Finding Eigenvalues of Integration Operator

Question

I'm solving a question from an Algebra Textbook:

Given a Linear Operator: $$ T: V \rightarrow V \space, \space where \space V=R[t] \space, \space R[t]\space is \space the \space pol. \space space\space with \space real \space coeff. $$ Defined as: $$ (Tf)(x) = \int_0^x f(t)dt\ $$

Find the eigenvalues and the eigenvectors of this operator.

What I already tried: Using the standard base of: $$ B={e _{1}, e _{2}, ..., e _{n}} $$ I constructed the matrix representation of the operator: $$ T(e _{1})=\int_0^x 1dt=x=e _{2} $$ $$ T(e _{2})=\int_0^x xdt=x^2/2=e _{2}/2 $$ And so on, getting: \begin{bmatrix} 0 & 0 & 0 & \dots & 0 \\ 1 & 0 & 0 & \dots & 0 \\ 0 & 1/2 & 0 & \dots & 0 \\ 0 & 0 & 1/3 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & 0 & \dots & 1/n \\ \end{bmatrix}

Solving for the eigenvalues I get: $$ \Lambda_{i}=0, \space for \space i=1:n $$ So the operator has n eigenvalues, all equal to zero?
Or does it mean that the operator doesn't have eigenvalues?

Answer 1

I suppose that $V$ is the set of all real polynomials. Let $\mu \in \mathbb R$ and let $f \in V \setminus \{0\}$ such that $Tf= \mu f$. Then we have

$\int_0^x f(t)dt\ = \mu f(x)$ for all $x$.

Taking derivatives we get

(*) $f(x)= \mu f'(x)$.

If $f$ has degree $n \ge 1$, then $f'$ has degree $n-1$. Thus (*) leads to a contradiction.

Conclusion: $T$ has no eigenvalues.

Answer 2

For $p(x) \in V = R[x]$, we may write

$p(x) = \sum_0^n p_i x^i, \tag 1$

where

$p_i \in R, \; 0 \le i \le n; \tag 2$

we then have

$Tp(x) = \displaystyle \int_0^x p(t) dt = \int_0^x \sum_0^n p_i t^i \; dt = \sum_0^n p_i \int_0^x t^i \; dt = \sum_0^n \dfrac{p_i}{i + 1} x^{i + 1}; \tag 3$

$p(x)$ is an eigevector with eigevalue $\lambda_{p(x)}$ if

$\displaystyle \sum_0^n \dfrac{p_i}{i + 1} x^{i + 1} = Tp(x) = \lambda_{p(x)}p(x) = \lambda_{p(x)} \sum_0^n p_i x^i = \sum_0^n \lambda_{p(x)} p_i x^i; \tag 4$

comparing coefficients on the left and right hand sides we find

$\dfrac{p_n}{n + 1} = \lambda_{p(x)} \cdot 0 = 0, \tag 5$

and

$\dfrac{p_i}{i + 1} = \lambda_{p(x)} p_{i + 1}, \; 0 \le i \le n - 1; \tag 6$

(5) implies that

$p_n = 0, \tag 7$

and using this in (6) and iterating we see that $p_n = 0$ forces

$p_i = 0, \; 0 \le i \le n -1, \tag 8$

or

$p(x) = 0; \tag 9$

but the definiton of eigenvector precludes $p(x) = 0$, so $T$ has no eigenvalues whatsoever.

Interestingly enough, this result generalizes to the case of $R$ any field of characteristic $0$, and perhaps even further, where we define $\int x^i dx = x^{p + 1}/p + 1$.