Let W less than or equal to R3 be the set of vectors [x1, x2, x3]T for which x1-
ID: 2945277 • Letter: L
Question
Let W less than or equal to R3 be the set of vectors [x1, x2, x3]T for which x1-x2-x3=0.Prove that W is a subspace. Find a basis for W.
Explanation / Answer
To prove that W is a subspace,it is sufficient to prove that W is closed under vector addition and scalar multiplication Let u=(x1,x2,x3) and v=(y1,y2,y3) belong to W u+v = (x1+y1,x2+y2,x3+y3) (x1+y1)-(x2+y2)-(x3+y3) = (x1-x2-x3)+(y1-y2-y3) = 0+0 (since u,v belong to W) = 0 => u+v belongs to W Hence W is closed under vector addition Let u=(x1,x2,x3) belongs to W and p be a scalar pu = (px1,px2,px3) px1-px2-px3 = p(x1-x2-x3) = p(0) = 0 (as u belongs to W) => pu belongs to W Hence W is closed under scalar multiplication Hence W is a subspace For finding basis, consider an arbitrary element u = (x,y,z) belongs to W x-y-z = 0 => z = x-y u = (x,y,z) = (x,y,x-y) = x(1,0,1)+y(0,1,-1) As every element in W can be written as linear combination of 2 linearly independent vectors (1,0,1) and (0,1,-1) B = {(1,0,1),(0,1,-1)} is a basis for W
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