- #1

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Find a basis for the span: [tex]\vec{a_{1}}=(1,\,-1,\,6,\,0),\,\vec{a_{2}}=(3,\,-2,\,1,\,4),\,\vec{a_{3}}=(1,\,-2,\,1,\,-2),\,\vec{a_{4}}=(10,\,1,\,7,\,3)[/tex]

- Thread starter ferry2
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- #1

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Find a basis for the span: [tex]\vec{a_{1}}=(1,\,-1,\,6,\,0),\,\vec{a_{2}}=(3,\,-2,\,1,\,4),\,\vec{a_{3}}=(1,\,-2,\,1,\,-2),\,\vec{a_{4}}=(10,\,1,\,7,\,3)[/tex]

- #2

- 131

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Make a matrix with a1, a2, a3, and a4 in separate rows, with each component of each vector in a separate column. Put it in row reduced echelon form. The nonzero rows of your new matrix are the vectors that form the basis.

Last edited:

- #3

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[tex]A=\left( \begin{array}{cccc}1 &-1 & 6 & 0\\ 3 &-2 & 1 & 4\\ 1 &-2 & 1 &-2\\ 10 & 1 & 7 & 3\\ \end{array} \right)[/tex] is [tex]\left( \begin{array}{cccc}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{array} \right)[/tex] so the basis are vectors [tex]\vec{e_1}(1,\,0,\,0,\,0),\,\vec{e_2}(0,\,1,\,0,\,0),\,\vec{e_3}(0,\,0,\,1,\,0)[/tex] and [tex]\vec{e_4}(0,\,0,\,0,\,1)[/tex] right?

- #4

- 131

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yeah that's what I got

- #5

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Thanks a lot for the tips :).

- #6

HallsofIvy

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Which says that the span of those four vectors is, in fact, all of [itex]R^4[/itex].

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