Express w 2 1 2 1 in the form w w1 w2 where w1 lies in th

Express w = (-2, -1, 2, 1) in the form w = w₁ + w₂, where w₁ lies in the subspace W spanned by the vectors u₁ = (2, 0, -2, 0) and u₂ = (1, -1, -2, 0), and w₂ is orthogonal to W.

Round your answers to 2 decimal places.

W₁=(_____,_____,_____,_____)

W₂=(_____,_____,_____,_____)

Solution

If w₁ lies in W = span{ u₁,u₂}, then w₁ is a linear combination of  u₁,u₂. Let w₁= au₁+bu₂ where a,b are scalars. Then w₁ = a( 2,0,-2,0) + b(1,-1,-2,0) = (2a,0,-2a ,0)+ (b,-b,-2b,0) = (2a+b, -b,-2a-2b,0).

Now, let w₂ = (p,q,r,s). Then w₂ .u₁ = 0 or,  (p,q,r,s).(2,0,-2,0) = 0 or, 2p-2q = 0 or, p-q= 0 or, p = q...(1)

Also, w₂. u₂ = 0 or,  (p,q,r,s). (1,-1,-2,0) = 0 or, p-q-2r = 0 ...(2).

Further, since p-q = 0, hence on substituting p-q = 0 in the 2nd equation, we get 0-2r = 0 or, r = 0. Then w₂ = (p, p, 0, s). Then w₁+w₂ = w or, (2a+b, -b,-2a-2b,0)+(p, p, 0, s) = (-2,-1,2,1) or,(2a+b+p, -b+p, -2a-2b,s) =(-2,-1,2,1) so that 2a+b+p = -2, -b+p= -1, -2a-2b= 2, s = 1. Since -b+p= -1 , therefore, b = p+1. Further, -2a-2b = 2 so that a+b = -1 or, a = -b-1 = -p-1-1 or, a = -p-2. On substituting these values of a and b in 2a+b +p = -2, we get (-2p-4)+p+1+p = -2 or, -3 = -2 which is incorrect. Hence w = (-2,-1,2,1) cannot be expressed as w₁+w₂