Problem
Let S={v1,v2,v3,v4,v5} where
v1=⎡⎣⎢⎢⎢122−1⎤⎦⎥⎥⎥,v2=⎡⎣⎢⎢⎢1311⎤⎦⎥⎥⎥,v3=⎡⎣⎢⎢⎢15−15⎤⎦⎥⎥⎥,v4=⎡⎣⎢⎢⎢114−1⎤⎦⎥⎥⎥,v5=⎡⎣⎢⎢⎢2702⎤⎦⎥⎥⎥.
Find a basis for the span Span(S).
Solution
We apply the leading 1 method.
Let A be the matrix whose column vectors are vectors in the set S:
A=⎡⎣⎢⎢⎢122−1131115−15114−12702⎤⎦⎥⎥⎥.
Applying the elementary row operations to
A, we obtain
A=⎡⎣⎢⎢⎢122−1131115−15114−12702⎤⎦⎥⎥⎥−→−−−−R4+R1R2−2R1R3−2R1⎡⎣⎢⎢⎢100011−1213−361−12023−44⎤⎦⎥⎥⎥−→−−−R4−2R2R1−R2R3+R2⎡⎣⎢⎢⎢10000100−23002−112−13−1−2⎤⎦⎥⎥⎥−→−−−−R4−2R3R1−2R3R2+R3⎡⎣⎢⎢⎢10000100−2300001012−10⎤⎦⎥⎥⎥=rref(A).A=[111122351721−140−115−12]→R4+R1R2−2R1R3−2R1[11112013−130−1−32−402604]→R4−2R2R1−R2R3+R2[10−22−1013−130001−10002−2]→R4−2R3R1−2R3R2+R3[10−201013020001−100000]=rref(A)See that the main, second, and fourth segment vectors of rref(A) contain the main 1 sections.
Thus, the main, second, and fourth section vectors of A shape a premise of Span(S).
⎪⎡⎣⎢⎢⎢122−1⎤⎦⎥⎥⎥,⎡⎣⎢⎢⎢1311⎤⎦⎥⎥⎥,⎡⎣⎢⎢⎢114−1⎤⎦⎥⎥⎥⎫⎭⎬⎪⎪⎪⎪⎪⎪
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