Eigenvalue of Polynomial Transformation

[PandaGunship]

Homework Statement

Let T:P₂→P₂ be defined by
T(a₀+a₁x+a₂x²)=(2a₀-a₁+3a₂)+(4a₀-5a₁)x + (a₁+2a₂)x²
1) Find the eigenvalues of T
2) Find the bases for the eigenspaces of T.

I believe the 'a' values are constants.

Homework Equations

None.

The Attempt at a Solution

The problem I am having is actually pulling out the matrix for T. I know how to find eigenvalues (by solving det(λI-A) - A being the matrix) and from that finding the bases of the eigenspaces comes from substituting the eigenvalues into (λI-A) and performing elementary row operations to find the eigenvectors which form the bases.

What I have tried to do is separate the basis (1,x,x²) from the rest and come up with the matrix :
2a₀ -a₁ 3a₂
4a₀ -5a₁ 0a₂
0a₀ a₁ 2a₂
Am I on the right track here or am I barking up the wrong tree? If so what method should i follow?

 

[Robert1986]

Well, sort of. You need to figure out which elements the polynomials 1,x,x^2 get sent to by T. Once you do that, put them into a matrix and proceed as you have described. So, if you put the polynomial p_1(x) = x into T what would come out "the other side"?

 

[tiny-tim]

Hi PandaGunship! Welcome to PF!

2a₀ -a₁ 3a₂
4a₀ -5a₁ 0a₂
0a₀ a₁ 2a₂

leave all the a's out!

 

[PandaGunship]

Well, sort of. You need to figure out which elements the polynomials 1,x,x^2 get sent to by T. Once you do that, put them into a matrix and proceed as you have described. So, if you put the polynomial p_1(x) = x into T what would come out "the other side"?

Just to check, my matrix is correct?
As per your direction, a₀ goes to 2a₀ + 4a₀x +0x² which I make into a column like the matrix I put it already?

Hi PandaGunship! Welcome to PF!

leave all the a's out!

Thanks Tim :) I thought of leaving the a's out, but what of their importance? Are they just to show what each constant of the basis is mapped into?

 

[tiny-tim]

Thanks Tim :) I thought of leaving the a's out, but what of their importance? Are they just to show what each constant of the basis is mapped into?

No, your vectors will have a's.

(The 1 x x² are just a basis like x y z …

you wouldn't expect x y z to appear in a vector or matrix, would you? )​

 

[Robert1986]

Well, I said to find out where p_1(x) = x goes, and what you did was find where p(x) = a_0 goes. I only mentioned to do it as an example, so its OK that you did it for p(x) = a_0.

So, the a_0, a_1, a_2 are there because you need to know what T does to a general quadratic polynomial, and that is the form of a general quadratic polynomial.

Now, the three basis polynomials we are using are: p_0(x) = 1, p_1(x) = x, p_2(x) = x^2. Now, for each of these basis polynomials, what is a_0, a_1, a_2? As an example, for p_0 we have that a_0 = 1, a_1 = 0, a_2 = 0. So, T(p_0(x)) = 2a_0 + 4a_0x = a + 4x. So, you got it correct, but as Tiny Tim said, you need to get rid of the a's because for the basis polynomials they are all 1 or 0.

 

[PandaGunship]

No, your vectors will have a's.

(The 1 x x² are just a basis like x y z …

you wouldn't expect x y z to appear in a vector or matrix, would you? )​

I suppose not :P Thanks :)

Well, I said to find out where p_1(x) = x goes, and what you did was find where p(x) = a_0 goes. I only mentioned to do it as an example, so its OK that you did it for p(x) = a_0.

So, the a_0, a_1, a_2 are there because you need to know what T does to a general quadratic polynomial, and that is the form of a general quadratic polynomial.

Now, the three basis polynomials we are using are: p_0(x) = 1, p_1(x) = x, p_2(x) = x^2. Now, for each of these basis polynomials, what is a_0, a_1, a_2? As an example, for p_0 we have that a_0 = 1, a_1 = 0, a_2 = 0. So, T(p_0(x)) = 2a_0 + 4a_0x = a + 4x. So, you got it correct, but as Tiny Tim said, you need to get rid of the a's because for the basis polynomials they are all 1 or 0.

Ok I understand properly now. In the beginning when I was going over this problem I thought that I must remove the a's but I didnt want to remove them 'just because'. I wanted to know that what I was doing was actually correct, even if my reasoning was a bit off the mark.
Thank you both for your help