Let W sp 1 6x 2x2 3 x 5 6x 4x2 5 5x 2x2 complement p

Let W = sp (1 + 6x + 2x^2, 3 + x, 5 + 6x + 4x^2, 5 + 5x + 2x^2) complement p_2 (z_n). Find a basis Beta for W. If dim W

Solution

(a) Let A =

1

3

5

5

6

1

6

5

2

0

4

2

To determine a basis for W, we will reduce A to its RREF as under:

Add -6 times the 1st row to the 2nd row; Add -2 times the 1st row to the 3rd row            

Multiply the 2nd row by -1/17; Add 6 times the 2nd row to the 3rd row

Multiply the 3rd row by 17/42; Add -24/17 times the 3rd row to the 2nd row

Add -5 times the 3rd row to the 1st row; Add -3 times the 2nd row to the 1st row

Then the RREF of A is

1

0

0

1/3

0

1

0

1/3

0

0

1

1/3

It is now apparent that 5+5x+2x² = (1/3)(1+6x+2x²) +(1/3)(3+x)+(1/3)(5+6x+4x²) and that the vectors 1+6x+2x², 3+x, 5+6x+4x² are linearly independent. Hence a basis for W is = { 1+6x+2x², 3+x, 5+6x+4x²}.

(b) We know that a standard basis for P₂(x) is {1,x,x²} and that P₂(Z₇) P₂(x). Now, let B =             

1

3

5

1

0

0

6

1

6

0

1

0

2

0

4

0

0

1

We will reduce B to its RREF as under:

Add -6 times the 1st row to the 2nd row; Add -2 times the 1st row to the 3rd row

Multiply the 2nd row by -1/17; Add 6 times the 2nd row to the 3rd row

Multiply the 3rd row by 17/42; Add -24/17 times the 3rd row to the 2nd row

Add -5 times the 3rd row to the 1st row; Add -3 times the 2nd row to the 1st row

Then the RREF of B is

1

0

0

-2/21

2/7

-13/42

0

1

0

2/7

1/7

-4/7

0

0

1

1/21

-1/7

17/42

Thus, it is apparent that 1,x,x² are all linear combinations of 1+6x+2x², 3+x, 5+6x+4x². Hence = { 1+6x+2x², 3+x, 5+6x+4x²} is a basis for P₂(Z₇)also.

1	3	5	5
6	1	6	5
2	0	4	2