Invariant space Prove or disprove the following statement: the integration operator $T: P(F) →
P(F),$
$$T(a_nz^n + a_{n−1}z^{n−1} + \cdots + a_0) = a_n{z^{n+1}\over n + 1}+ a_{n−1}{z^n\over n}+ \cdots + a_0z$$
has an invariant one-dimensional space.
Solution:
Here are my scratch ideas: dimension of the polynomial is $n+1$ and $P(F)$ is invariant under T if $z^n$ is element of $P(F)$ implies $Tz^n$ is element of U.
Here is my approach: Polynomial of zero degree has 1 dimension. 
Counter example: take a polynomial from $P(F)$ such that $a_0+a_1z^1$, then integrate term so we have $a_0 z + {a_1\over 2} z^2$ but we lost the constant term after integration. Thus, T does not have an invariant one-dimension space.
Any hints and suggestions are welcome and appreciated ! 
Thanks in advance!
 A: Hints:


*

*If $U$ is an invariant subspace of $P(F)$ for the operator $T$ and $f \in U$, then $T(f) = bf$ for some scalar $b \in F$.

*Every polynomial $f = a_n z^n + \cdots + a_1 z + a_0$ has a leading term.

*For $f \in U$, what is the leading term of $T(f)$?



Suppose that $f = a_n z^n + \cdots + a_1 z + a_0$ generates a one-dimensional subspace $U \subset P(F)$.  In other words, $U = \{ bf \mid b \in F \}$.  (So $f \ne 0$; otherwise $\dim U = 0$).  Further assume that $\deg(f) = n$.  This means that $a_n \ne 0$ and that $a_k = 0$ for all $k > n$.
Now, the leading term of $T(f)$ is $\dfrac{a_n}{n+1} z^{n+1}$, and so $\deg \bigl( T(f) \bigr) = n+1$, showing that $T(f) \notin U$.
What have we shown?  $T(U) \nsubseteq U$ or that $U$ is not invariant.  Since $U$ was an arbitrary one-dimensional subspace of $P(F)$, this shows that a one-dimensional subspace is not possible.

A slight variation on this argument actually shows that any finite-dimensional subspace cannot be invariant under $T$.  (Consider the degree of $T(f)$, where $f$ is a polynomial of maximal degree in the proposed finite-dimensional subspace.)
