Question

[1] (a) Verify that vectors ul 2 | ,u2 -1 . из 0 | are pairwise orthogonal (b) Prove that ũi,u2Ф are linearly independent and hence form a basis of R3. (c) Let PRR3 be the orthogonal projection onto Spansüi, us]. Find bases for the image and kernel of P, without using the matrix of P. Find the rank and nullity (d) Find Pul, Риг, and Риз in a snap. Find the matrix of P with respect to the basis (e) Let S : R3- R3 be a linear transformation whose matrix with respect to the basis {иъщ2, из } is | 0 1 0 | . Give a geometric description of the transformatrion of P U1, U2, U3f in a snap (f) Let T :R3R3 be a linear transformation whose matrix with respect to the basis {ũ,, иг, из} is | 1 1 | . Find the natrix of T with respect to the standard ー1 1 basis.

Answer 1

(a)

101 2 2

2 =121.1 01 = 1-1 = 0 241 .213

Hence is orthogonal set.

U1, u2, U3

(b)

To show that is linearly independent ,let

U1, u2, U3

CI (ul.ul) + c2(ul.U2) + C3(ul.U3) = 0 C10

c20

C30

$c_1=c_2=c_3=0\\ \Rightarrow$

is linearly independent.

U1, u2, U3

(c)

$P:\mathbb{R}^3\rightarrow \mathbb{R}^3\\ P(v)=\frac{(v_1,v_2,v_3).(1,2,-1)}{(1,2,-1).(1,2,-1)}(1,2,-1)+\frac{(v_1,v_2,v_3).(1,0,1)}{(1,0,1).(1,0,1)}(1,0,1)$

            

1+4+1 1+1

               $=\frac{v_1+2v_2-v_3}{6}(1,2,-1)+\frac{v_1+v_3}{2}(1,0,1)$

               

, 2 6 6 6

                $=(\frac{v_1+2v_2-v_3+3v_1+3v_3}{6},\frac{v_1+2v_2-v_3}{3},\frac{-v_1-2v_2+v_3+3v_1+3v_3}{6})$

                

Plul, U2,U3) = ( 6 6

$(v_1,v_2,v_3)\in Null(P)\\ \Leftrightarrow \\ (v_1,v_2,v_3)\in R^3\\ P(v_1,v_2,v_3)=0\\ (\frac{4v_1+2v_2+2v_3}{6},\frac{v_1+2v_2-v_3}{3},\frac{2v_1-2v_2+4v_3}{6})=0\\$

$4v_1+2v_2+2v_3=0\\v_1+2v_2-v_3=0\\2v_1-2v_2+4v_3=0$

$4v_1+2v_2+2v_3=0\\v_1+2v_2-v_3=0\\2v_1-2v_2+4v_3=0 \Rightarrow \\ v_2=v_3,v_1=-v_3\\ v=(-1,1,1)v_3\\ NullP=\{(-1,1,1)\}\\ Nullity(P)=1$

By Rank Nullity theorem ,

Plul, U2,U3) = ( 6 6

$P(u_1)=P(1,2,-1)=(\frac{4+4-2}{6},\frac{1+4+1}{3},\frac{2-4-4}{6})$

                                               $=(1,2,-1)$

$P(u_2)=P(1,-1,-1)=(\frac{4-2-2}{6},\frac{1-2+1}{3},\frac{2+2-4}{6})$

                                                 $=(0,0,0)$

$P(u_2)=P(1,0,1)=(\frac{4+2}{6},\frac{1+0-1}{3},\frac{2-0+4}{6})$

                                           $=(1,0,1)$