Calculate coefficients of expansion for vector y

[nacreous]

Homework Statement

Let v⁽⁰⁾ = [0.5 0.5 0.5 0.5]^T, v⁽¹⁾ = [0.5 0.5 -0.5 -0.5]^T, v⁽²⁾ = [0.5 -0.5 0.5 -0.5]^T, and z = [-0.5 0.5 0.5 1.5]^T.

a) How many v⁽³⁾ can we find to make {v⁽⁰⁾, v⁽¹⁾, v⁽²⁾, v⁽³⁾} a fully orthogonal basis?

b) What are z's coefficients of expansion α_k in the basis found in part a)?

Homework Equations

See attempt at solution. I thought I had the answers, but according to my online test, they are wrong.

The Attempt at a Solution

a) Row reduction:
0.5a + 0.5b + 0.5c + 0.5d = 0 → a - d = 0
0.5a + 0.5b - 0.5c - 0.5d = 0 → b + d = 0
0.5a - 0.5b + 0.5c - 0.5d = 0 → c + d = 0 so ±a = ±d = ∓b = ∓c.

Then v⁽³⁾ must take the form [t -t -t t]^T. There are two if t = ±0.5, so the answer to a) is 2. (Marked wrong.)

b) I know the answer is asking me to find α₀, α₁, α₂, α₃ such that z = [-0.5 0.5 0.5 1.5]^T = v⁽⁰⁾α₀ + v⁽¹⁾α₁ + v⁽²⁾₂ + v⁽³⁾α₃. My notes talk about the change of basis in 2 dimensions but not 4 and I'm having trouble translating the concept to 4D...

I have [x₀ x₁]^T = α₀[1 0]^T + α₁[1 1]^T; α₀ = x₀ - x₁ and α₁ = x₁. So I assumed that I can do row reduction here as well:

z = v⁽⁰⁾α₀ + v⁽¹⁾α₁ + v⁽²⁾α₂ + v⁽³⁾α₃ using v⁽³⁾ = [0.5 -0.5 -0.5 0.5]^T from part a)

multiply the following {} by 0.5:
{α₀ +α₁ +α₂ + α₃ = -1
α₀ +α₁ - α₂ - α₃ = 1
α₀ - α₁ + α₂ - α₃ = 1
α₀ - α₁ - α₂ + α₃ = 3}

getting α₀ = 1, α₁ = -1, α₂ = -1, and α₃ = 0. (Marked wrong).

I'm so stuck on this answer that I don't know how to proceed correctly. Any help is appreciated.

 

[Orodruin]

a) Why should t be +-0.5?

b) Seems fine to me.

 

[RUber]

For part a, I agree with Orodruin, there seems to be no such restriction on what t should be, other than non-zero.
For part b, your development with the vectors you used was right. I suspect if you get the correct form for part a, you will get the correct solution for b. Try solving for the last coefficient as a function of t, where t represents your constant coefficient on the vector [1,-1,-1,1].
**edit, it would still be zero -- never mind**

 

[nacreous]

a) Why should t be +-0.5?

Good point. I thought that t would have to have a value of 0.5 if it was to be fully orthogonal with the other three v-vectors. I suppose then that there would be more than two solutions (my options were 0, 1, 2, >2). Which makes solving for the variables in b easier as I have more vectors to choose from to plug into the problem.

Thanks! At least I have the technique down so I'm satisfied. Now to earn the exam points...

 

[nacreous]

For part a, I agree with Orodruin, there seems to be no such restriction on what t should be, other than non-zero.
For part b, your development with the vectors you used was right. I suspect if you get the correct form for part a, you will get the correct solution for b. Try solving for the last coefficient as a function of t, where t represents your constant coefficient on the vector [1,-1,-1,1].
**edit, it would still be zero -- never mind**

a) I thought that t would have to have a value of 0.5 if it was to be fully orthogonal with the other three v-vectors. But I suppose that defeats the purpose of being able to transform the vectors in the first place. The options I had for this problem were 0, 1, 2, and >2, so there would be more than two solutions for v⁽³⁾.

b) With part a) in mind, I guess there would be more than just one variable instead of just one t. At least I used the correct method for this problem. If t is nonzero, though, I don't understand why t = 0.5 wouldn't work. As for your solving for an f(t) instead of a constant t, I guess over the entire signal it would sum to zero. Don't quote me on that though!

 

[Orodruin]

I thought that t would have to have a value of 0.5 if it was to be fully orthogonal with the other three v-vectors.

Orthogonality does not depend on the vector normalisation.

If t is nonzero, though, I don't understand why t = 0.5 wouldn't work.

$t$ has to be non-zero. Otherwise it is linearly dependent on the other vectors and does not form a complete basis. $t = 0.5$ works perfectly well.