Linear map triangulizable

Question

What's the definition of a linear map that is triangulazible? I can't find it anywhere.

In addition, I was asked to find a linear map that doesn't have any invariant sub-spaces. I know that if a map is triangulazible it does have invariant sub-spaces, from there my request on the exact definition.

Do you know of a linear map that doesn't have invariant subspaces?

Thank so much!

Answer 1

Here's a definition:

Let $V$ be a finite-dimensional vector space over a field $\Bbb{F}$, and let $T:V \to V$ be linear. We say $T$ is triangualrizable over $\Bbb{F}$ if there is a basis $\beta$ of $V$ such that the matrix representation $[T]_{\beta}$ is a triangular matrix.

By the way, every linear map has invariant subspaces, namely $\{0\}$ and $V$. You're probably interested in non-trivial invariant subspaces, i.e a subspace $W$ such that $\{0\}\subsetneq W \subsetneq V$ and $T[W] \subset W$. The following theorem addresses this question:

Let $V$ be a non-zero finite-dimensional vector space over a field $\Bbb{F}$, and $T:V \to V$ be linear. Then, $\{0\}$ and $V$ are the only $T$-invariant subspaces if and only if the characteristic polynomial of $T$ is irreducible over $\Bbb{F}$.

So, this tells you exactly when a linear map has no (non-trivial) invariant subspaces.

Answer 2

Consider the matrix $R(\theta)\in M_2(\mathbb{R})$

$$R (\theta)=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$$ for $\theta\in(0,\pi)$, this matrix corresponds to rotation about the origin, you may check it has no invariant subspace except for $\{0\}$ and $\mathbb{R}^2$, which we regard as a trivial subspaces.

For the definition of triangulazible:

$T:V \to V$ a linear map over $V$, a finite-dimensional vector space over a field $\Bbb{F}$. We say $T$ is triangualrizable over $\Bbb{F}$ if there exists a basis $\mathcal{B}$ of $V$ such that the matrix representation of $T$ with respect to $\mathcal{B}$ is a triangular matrix.