Question

6.20 Below, you are given two sets of vectors B, and B, in R. B = {[-1. 1. -11', [1,3,21') B2 = {[3, 1, 41', (-4,16,-1]'}

Answer 1

Given,
Bi = 311
and
3 B2 = 1

Let u =
11
, v = $\begin{bmatrix} 1\\ 3\\ 2 \end{bmatrix}$ , p = $\begin{bmatrix} 3\\ 1\\ 4 \end{bmatrix}$ , q = $\begin{bmatrix} -4\\ 16\\ -1 \end{bmatrix}$ .

(e) Now projection of X = [x,y,z]^t on W using the basis B₁ is :

Prow (A) = Projw (X) = (X, u), _ (X, v) " Tv.)

i.e.,

Prow(x) = Projw(X) = x + 3y + 2 -+y-2 1+1+1 1 **+ 2 + 14

i.e.,

(x – y + 2)/3] Projw(X) = |(-2+y - 2)/3 L (x – y + 2)/3] U + (x + 3y +22)/14 3(1 + 3y +22)/14 (2 + 3y + 2)/7

i.e.,

Projw (X) = (172 – 5y +202)/42 (-5.6 +41y + 4z)/42 (10.+ 2y + 132)/21

Now,

(17.x - 5y +202)/42 (-5.6 + 41y + 4z)/42 (10.x + 2y + 13z)/21 = 17/42 -5/42 10/21 -5/42 41/42 2/21 10/212/21 13/21

Therefore, R = .

17/42 -5/42 10/21 -5/42 41 42 2/21 | 10/21 221 13/21

(f) Now projection of X = [x,y,z]^t on W using the basis B₂ is :

Projw(X) = (X,p)p+ (8,9), Prow(A) = (p.p) (9,9)

i.e.,

3.0 + y +4: Projw (X) = 9 +1 +16 - 4.2 + 164 - 2 16 +256 +1

i.e.,

3.0 + y + 4z Projw(X) = -4.r + 16y - 2 273 26

i.e.,

Projw (X) = (172 – 5y +202)/42 (-5.6 +41y + 4z)/42 (10.+ 2y + 132)/21

Now,

(17.x - 5y +202)/42 (-5.6 + 41y + 4z)/42 (10.x + 2y + 13z)/21 = 17/42 -5/42 10/21 -5/42 41/42 2/21 10/212/21 13/21

Therefore, R = .

17/42 -5/42 10/21 -5/42 41 42 2/21 | 10/21 221 13/21

(c) Now, R² = R*R

i.e., R² =
17/42 -5/42 10/21 -5/42 41 42 2/21 | 10/21 221 13/21

17/42 -5/42 10/21 -5/42 41 42 2/21 | 10/21 221 13/21

i.e., R² = [Using Calculator]

17/42 -5/42 10/21 -5/42 41 42 2/21 | 10/21 221 13/21

i.e., R² = R

Geometric meaning of this equality is :

If we apply this transformation twice on a vector, then it will return the effect of the first transformation.