Eigenvalues of a polynomial transformation

In summary, the linear space of real polynomials with degree less than n has a transformation T that results in q(t) = p(t + 1). The eigenvalue of T is 1 and the only eigenfunction is a constant.
  • #1
zwingtip
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Homework Statement


Let V be the linear space of all real polynomials p(x) of degree < n. If p [tex]\in[/tex] V, define q = T(p) to mean that q(t) = p(t + 1) for all real t. Prove that T has only the eigenvalue 1. What are the eigenfunctions belonging to this eigenvalue?


Homework Equations


Not sure there are any.


The Attempt at a Solution


I am completely lost on this. I know how to solve this for polynomials of a specific degree, but n is unspecified. Here's an example I worked for a polynomial of degree 2:

[tex]p(t)=a_{0}+a_{1}t+a_{2}t^2[/tex]

can be written as

[tex]\left[ \begin{array}{ccc} a_0 & 0 & 0 \\ 0 & a_1 & 0 \\ 0 & 0 & a_2 \end{array} \right]
\left[ \begin{array}{c} 1 \\ t \\ t^2 \end{array} \right] = \left[ \begin{array}{c} a_0 \\ a_{1}t \\ a_{2}t^2 \end{array} \right][/tex]

then
[tex]T(p)=a_{0}+a_{1}(t+1)+a_{2}(t+1)^2

\left[ \begin{array}{ccc} a_0 & 0 & 0 \\ a_1 & a_1 & 0 \\ a_2 & 2a_2 & a_2 \end{array} \right]
\left[ \begin{array}{c} 1 \\ t \\ t^2 \end{array} \right] = \left[ \begin{array}{c} a_0 \\ a_{1}(t+1) \\ a_{2}(t+1)^2 \end{array} \right][/tex]

But taking the characteristic polynomial of T(p) for p of degree 3, I'm getting distinct eigenvalues equal to the coefficients [tex]a_0, a_1, a_2[/tex] and not 1. Unless this has to do with the eigenfunction somehow. Any ideas?
 
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  • #2
so for a general polynomial you can write the requirement as
[tex]
q(t) = \sum_n a_n t^n = \sum_n b_n (t+1)^n = p(t+1)
[/tex]

now consider given powers of t

another hint is that any constant is clearly an eigenfunction - what is it eigenvalue?
 
  • #3
Thanks, that made sense. Question, though. I found that only a constant can be the eigenfunction. Is this true?
 
  • #4
yeah i agree, i haven't worked it thorugh, but i think it was heading that way

P(t+1) is basically translating the function by -1, so as no basis polynomials will overlie themselves after the translation, the constant is the only eigenfunction
 

What are eigenvalues of a polynomial transformation?

Eigenvalues of a polynomial transformation are the values that satisfy the characteristic equation of a polynomial matrix. They represent the scaling factor by which the corresponding eigenvectors are stretched or compressed under the transformation.

How are eigenvalues of a polynomial transformation calculated?

The eigenvalues of a polynomial transformation can be calculated by first finding the characteristic polynomial of the matrix. This is done by subtracting a scalar from the diagonal entries and finding the determinant of the resulting matrix. The roots of the characteristic polynomial are the eigenvalues of the transformation.

What is the significance of eigenvalues in polynomial transformations?

Eigenvalues play a crucial role in understanding the behavior of polynomial transformations. They determine the stability and type of critical points, the existence of periodic solutions, and the overall behavior of the transformation. They also provide insights into the geometric properties of the transformation.

Can a polynomial transformation have complex eigenvalues?

Yes, a polynomial transformation can have complex eigenvalues. This is because the eigenvalues are the roots of the characteristic polynomial, which can have complex coefficients. This often occurs when dealing with matrices with non-real entries or when studying complex systems.

How can eigenvalues be used to solve polynomial equations?

Eigenvalues can be used to solve polynomial equations by converting the polynomial equation into a matrix equation. The eigenvalues of this matrix can then be calculated, and the solutions to the polynomial equation can be found by solving for the corresponding eigenvectors. This approach is particularly useful in solving higher-order polynomial equations with complex roots.

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