Lets start with the basis omega1 5 2 and omega2 1 12 for R

Let\'s start with the basis omega_1 = (5, 2) and omega_2 = (1, 12) for R^2, and let omega be the vector omega = (14, 23). (a) Write to as a linear combination of omega_1 and omega_2. The dot product rule doesn\'t work to find the coefficients here - why not? (b) Use the Gram-Schmidt process to find an orthogonal basis upsilon_1, upsilon_2 for R^2 with upsilon_1 = omega_1. (c) Write omega as a linear combination of upsilon_1 and upsilon_2. (d) Write upsilon_1 and upsilon_2 as linear combinations of omega_1 and omega_2. (e) Use the answers in (c) and (d) to recompute the answer for (a). Is this process easier than solving the equations?

Solution

(a) Let A =

5

1

14

2

12

23

We will reduce A to its RREF as under:

multiply the 1st row by 1/5

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

multiply the 2nd row by 5/58

add -1/5 times the 2nd row to the 1st row

Then the RREF of A is

1

0

5/2

0

1

3/2

Thus, w = 5/2 w₁ +3/2 w₂

We can also use the dot product to determine the coeffients as under:

Let w = aw₁ +bw₂. Then w.w₁ = aw.w₁ +bw₂.w₁ or, (70+46) = (25+4)a + (5+24)b or, 29a+29b = 116 or, a+b= 4…(1). Also, w.w₂ = aw₁.w₂ +b bw₂.w₂ or, (14+276) = a(5+24)+b(1+144) or, 29a+145b = 290 or, a+5b = 5…(2). On solving these equations, we get a = 5/2 and b = 3/2.

(b) Let v₁ = w₁ = (5,2)^T and v₂ = w₂-proj_v1(w₂) = w₂-[(w₂.v₁)/(v₁.v₁)]v₁ = (1,12)^T –[ (5+24)/(25+4)](5,2)^T = (1,12)^T - (5,2)^T = (-4,10)^T.Then {v₁,v₂} is an orthogonal basis for R².

(c ) Let B =

5

-4

14

2

10

23

The RREF of B is

1

0

4

0

1

3/2

Hence w = 4v₁ +3/2 v₂.

(d) v₁= w₁ = w₁+0w₂. Further let C =

5

1

5

-4

2

12

2

10

The RREF of C is

1

0

1

-1

0

1

0

1

Then v₂ = -w₁ +w₂.

(e ) We have w = 4v₁ +3/2 v₂. Also, v₁= w₁, and v₂ = -w₁ +w₂.Hence w = 4w₁ +3/2(-w₁ +w₂) = (4-3/2)w₁ +3/2 w₂ = 5/2w₁ +3/2 w₂. The process of solving 2 linear equatoions in 2 unknowns is easier than this process.

5	1	14
2	12	23