The vectors u1 12 1 1 1 1 u2 12 1 1 1 1 u3 12 1 1 1 1 u4

The vectors u_1 = 1/2 [1 1 1 1], u_2 = 1/2 [1 1 -1 -1], u_3 = 1/2 [1 -1 1 -1], u_4 = 1/2 [1 -1 -1 1] form an ordered, orthogonal basis B = {u_1, u_2, u_3, u_4} of R^4. a: Compute the length (norm) of the vector u_3. b: Find the B-coordinates [v]_B = [a_1 a_2 a_3 a_4] of the vector v = [2 4 6 8] (that is, find the coefficients the linear combination v = a_1 u_1 + a_2 u_2 + a_3 u_3 + a_4 u_4).

Solution

4(a). The norm of the vector u₂ =||u₂||= [ (1/2)²+(1/2)²+(-1/2)²+(-1/2)²] =(1/4 +1/4+1/4+1/4) = 1 = 1

(b) Let A = [u₁,u₂,u₃,v] . To determine [v]_B, we will reduce A to its RREF as under:

Multiply the 1st row by 2; Add -1/2 times the 1st row to the 2nd row

Add -1/2 times the 1st row to the 3rd row; Add -1/2 times the 1st row to the 4th row

Interchange the 2nd row and the 3rd row; Multiply the 2nd row by -1

Add 1 times the 2nd row to the 4th row; Multiply the 3rd row by -1

Add 1 times the 3rd row to the 4th row; Multiply the 4th row by ½

Add -1 times the 4th row to the 3rd row                ; Add -1 times the 4th row to the 2nd row

Add -1 times the 4th row to the 1st row; Add -1 times the 3rd row to the 1st row

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

Then the RREF of A is

1

0

0

0

10

0

1

0

0

-4

0

0

1

0

-2

0

0

0

1

0

Now, it is apparent that v = 10u₁-4u₂-2u₃+0u₄ . Thus, [v]_B= (10,-4,-2,0)^T.

1	0	0	0	10
0	1	0	0	-4
0	0	1	0	-2
0	0	0	1	0