1\end{array}\right],\left[\begin{array}{r}-2\\ 1\\ -1\\ 1\end{array}\right],\left[\begin{array}{r}6\\ -1\\ 2\\ -1\end{array}\right]$
Result: Basis is $\left[\begin{array}{l}1\\ 0\\ 0\\ 1\end{array}\right],\left[\begin{array}{r}-2\\ 1\\ -1\\ 1\end{array}\right],\left[\begin{array}{r}6\\ -1\\ 2\\ -1\end{array}\right]$