Question

I am looking for how to explain #4 part b. I have gotten the matrix A and I believe the answer is W = span{ v1 u2 u3 } however I'm not really sure if that is correct or not. Please give a small explanation. Also im not sure if I need to represent the vectors in A as columns or rows, or if either one works.

For the next two problems, W is the subspace of R4 given by 0 W = SpanV1 = Problem 3 a) If W; = Span{vi}, i = 1, 2, 3, compute the projections wų := prw,(v2) and W2 := prw,(v1). How does Span{w1,W2} compare to Span{V1, V2}? Are they the same subspace, or are they different? (You should justify your answer) b) Let uz = V2 - W1, u = V1 - W2. Compute the dot products u;. V1, i = 1,2. Is either {V1, uz} or {V1, U2} an orthogonal set of vectors? (You should justify your answer) Problem 4 a) Using the vectors vi,Wi, u, as defined in the previous problem, compute the projections prw. (v3), pru.(v3) where U2 = Span{u2}. b) Let uz = V3 – prw, (v3) – pruz (v3). Show that {V1, 42, 43} is an orthogonal set of vectors. Does W = Span{V1, U2, U3}? (You should justify your answer. Hint: form the matrix A = [v1 V2 V3 V1 u2u3] and compute its rref.)

Answer 1

Question 4 b)

find U₂ = V₂ - w - prw (2) W, = span {v} Now, prw (12) = &.vv VV Vzovy- (2)(0) + (32612 + (-13(-3) + (57(4) = 26 VA VE (2)2 + 43)2 + (-102 + (5)2 = 39 26 v, - 2 39 0 2 -4/3 3 - - -1/3 -3 5 2/3 V₂ - prw, (V3) - pru (3) 13:42 42 - v. V Vri 42 42 since, z. v = (-3) (2) + (0)(3) + (4) (-1) + (-2)(3)= -20 +3)()+ (0) {=})+ (444-7)+(-2)(3) -20 (412+ 323

4₂ - (-4) + (-132 + 1-392 + (i = 26 Then 2 O 4 = (-202 39 3 -4/3 (-20/3) (26/3) -13 43 4 - 3 4 10/13 22/13 14/13 Now, Conside the mateix A A = [v, V3 V "₂ uz Represent the rectors of columns. ( Roues does not works) 2 2 -3 -4/3 -3 0 - 3 0 3 A = -1 -3 4 - 1 -713 2/3 10/13 22/13 T4/13 4 5 5 1 O 1 we get the -23 0 10/13 Apply row operations 0 0 ref (A)= 0 - 0 o - 0 D 0 0 0 0 6 Here, v. 42 and 4 can be msitten as a linear combpation the system in consistent sot, ka w = span {vet₂, 4}