TITLE: When is a matrix triangularisable?
QUESTION [9 upvotes]: When is a matrix triangularisable? I don't seem to find much online regarding the triangularisability of matrices. 
What should I look for to prove if a matrix is triangularisable? 
What are the implications on its eigenvalues eigenvectors and eigenspaces?
An example a simple triangularisable and one of a non-triangularisable matrix in $\mathbb{R}$ would also help.

REPLY [12 votes]: Say you have a vector space $V = k^n$ over the field $k$. Then a matrix $A \in k^{n \times n}$ is triangularizable exactly when its characteristic polynomial (or minimal polynomial) splits in $k$ (can be written as a product of linear factors). Equivalently: $A$ has all its eigenvalues in $k$.
If $A$ has all its eigenvalues in $k$, we can put the matrix in Jordan normal form, which is triangular.
Conversely, suppose that $A = (a_{ij})$ is upper triangular. Then the characteristic polynomial of $A$ is
$$ \prod_{j = 1}^n (x - a_{jj}) $$
which is a product of linear factors. You may also note that this implies that $a_{11},\dots,a_{nn}$ are eigenvalues for $A$.
An example of a non-triangularizable matrix over $\mathbf{R}$ is
$$ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$
since its characteristic polynomial, $x^2 + 1$, cannot be written as a product of linear polynomials over $\mathbf{R}$.
Every matrix over $\mathbf{C}$ is triangularizable.