Task Find the Kernel of a Linear ransformation Note All entr

Task: Find the Kernel of a Linear ransformation Note: All entries must be integers or rational numbers in \"p/q\" form Given near transformation L: Pa(z) 3,1 described by L (1) 16 13 L(1 z) 10 L(1 2) 18 14 17 L(1 z z 10 What is the dimension of ker(L) dim(ker (L)) 1 dim (ker(L)) 1 CHECK Thats right. Continue Give a basis for ker(L) A decimal approximation to a basis is km (1.013 0.013a 0.461 z2 0.68Ar3)

Solution

We have L(1) = (-6,16,-11)^T

L(x) = L(1+x -1) = L(1+x)-L(1) = (-13,5,-1)^T-(-6,16,-11)^T = ( -7,-11,10)^T

L(x²) =[ (1+x+x²)-(1+x)] = L(1+x+x²)- L(1+x)= (-10,-18,14)^T –(-13,5,-1)^T = (3,-23,15)^T

L(x³) = L[(1+x+x²+ x³)-(1+x +x²)] = (-17,-10,8)^T- (-10,-18,14)^T = (-7,8,-6)^T.

Now, if A is the standard matrix of T, then the column=s of A are L(x), L(x²), L(x³) and L(x³). Hence A =

-6

-7

3

-7

16

-11

-23

8

-11

10

15

-6

Now, Ker(L) is the set of solutions to the equation AX = 0. To solve this equation, we will reduce A to its RREF as under:

Multiply the 1st row by -1/6

Add -16 times the 1st row to the 2nd row

Add 11 times the 1st row to the 3rd row

Multiply the 2nd row by -3/89

Add -137/6 times the 2nd row to the 3rd row

Multiply the 3rd row by -89/182

Add -45/89 times the 3rd row to the 2nd row

Add 1/2 times the 3rd row to the 1st row

Add -7/6 times the 2nd row to the 1st row

Then the RREF of A is

1

0

0

77/52

0

1

0

1/52

0

0

1

35/52

Then ,if X = (x,y,z,w)^T, the equation AX = 0 is equivalent to x+77w/52 = 0, y + w/52 = 0, z +35w/52= 0,so that X=(-77w/52,-w/52-35w/52,w)^T=w(-77/52,-1/52,-35/52,1)^T. Thus Ker(L) =                                                       span{(-77/52,-1/52,-35/52,1)^T } .Hence , dim(Ker(L)) = 1.

A basis for Ker (L) is {(-77/52,-1/52,-35/52,1)^T }.

k₁ = (-77/52)-(1/52)x–(35/52)x²+x³. A decimal approximation to a basis is k₁ = -(1.481+0.019x+0.673x² –x³) or, (1.481+0.019x+0.673x² –x³)

-6	-7	3	-7
16	-11	-23	8
-11	10	15	-6