Suppose $V$ is finite-dimensional with dim $V>1$ and $T\in \mathscr L (V)$. Prove that $\{p(T):p\in \mathscr P(\Bbb F)\}\neq \mathscr L(V)$.
I believe that I can prove this if $V$ is a complex vector space, but I need assistance if $V$ is a real vector space.
My proof: Suppose $V$ is a finite-dimensional complex vector space. Then $T$ has an upper-triangular matrix with respect to some basis of $V$: $v_1,\ldots, v_n$. Since dim $V>1$, then $n>1$. Thus, span $(v_1,\ldots,v_j)$ is invariant under $p(T)$ for $j=1,\ldots,n$. Consider some $S\in \mathscr L(V)$ such that $S(v_1)=v_2$. Note that $v_1\in$ span$(v_1)$, but $v_2 \notin$ span $(v_2)$ since the basis elements are linearly independent. Then $S(v_2)\notin$ span$(v_2)$, this implies span$(v_1)$ is not invariant under S. Thus, S is not in $\{p(T):p\in \mathscr P(\Bbb F)\}$
If the vector space is over the real numbers and $T$ has and eigenvalue, then the argument above holds.
So I would like some clarification on the part I bolded, as well as some insight for if $V$ is a vector space over the reals when $T$ does not have an eigenvalue.
Note that polynomials in $T$ commute with $T$, so if every endomorphism of $V$ was a polynomial on a single $T$, $\text{End}(V)$ would be commutative as an algebra. However, $\text{End}(V)$ is non-commutative as soon as $\dim V >1$, for it is (isomorphic to) a matrix ring. To expand, if $V$ has dimension $n>1$ over a field $\mathbb k$, then $\text{End}(V)$ is isomorphic to $M(n,\mathbb k)$, and this contains as a subalgebra $M(2,\mathbb k)$. This subalgebra is not commutative, since the following matrices do not commute: $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix},B=\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}.$$