Question

Let T: R3 → R3 be the linear transformation that projects u onto v = (9, -1, 1). (a) Find the rank and nullity of T. rank nullity (b) Find a basis for the kernel of T.

Answer 1

TF1RDIR T(4)= proj u where v=(9,-1,1) let us C 21:34,3) T(U)= proju= Luv) <V,V? V = C2,4,3).(9,7,12 . (9,1,1). C9,-),1) v ga-Y+8 81+iti v = ga-y+3 (9,-1,1 4,1) 83 83 T(u) = ( 812-9y+9,-g24+y-2, 924-y+3) o i.e. T 81a-gy+g2 81 -gaty-3 1 -1 -ga.-Y+3 OL -g - - Y Y 83 Lg -g 83 g lo W - 81 og :. marrin of T is Cr=1 83 -g os

Ri → 83, Ri 1 "/g / 81 R2 » 83 R2 R₂ 83 R3 -g -1 g - 1 R2 » R2tgRi R3 R₃-9 RI -yg. Yoj 0 Ereduced roue echelon form of CT] 0 0 .. rank (T) = number of non-zero rows in redicced roue echelon form 1 Noritt of (t) = Number of columny of T rank(T) 3-1 =2 RankCT)=1 Nullity (t)=2 Basis of kernel of T= Basis of null space of CT] from reduced rou echelon form of CT). consider CDx=52 I 1975 H = he = auto 2) =) 8 - 11 2 2 + 1 2 350 f ks, ez are free variables sa= 1/222-15 23'; d., k3 are free

sel i', X= 1 se- 1 2 3 Vgre -Yg X3 = er dez + o 13 23 rez وا وا۔ - Hit Hz, az are o free ,, Basis of ker(t)= ,(ه را رو) 65,0,1)