Calculate coefficients of expansion for vector y

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1. Feb 23, 2017

nacreous

1. The problem statement, all variables and given/known data

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 αk in the basis found in part a)?

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

3. 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:
{α012 + α3 = -1
α01 - 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.

Last edited: Feb 23, 2017
2. Feb 23, 2017

Orodruin

Staff Emeritus
a) Why should t be +-0.5?

b) Seems fine to me.

3. Feb 23, 2017

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**

4. Feb 23, 2017

nacreous

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...

Last edited: Feb 23, 2017
5. Feb 23, 2017

nacreous

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!

6. Feb 23, 2017

Orodruin

Staff Emeritus
Orthogonality does not depend on the vector normalisation.

$t$ 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.