Lets start with the basis w1 5 2 and w2 1 12 for R2 and le

Let\'s start with the basis w_1 = (5, 2) and w_2 = (1, 12) for R^2, and let w be the vector w = (14, 23). (a) Write was a linear combination of w_1 and w_2. The dot product rule doesn\'t work to find the coefficients here - why not? (b) Use the Grain-Schmidt process to find an orthogonal basis upsilon_1, upsilon_2 for R^2 with upsilon_1 = w_1. (c) Write was a linear combination of w_1 and w_2. (d) Write upsilon_1 and upsilon_2 as linear combinations of w_1 and w_2. (e) Use the answers in (c) and (d) to recomputed the answer for (a). Is this process easier than solving the equations?

Solution

a) Let A =

5

1

14

2

12

23

In order to check whether w is a linear combination of w₁ and w₂, 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 + 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

In order to check whether w is a linear combination of v₁ and v₂, we will reduce B to its RREF. The RREF of B is

1

0

4

0

1

3/2

Hence w = 4v₁ +(3/2) v₂.

(d) We have v₁= w₁= w₁+0w₂. Further let C =

5

1

5

-4

2

12

2

10

In order to check whether v1 and v2 are linear combinations of w1 and w2, we will reduce C to its RREF. 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