# 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.