Matrices problem : V define a vector space. Determine whether the following sets
ID: 3032229 • Letter: M
Question
Matrices problem : V define a vector space.
Determine whether the following sets V define a vector space. V is the set of all vectors [x y z]^T such that x + y = 2. The superscript T denotes the 'transpose'. V is the set of all vectors [x_1 x_2 x_3 x_4]^T such that x_1 + 2x_2 = x_3 + x_4. V is the set of all vectors [x_1 x_2 x_3]^T such that x^2_1 + x^2_2 = x^2_3. V is the set of all vectors in R^3 such that x_1le 0 and X_2 greaterthanorequalto 0. V is the set of all vectors x in R^n] such that Ax = lambda x, where A is an n times n matrix and lambda is a constant.Explanation / Answer
(a). Let X = [ x1 , y1 , z1 ]T and Y = [ x2 , y2 , z2 ]T be two arbitrary vectors in V and let c be an arbitrary scalar. Then, x1 + y1 = 2 and x2 +y2 = 2 . Further, (x1 +x2 ) + (y1 +y2 ) = (x1 + y1) +( x2 +y2) = 2+2 = 4 2. Hence X+Y V. Hence V, not being closed under vector addition, is not a vector space.
(b). let X = ( x1 , x2 , x3, x4 )T and Y = ( y1 , y2 , y3 , y4 )T be two arbitrary vectors in V and let c be an arbitrary scalar. Then x1 +2x2 = x3 +x4 and y1 +2y2 = y3 +y4. Further (x1 +y1 ) + 2( x2 + y2 ) = (x1 +2x2) + ( y1+2y2) = (x3 +x4 )+(y3 +y4) = (x3 +y3) +(x4 +y4). Hence X +Y V. Further, (cx1+2cx2 ) = c(x1 +2x2) = c(x3 +x4) = (cx3 + cx4) . Hence cX X. Also, the zero vector apparently belongs to V. Hence V is a vector space.
(c) Let X = (x1 , x2 , x3 )T and y = ( y1, y2 , y3 )T be two arbitrary vectors in V. Then x12 +x22 = x32 and y12 +y22 = y32. Further( x1 +y1 )2 +(x2 +y2)2 = x12 + y12 + 2x1 y1 + x22 + y22 + 2x2 y2 = x12 + x22 + y12+ y22 + 2x1 y1+2x2 y2 = x32 + y32 + 2x1 y1+2x2 y2 which is not necessarily equal to(x3 +y3 )2 . Hence V, being not closed under vector addition , is not a vector space.
(d) Let c be an arbitrary scalar and let X = ( x1 , x2 , x3 ) be an arbitrary vector in V. Then x1 0 and x2 0. If x1 < 0 and c is a negative scalar, then cx1 > 0 so that cX V. Hence V is not a vector space.
(e) Let X and Y be 2 arbitrary vectors in V and let c be an arbitrary scalar. Then A(X+Y)= AX +AY = X +Y = (X+Y) so that X+Y V. Further A(cX) = c(AX) = cx = (cX) . Hence cX V. Apparently, the zero vector also belongs to V. Hence V is a vector space.