# Linear Algebra and polynomials

1. Nov 21, 2005

Let U and V denote, respectively, the spaces of even and odd polynomials in Pn. Show that dimU + dimV = n+1 [Hint: Consider T: Pn ---> Pn where T[p(x) - P(-x) ]

So where to begin?

I thought that i should let p(x) = a + a0x + a2x^2 + ... + anX^n

So if U is the space of even polynomials, and let n be even, then U is the set of polynomials p(x) = a + a0x + a2x^2 + ... + anX^n then dimU = n

Then im lost here, however, I thought id use the transformation... T: Pn ---> Pn where T[p(x) - P(-x) ] (by the way, is that transformation T: U ---> V well if it is.. i assumed it as and continuted)

Applying that transformation, you get

a + a0x + a2x^2 + ... + anX^n - [ a + a0(-x) + a2(-x)^2 + ... + an(-X)^n] = 2a0x + ... + 2anX^n

Can somebody help me?

Thanks

2. Nov 22, 2005

### matt grime

A polynomial is even if p(x)=p(-x), it is odd if p(x)=-p(-x). What does T defined above do to the space of even and odd polys as a linear map? What about the rank nullity theorem or other such things?

Personally, I don't see why you should do that method of attack since I can easily think of enough linearly independent even and odd polynomials (vectors) That make it clear that dim(U)+dim(V)=>n+1, and we already know that dim(U)+dim(V)-dim(UnV)<=n+1 and obivously the intersection is trivial.

3. Nov 22, 2005

### HallsofIvy

Staff Emeritus
No, you can't do that. n is "given" and can be any positive integer.
No, that's not true either. U is the set of polynomials with even powers only. $$p(x)= a_0+ a_2x^2+ ....$$. If n is even, U has dimension n/2. If n is odd, U has dimension (n-1)/2.
First what is the dimension of Pn? Given any polynomial in Pn, can you write it as a sum of a polynomial in U and a polynomial in V (That's where the hint comes in)? What does that tell you?

Last edited: Nov 22, 2005
4. Nov 22, 2005

the dimension of Pn is just n+1

What do you mean write it as a sum of a polynomial in U and a polynomial in V?

So if i take the transformation of even polynomials...T: Pn ---> Pn where T[p(x) - P(-x) ] i get:

a0 + a2x^2 +... anX^n - [a0 + a2(-x)^2 + ... an(-x)^n]
= 0?

and if do the same with odd powered polynomials i get

ax + a3x^2 + .. + anx^n - [ -ax - a3x^2 - .. - anx^n]
= 2ax + 2a3x^2 + .. +2anx^n

5. Nov 23, 2005

### matt grime

Checking back, your defintion of T makes no sense (I read it as T(p(x)) =p(x)-p(-x), perhaps I ought to have read it as T(p(x) = -p(-x)))

How come both odd and eve polys have potentially x^n in them? What's going on with your notation?

6. Nov 23, 2005

### HallsofIvy

Staff Emeritus
Matt, I see no problem with his definition of T (although it would make more sense for this problem if it were T(p)= (1/2)p(x)- (1/2)p(-x)). T(p) is twice the "odd part" of p.

rad0786: What do you get if you apply T (or, better (1/2)T) to a general nth degree polynomial? What is p- (1/2)T(p) for polynomial p?

If you are not sure, try it on some simple polynomial:

What is (1/2)T(p) if p= 3x3+ 5x2- 4x+ 7?

What is p- (1/2)T(p)?

How would you write p(x) as a member of U plus a member of V?

Last edited: Nov 23, 2005
7. Nov 23, 2005

### matt grime

You see no problem with:

T:Pn --> Pn T[p(x)-p(-x)]?

is this some kind of new notation I'm not aware of? Perhaps you like me autocorrected this to something we were expecting to see based on the question? Ie T(p(x)) = p(x)-p(-x), from which we can apply the rank nullity theorem and get the answer.

8. Nov 23, 2005

### HallsofIvy

Staff Emeritus
Oh, I see your point. Yes, you were right, the notation was wrong. My mind automatically interpreted what it should be!