Is the set of all x, y, z such x+ 3y + 2z = 0 a subspace of R^3 ? If so find a basis for the space.
Let X1 = (x1,y1,z1)T and X2 = (x2,y2,z2)T be 2 arbitrary vectors in V (say), the given subset of R3 and let k be an arbitrary scalar. Then x1+ 3y1 + 2z1 = 0 and x2+ 3y2 + 2z2 = 0. Further, X1+X2 = (x1,y1,z1)T +(x2,y2,z2)T = (x1+x2,y1+y2,z1+z2)T and (x1+x2)+ 3(y1 +y2)+ 2(z1+z2) = (x1+ 3y1 + 2z1)+( x2+ 3y2 + 2z2)=0+0 = 0. This implies that X1+X2 ∈ V so that V is closed under vector addition. Also kX1 = k(x1,y1,z1)T = (kx1,ky1,kz1)T and kx1+ 3ky1 + 2kz1 =k(x1+ 3y1 + 2z1 ) =k.0 = 0. This implies that kX1 ∈ V so that V is closed under scalar multiplication. Also, since 0+3.0+2.0 = 0, hence the zero vector (0,0,0)T ∈ V so that V is a vector space, and , therefore, a subspace of R3.
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