Question

Suppose T: M22-R3 is a linear transformation whose action on a basis for M2.2 is as follows: 6 1 -3 -3 0 1 1 1 T T T T -3 -3 0 1 1 2 1 Give a basis for the kernel of T and the image of T by choosing which of the original vector spaces each is a subset of, and then giving a set of appropriate vectors. Basis of Kernel is a Subset of M2,2 Number of Matrices: 1 Bker Coo) Basis of Image is a Subset of M2,2 Number of Matrices: 1 Bim (oo

Answer 1

SA!? we have 1 1-3 -3 -3 -3 6 -6 6 T 0 JA T-6 6 -6 2 Now, we hare. -3 - 3 -3 + 0 3 2 0 0 1 -3 -3 -3 - 3 a 3 = 0 0 3 o 3 6 T O 12 0 3 -12 -12 12 12 -4 14+ [: :])- (I (cuaC1w) [ ] ti ] - [ ] [ ] 4 Nent

2 [•]-[8:1=[: :)-[::] [:] +2 + - 0 vent! 5 -S - I - 5 Lo IT: O + O

1 72T - [i: ] ==||-]"[• ] + +[:] 13:] 4:10 -2 T 47 Ч 4 2 -2 Now , we have т + b T Lag] = a TFó ;]* to! :]+Ctto] dil O o 0 -

IS 1-4) a b al + b -5 t.d 5 -4 -4 a + 5b-4c-4d -a - 5 b +40 +4d a +5b 4c 4d Kernel of I 0 a b I cd o 0 a + 5b-4c- yd -a - 55 +40 +40 a + 5b-4C-44 a56-40 - 4450 1: we get a a = -56 + 40 +40

-5b +40 +48 b d a a b -5 +4 0 :7+1 abl :7+ cd O - 0 1 Basts of keinee شی a subset of M2x2 Numbu of matrice : 3 -5 +4 - 4 0 kee st - O -

for image of I contider 5 -4 05 -4 -1 -4 4 4 o 4 -4 0 D --- 5 O 0 As can be seen, only columm I has a fivot element. There in a subsit if R. of Image i of matrice : Number Bim-