Are Trigonometric Polynomials a Basis for a Complex Vector Space?

[albega]

I'm having trouble with a couple of things written in some notes I'm reading.

Firstly, in stating examples of vector spaces, they say
Trigonometric polynomials - Given n distinct (mod 2π) complex constants λ₁,...,λ_n, the set of all linear combinations of e^iλ_nz forms an n-dimensional complex vector space.

Now does this even make sense? I feel as though it is trying to say that we can have e^iλ_nz for integer λ_n as the basis for the vector space of functions, i.e like Fourier series, but I'm not sure really.

Secondly, for the complex vector space defined by the vectors in the 2D plane
|r,φ>=(rcosφ,rsinφ)
where vector addition is as usual and scalar multiplication follows the rule
α|r,φ>=||α|r,φ+argα>
I'm having trouble proving that
α(|a>+|b>)=α|a>+α|b>
and
(α+β)|a>=α|a>+β|a>

For the second one, I do
(α+β)|a>=(α+β)|r,φ>
=||(α+β)|r,φ+arg(α+β)>
and I have no idea what to do next, as expanding the modulus/argument doesn't seem possible. Not sure about the second either.

 

[Mark44]

I'm having trouble with a couple of things written in some notes I'm reading.

Firstly, in stating examples of vector spaces, they say
Trigonometric polynomials - Given n distinct (mod 2π) complex constants λ₁,...,λ_n, the set of all linear combinations of e^iλ_nz forms an n-dimensional complex vector space.

Now does this even make sense? I feel as though it is trying to say that we can have e^iλ_nz for integer λ_n as the basis for the vector space of functions, i.e like Fourier series, but I'm not sure really.

The λ_i's are complex, not integer.

Secondly, for the complex vector space defined by the vectors in the 2D plane
|r,φ>=(rcosφ,rsinφ)
where vector addition is as usual and scalar multiplication follows the rule
α|r,φ>=||α|r,φ+argα>
I'm having trouble proving that
α(|a>+|b>)=α|a>+α|b>
and
(α+β)|a>=α|a>+β|a>

Show us where you've gotten on these. Start with the left side and expand it using the rule above for scalar multiplication.

For the second one, I do
(α+β)|a>=(α+β)|r,φ>
=||(α+β)|r,φ+arg(α+β)>
and I have no idea what to do next, as expanding the modulus/argument doesn't seem possible. Not sure about the second either.

 

[RUber]

For thinking of them as a basis set, first consider what it would take to normalize them.
Then, since the lambdas are distinct, mod 2pi, it can be shown that they are also orthogonal.
This is similar in concept to the Fourier basis, but with complex lambdas.
For the last part, you should be able to put it back into complex exponential form and demonstrate the properties you are looking for.

 

[albega]

The λ_i's are complex, not integer.
Show us where you've gotten on these. Start with the left side and expand it using the rule above for scalar multiplication.

Yeah, I know that's what it says, but I just don't understand how that is the case... A quick google of trigonmetric polynomials shows them to be the functions
sin(mx), cos(mx) or exp(imx) for integer m...

I've started off one of them at the bottom, but can't progress...

 

[albega]

For thinking of them as a basis set, first consider what it would take to normalize them.
Then, since the lambdas are distinct, mod 2pi, it can be shown that they are also orthogonal.
This is similar in concept to the Fourier basis, but with complex lambdas.
For the last part, you should be able to put it back into complex exponential form and demonstrate the properties you are looking for.

Hmm I think I'm having trouble understanding what it means by mod(2pi), especially with the lambdas being complex. Could you explain please?

 

[Mark44]

Hmm I think I'm having trouble understanding what it means by mod(2pi), especially with the lambdas being complex. Could you explain please?

$\pi/6$ and $13\pi/6$ are in the same congruence class, modulo $2\pi$

 

[albega]

$\pi/6$ and $13\pi/6$ are in the same congruence class, modulo $2\pi$

Ok, so what happens if we have (13π/6)+2i, as we are allowed complex numbers? Is this (π/6)+2i?

