x1+x2+x3-x4=0 find a basis for the subspace S in R4 spanned by all thesolution t

Question

x1+x2+x3-x4=0 find a basis for the subspace S in R4 spanned by all thesolution to above equation, find b1 in S and b2 in S so that b1+b2=b=(1,1,1,1) x1+x2+x3 - x4=0 find a basis for the subspace S in R4 spanned by all thesolution to above equation, find a basis for the orthogonal complement S , find b1 in S and b2 in S so that b1+b2=b=(1,1,1,1)

Explanation / Answer

x1 + x2 + x3 - x4 = 0 implies x1 + x2 + x3 = x4 Basis for S = {(1,0,0,1), (0,1,0,1), (0,0,1,1)} Orthogonal complement of S (S perp) = {(1,1,1,-1)} Now we find b1 in S and b2 in S perp so that b1 + b2 =(1,1,1,1) Consider (1,1,1,1) = b + c 1 = b1 + c1, 1 = b1 + c2, .... c = u(1,1,1,-1) b = v(1,0,0,1) + w(0,1,0,1) + x(0,0,1,1) so 1 = b1 + c1 = v + u (1) 1 = u + w (2) 1 = u + x 1 = -u + v + w + x now (1) - (2) -> v - w = 0 -> v = w similarly w= x and x = v thus 1 = -u + 3x and 1 = u + x, and adding gives us 2 = 4x or x = 1/2 so u = 1/2, indeed 1/2(1,0,0,1) + 1/2(0,1,0,1) + 1/2(0,0,1,1) + 1/2(1,1,1,-1) =(1,1,1,1)

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