I know of a theorem from Axler's Linear Algebra Done Right which says that if $T$ is a linear operator on a complex finite dimensional vector space $V$, then there exists a basis $B$ for $V$ such that the matrix of $T$ with respect to the basis $B$ is upper triangular.
The proof of this theorem is by induction on the dimension of $V$. For dim $V = 1$ the result clearly holds, so suppose that the result holds for vector spaces of dimension less than $V$. Let $\lambda$ be an eigenvalue of $T$, which we know exists for $V$ is a complex vector space.
Consider $U = $ Im $(T - \lambda I)$. It is not hard to show that Im $(T - \lambda I)$ is an invariant subspace under $T$ of dimension less than $V$.
So by the induction hypothesis, $T|_U$ is an upper triangular matrix. So let $u_1 \ldots u_n$ be a basis for $U$. Extending this to a basis $u_1 \ldots u_n, v_1 \ldots v_m$ of $V$, the proof is complete by noting that for each $k$ such that $1 \leq k \leq m$, $T(v_k) \in $ span $\{u_1 \ldots u_n, v_1 \ldots v_k\}$.
The proof of this theorem seems to be only using the hypothesis that $T$ has at least one eigenvalue (in a complex vector space). So I turned to the following example of a linear operator $T$ on a real vector space $(\mathbb{R}^3)$ instead that has one real eigenvalue:
$T(x,y,z) = (x, -z ,y)$.
In the standard basis of $\mathbb{R}^3$ the matrix of $T$ is
$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array}\right]$
The only real eigenvalue of this matrix is $1$ with corresponding eigenvector
$\left[\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right]$.
This is where I run into trouble: If I just extend this to any basis of $\mathbb{R}^3$, then using the standard basis again will not put the matrix of $T$ in upper triangular form. Why can't the method used in the proof above be used to put the matrix of $T$ in upper triangular form?
I know this is not possible for if $T$ were to be put in upper triangular form, then this would mean all its eigenvalues are real which contradicts it having eigenvalues $\pm i$ as well.