Consider the subspace V and W as memers of R4 V span 110100

Consider the subspace V and W as memers of R⁴

V = span {(1,-1,0,1);(0,0,1,2);(1,0,0,1)}

W = { x + y - 2z + w = 0

{2x - y - 3z + 2w = 0

(a) Express V as the set of solutions of a homogeneous linear system.

(b) Can a vector (x, y, z, w) where x,y,z,w are members of R⁴ be orthogonal to itself?

(c) Find a subspace W^? such that all its vectors are orthogonal to all the vectors of W.

Solution

(a) Let A =

1

0

1

-1

0

0

0

1

0

1

2

1

If X = (x,y,z)^T, then the required linear system is AX= 0 or,

x+z = 0; -x = 0, y = 0 and x +2y +z = 0

(b) Only the zero vector in R⁴ is orthogonal to itself. If X= (x,y,z,w)^T in R⁴ is orthogonal to itself, then X.X = 0 or, x² +y² + z²+ w²= 0. Since the square of a number is always 0 or positive, this equation has a solution only if x = y = z = w = 0.

(c) We have x + y - 2z + w = 0 and 2x - y - 3z + 2w = 0 or, x + y = 2z-w and 2x - y = 3z-2w. On solving these equations for x and y, we get x= 5z/3–w and y= z/3 so that X = (x,y,z,w)^T = (5z/3 –w, z/3,z,w)^T= z(5/3, 1/3,1,0)^T +w (-1, 0,0,1)^T. Thus W = span{(5/3, 1/3,1,0)^T , (-1, 0,0,1)^T} = span{(5,1,3,0)^T, (-1, 0,0,1)^T}. Let u = (a,b,c,d)^T be an arbitrary vector in W. Then (a,b,c,d)^T. (5,1,3,0)^T = 0 or, 5a+b+3c = 0. Also, (a,b,c,d)^T. (-1, 0,0,1)^T= 0 or, -a+d= 0 or, a = d. Then 5d +b+3c = 0 so that b = -3c-5d so that (a, b,c,d)^T = ( d, -3c-5d,c,d)^T = c(0,-3,1,0)^T +d(1,-5,0,1)^T. Hence W = span{(0,-3,1,0)^T , (1,-5,0,1)^T}.

(d) Since W is the orthogonal complement of W, hence W W=