# Find a basis

by ferry2
Tags: basis
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 P: 15 I don't wan't a solution I wan't only instructions how to solve this problem: Find a basis for the span: $$\vec{a_{1}}=(1,\,-1,\,6,\,0),\,\vec{a_{2}}=(3,\,-2,\,1,\,4),\,\vec{a_{3}}=(1,\,-2,\,1,\,-2),\,\vec{a_{4}}=(10,\,1,\,7,\,3)$$
 P: 129 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.
 P: 15 So under your guidance the row reduced eshelon form of the matrix: $$A=\left( \begin{array}{cccc}1 &-1 & 6 & 0\\ 3 &-2 & 1 & 4\\ 1 &-2 & 1 &-2\\ 10 & 1 & 7 & 3\\ \end{array} \right)$$ is $$\left( \begin{array}{cccc}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{array} \right)$$ so the basis are vectors $$\vec{e_1}(1,\,0,\,0,\,0),\,\vec{e_2}(0,\,1,\,0,\,0),\,\vec{e_3}(0,\,0,\ ,1,\,0)$$ and $$\vec{e_4}(0,\,0,\,0,\,1)$$ right?
 P: 129 Find a basis yeah that's what I got
 P: 15 Thanks a lot for the tips :).
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,491 Which says that the span of those four vectors is, in fact, all of $R^4$.
 P: 49 Because the span of those vectors is all of R^4, both the 4 standard R^4 basis vectors and the column vectors of your matrix form a basis. What's more interesting is the case when the column vectors do not span the entire space. Then you take the column vectors from your matrix that correspond to the columns with pivots in reduced form, and those form a basis for the column space.

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