Question

Exercise 1. In the inner product space R3 (with the usual dot product), consider the subspace a S { ER3 : a+b+c=0}. (i) Find an orthonormal basis {ū1, ū2} for S (ii) Consider the function f : R3 → R3 that to each vector ū assigns the vector of S given by f(ū) = (ū1, ū)ū1 + (ū2, ū)ū2. Show that it is a linear function. (iii) What is the matrix of f in the standard basis of R3? (iv) What are the null space and the column space of the matrix that you computed in the previous point?

(i) Find an orthonormal basis {~u1, ~u2} for S
(ii) Consider the function f : R3 -> R3 that to each vector ~v assigns the vector of S given by
f(~v) = <~u1, ~v>~u1 + <~u2; ~v>~u2. Show that it is a linear function.
(iii) What is the matrix of f in the standard basis of R3?
(iv) What are the null space and the column space of the matrix that you computed in the
previous point?

Answer 1

(1) Let a subspace of R3 Suppose tes is a D Given that S = {(a, b, c EAR? la btC=of is a=(1,0,1) be a vector in s. Then vertor orthogonal to if a b=0 and 5=(bi, bz, b.-bz) (1,0, aD (61,62,-b762) = bitbit b=0 +2=-2bi. a choice of b es (1, -2, 1), Hence a 5=0. Now, lal=5+1=52 1561=51+4+1=56 Hence а b vectors with norm I and gives J2 orthogonal and ūz= (to 7 / 7 Jo tence het u,= (tz Thin ū, and to form a ban bor s. Hence and and 17 / orthonormal

ii) Define 6. R²R by f(G) = (h, uu, + <ūs, ſūza, VER? Claim f is lineon function 2=2 and het V, U2ER² L, BER. and B =B, < 5 < 5) Blocüi + BūZ) = sū dvi + Biz Yü, + sữa ofūst Biz]ūz =&am qui) + Kü Buz ür, + Kūr, & 5, 7+ <hr Bre) in = Kno, vi) un + B <ü, uz) un ta<üz Vizur + B rus uz zu & (Kü, uyu + <ür Uz) ur) + B Kuzhi) unt <ūzł z in) ER $10) + BLOG) Thunt is lineon function I= (2,4, 2) ER? (ii) for u= (t/ we have 2-7 ,0) 2 6 f 1,0, and r = (to - (to to, bolxy, z) = (-3) (W207a)+(** **)(*-* to) - **2)+ ( x=24+2 ; 2 (x-2y2) , x-2y+3) 4x +2y-22, -214-2y+z), -3x - 2y +42 ) (2x=y=2, 2+2y=2 2-x_y) €19,9; 2) = 2x+y=2(1,-1,-) 6 Zn-y > 6 (0,9, 2)

The matrix of the function ban be calenlated as follows. $(e) = $(1,010) = () - }e + e + 27 + 27 ez f(02) = f (0o,0) =( 4 7 - 3 ž) = le+ te the 6103) = f (0,0,1)= (- į I se,+ e + 3 ez - Hence, the matrix is of b 3 - M[8] WINWINWIN ابعا 4) Nall space of M[6] =? and column space of M6b] = ? Column Space of a matrx š the vesta space made up of all linen Combination of the Column vectors of A. Now, şX (22,- 7 -7/3) + Blot de solumn space of MC6] +86 13 3 / 3 بي =2&+B+0) Thin i {«C-1,1,1) E TRILER] the column space of M [6] = ) Null space of a r {XER / Ax=0} Null space of MEB] = { NER I M[b]u=0}, - - 2 x - 1 34-42=0. 2x -y - Z=0 → 2X-=2. Null space of M[8] = {(1,9, ZDER? / 22-y=z}= {(219, 22-y) |2,90 R$ M13 Z -2/3