Problem 709

Let \$S=\mathbfv_1,mathbfv_2,mathbfv_3,mathbfv_4,mathbfv_5\$ whereFind a basis for the span \$Span(S)\$.

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We will certainly offer 2 solutions.

Solution 1.

We apply the leading 1 technique.Let \$A\$ be the matrix whose column vectors are vectors in the collection \$S\$:Applying the elementary row operations to \$A\$, we obtaineginalign*A=eginbmatrix1 & 1 & 1 & 1 & 2 \2 & 3 & 5 & 1 & 7 \2 & 1 & -1 & 4 & 0 \-1 & 1 & 5 & -1 & 2endbmatrixxrightarrowsubstackR_2-2R_1 \ R_3-2R_1eginbmatrix 1 & 1 & 1 & 1 & 2 \ 0 & 1 & 3 & -1 & 3 \ 0 & -1 & -3 & 2 & -4 \ 0 & 2 & 6 & 0 & 4 endbmatrix\<6pt> xrightarrowsubstackR_1-R_2 \ R_3+R_2 eginbmatrix 1 & 0 & -2 & 2 & -1 \ 0 & 1 & 3 & -1 & 3 \ 0 & 0 & 0 & 1 & -1 \ 0 & 0 & 0 & 2 & -2 endbmatrix xrightarrowsubstackR_1-2R_3 \ R_2+R_3 eginbmatrix 1 & 0 & -2 & 0 & 1 \ 0 & 1 & 3 & 0 & 2 \ 0 & 0 & 0 & 1 & -1 \ 0 & 0 & 0 & 0 & 0 endbmatrix= ref(A).endalign*Observe that the initially, second, and also fourth column vectors of \$ ref(A)\$ contain the leading 1 entries.Hence, the first, second, and fourth column vectors of \$A\$ develop a basis of \$Span(S)\$.Namely, is a basis for \$Span(S)\$.

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LetThen \$Span(S)\$ is the column space of \$A\$, which is the row room of \$A^T\$. Using row operations, we have< oeginbmatrix1 & 0 & 0 & -13 \0 & 1 & 0 & 4 \0 & 0 & 1 & 2 \0 & 0 & 0 & 0 \0 & 0 & 0 & 0endbmatrix.>Therefore, the collection of nonzero rowsis a basis for the row room of \$A^T\$, which amounts to \$Span(S)\$.