 

[Mark44]

Secondly, for the complex vector space defined by the vectors in the 2D plane
|r,φ>=(rcosφ,rsinφ)
where vector addition is as usual and scalar multiplication follows the rule
α|r,φ>=||α|r,φ+argα>
I'm having trouble proving that
α(|a>+|b>)=α|a>+α|b>

It might be easier to start with the right side, α|a>+α|b>.
You could write the vectors a and b as (r₁cosφ₁) and (r₂cosφ₂). According to the scalar multiplication rule above, what is αa? What is αb?

 

[Mark44]

Ok, so what happens if we have (13π/6)+2i, as we are allowed complex numbers? Is this (π/6)+2i?

I believe so.

 

[RUber]

Notice that the imaginary part of $\lambda$ will turn into your real part in the exponential, so the mod 2pi only needs to apply to the real part of $\lambda$ to prevent duplication in the imaginary part, since sin(x)=sin(x+2pi).

 

[albega]

Notice that the imaginary part of $\lambda$ will turn into your real part in the exponential, so the mod 2pi only needs to apply to the real part of $\lambda$ to prevent duplication in the imaginary part, since sin(x)=sin(x+2pi).

Ok, so are these actually trigonometric polynomials - wikipedia seems to claim they are just sin(mx), cos(mx), exp(imx) for integer m.

I'm confused about the whole mod(2pi) thing - why don't we just say we have all these lambdas and they must be between zero and 2pi?

So the vector space produced by linear combinations of these will be any function?

 

[albega]

It might be easier to start with the right side, α|a>+α|b>.
You could write the vectors a and b as (r₁cosφ₁) and (r₂cosφ₂). According to the scalar multiplication rule above, what is αa? What is αb?

αa=[(|α|r1)cos(φ1+argα),(|α|r1)sin(φ1+argα)],
αb=[(|α|r2)cos(φ2+argα),(|α|r2)sin(φ2+argα)]
I did try going down this route and things got very messy...

Anyway,
αa+αb=[(|α|r1)cos(φ1+argα)+(|α|r2)cos(φ2+argα),(|α|r1)sin(φ1+argα)+(|α|r2)sin(φ2+argα)]

 

[Mark44]

I'm having some trouble with your notation. Apparently, you're using this notation for a vector: |r, φ>.

|r,φ>=(rcosφ,rsinφ)
where vector addition is as usual and scalar multiplication follows the rule
α|r,φ>=||α|r,φ+argα>
I'm having trouble proving that
α(|a>+|b>)=α|a>+α|b>

Here, you're writing vectors as |a>. Where is the angle?

 

[albega]

I'm having some trouble with your notation. Apparently, you're using this notation for a vector: |r, φ>.

Here, you're writing vectors as |a>. Where is the angle?

Well |a> is just the vector, but it has an associated modulus and argument so I can write |a>=|r,φ>.

 

[Mark44]

αa=[(|α|r1)cos(φ1+argα),(|α|r1)sin(φ1+argα)],
αb=[(|α|r2)cos(φ2+argα),(|α|r2)sin(φ2+argα)]
I did try going down this route and things got very messy...

Anyway,
αa+αb=[(|α|r1)cos(φ1+argα)+(|α|r2)cos(φ2+argα),(|α|r1)sin(φ1+argα)+(|α|r2)sin(φ2+argα)]

This stuff would be easier to read if you put in some spaces. The expressions you have written don't have a single space anywhere.

The natural thing to do would be to use the sum formulas for sine and cosine, and see if you can get some simplification from doing that.

 

[RUber]

I'm confused about the whole mod(2pi) thing - why don't we just say we have all these lambdas and they must be between zero and 2pi?

So the vector space produced by linear combinations of these will be any function?

Since pi is irrational the the set of integers times x is equivalent to the set of distinct points in the range 0 to 2pi.
In any case, whether you use mx or lambda x, you end up with the same vector space.