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.
Best Answer
The problem is that it doesn't just use the fact that T has an eigenvalue -- it uses that, plus the inductive hypothesis. And in order to prove it for this smaller invariant subspace, you need the fact that T has an eigenvalue on that as well. So there's the problem -- it's not enough to have an eigenvalue, you need an eigenvalue on every invariant subspace. (Well, perhaps you don't need one on every invariant subspace.) In the complex case you have this, since any linear transform on any finite-dimensional vector space has an eigenvalue. Here you don't -- the image of T-1 is the plane x=0, and though T preserves that plane, it acts as a quarter-turn rotation on it and thus has no eigenvalue when restricted.