Finding eigenvalues and eigenvectors for the polynomial equation

Question

I am struggling to find the eigenvectors and eigenvalues for $T: \mathcal{P}(\mathbb{R}) \rightarrow \mathcal{P}(\mathbb{R})$ defined by $Tp=p+p'$

I started by equating

$\lambda a_0+\dots+\lambda a_nx^n=(a_0+a_1)+(a_1+2a_2)x^2+\dots+(a_{n-1}+na_n)x^{n-1}+a_nx^n$

This gives the system of equations:

$\lambda a_0=a_0+a_1$

$\lambda a_1=a_1+2a_2$

$\cdot$ $\cdot$ $\cdot$

$\cdot$ $\cdot$ $\cdot$

$\cdot$ $\cdot$ $\cdot$

$\lambda a_{n-1}=a_{n-1}+na_n$

$\lambda a_n=a_n$

Would this last equation imply $\lambda=1$? How would I solve this system of equation?

Would the eigenvalue $\lambda=1$ ?

Answer 1

The key to this question is realizing that you know something about $p'$ : it always has degree one less than $p$, if $p$ is non-constant.

Therefore, if $p+p' = \lambda p$ then $(\lambda - 1)p = p'$. If $\lambda \neq 1$, then the only way for this equation to work is for $p = p' = 0$, because the degrees must be equal.

On the other hand, if $\lambda = 1$ then we have $p' = 0$, so $\lambda = 1$ works with any constant polynomial $p =a_0$.

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If you did not know this information but wanted to proceed purely from the equations, then indeed you look at the last equation.

Suppose $\lambda a_n = a_n$. There are two possibilities : $\lambda = 1$ and $\lambda \neq 1$.

If $\lambda \neq 1$ then $a_n = 0$. Putting this in the second last equation, $na_n = 0$ so $\lambda a_{n-1} = a_{n-1}$, again giving $a_{n-1} = 0$. We keep going from the end, till we get to $\lambda a_0 = a_0$, where now $a_0 = 0$ is also forced. Thus, $\lambda \neq 1$ is not an eigenvalue.

On the other hand, if $\lambda = 1$ then from the second last equation, $na_n = 0$ so $a_n = 0$. SImilarly from the third last equation, $(n-1)a_{n-1} = 0$ so $a_{n-1} = 0$. Proceeding like this till $\lambda a_0 = a_0 + a_1$ gives $a_1 = 0$ but $a_0$ can be anything.

Consequently, the answer is : the only eigenvalue is $+1$, and it has constant polynomials as eigenvectors.

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In fact, the matrix corresponding to $T$ is a single Jordan block with $1$s on the diagonal and $1$s on the upper diagonal. Therefore the same conclusion could have been deduced just from there, provided you know the JCF of course.

Answer 2

Yes, you are on the right track.

Suppose $a_n \neq 0$ ($\because$ nonzero eigenvector). Then $\lambda a_n = a_n$ implies $\lambda = 1$. If $n\geq 1$, then $\lambda a_{n-1} = a_{n-1} + n a_n$ yields $na_n = 0$. This is absurd because $n >0$ and $a_n \neq 0$. Thus $n =0$.

To sum up, the only eigenvalue of the linear operator $T$ is $1$ (with multiplicity $1$) and its eigenspace consists of constant functions.