(adsbygoogle = window.adsbygoogle || []).push({}); Question:In R^{3}, show that (1,-1,0) and (0,1,-1) are a basis for the subspace V={(x,y,z) [tex]\in[/tex] R^{3}: x+y+z=0}

Attempt:By def of a basis, the vectors (1) must be linearly independent and (2) must span V.

1. For LI, show that if a(1,-1,0) + b(0,1,-1) = (0,0,0), then a=b=0.

(a,-a,0)+(0,b,-b)=(0,0,0)

(a,b-a,-b)=(0,0,0)

a=0, b-a=0, -b=0

Since the zero vector can only be expressed as a combination of a=0 and b=0, the two vectors are linearly idependent.

2. I'm not really certain how to show if the two vectors span V. If I was to guess, I'd say plug in the vectors into V, so (1,-1,0)((1,1,1)^{t})=0 and (0,1,-1)((1,1,1)^{t})=0, but I could be wrong.

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# Show that two vectors are a basis of a subspace

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