Find a basis for the space spanned by the given vectors, v1…,v5 [1−265001−1−330−123−111−1−41]∼[100−1−2010−350010200000]

This problem is equivalent to finding a basis for Col A, where [1−265001−1−330−123−111−1−41]∼[100−1−2010−350010200000] Since the reduced echelon form A is [1−265001−1−330−123−111−1−41]∼[100−1−2010−350010200000] 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 [1001],[−21−11],[6−12−1] Result: Basis is [1001],[−21−11],[6−12−1]

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College
Algebra
Linear algebra
Vectors and spaces

geduiwelh

Answered question

2021-09-18

Find a basis for the space spanned by the given vectors, $v_{1} \dots, v_{5}$
$[\begin{matrix} 1 & - 2 & 6 & 5 & 0 \\ 0 & 1 & - 1 & - 3 & 3 \\ 0 & - 1 & 2 & 3 & - 1 \\ 1 & 1 & - 1 & - 4 & 1 \end{matrix}] \sim [\begin{matrix} 1 & 0 & 0 & - 1 & - 2 \\ 0 & 1 & 0 & - 3 & 5 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]$

Answer & Explanation

Arnold Odonnell

Skilled2021-09-19Added 109 answers

This problem is equivalent to finding a basis for Col A, where
$[\begin{matrix} 1 & - 2 & 6 & 5 & 0 \\ 0 & 1 & - 1 & - 3 & 3 \\ 0 & - 1 & 2 & 3 & - 1 \\ 1 & 1 & - 1 & - 4 & 1 \end{matrix}] \sim [\begin{matrix} 1 & 0 & 0 & - 1 & - 2 \\ 0 & 1 & 0 & - 3 & 5 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]$
Since the reduced echelon form A is
$[\begin{matrix} 1 & - 2 & 6 & 5 & 0 \\ 0 & 1 & - 1 & - 3 & 3 \\ 0 & - 1 & 2 & 3 & - 1 \\ 1 & 1 & - 1 & - 4 & 1 \end{matrix}] \sim [\begin{matrix} 1 & 0 & 0 & - 1 & - 2 \\ 0 & 1 & 0 & - 3 & 5 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]$
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
$[\begin{array}{l} 1 \\ 0 \\ 0 \\ 1 \end{array}], [\begin{array}{r} - 2 \\ 1 \\ - 1 \\ 1 \end{array}], [\begin{array}{r} 6 \\ - 1 \\ 2 \\ - 1 \end{array}]$
Result: Basis is $[\begin{array}{l} 1 \\ 0 \\ 0 \\ 1 \end{array}], [\begin{array}{r} - 2 \\ 1 \\ - 1 \\ 1 \end{array}], [\begin{array}{r} 6 \\ - 1 \\ 2 \\ - 1 \end{array}]$

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