TITLE: sufficient and necessary conditions for matrix to be diagonalize or triangular
QUESTION [8 upvotes]: There are many sufficient and necessary conditions that derive from the fact that a matrix can be diagonalize means it is similar to a diagonal matrix.
Is there a list of all of conditions?

REPLY [17 votes]: No list of necessary and sufficient conditions for a linear operator (or equivalently, for a square matrix) to be diagonalizable will be complete, but a list of my favorites is below. In the result below, $E(\lambda, T)$ denotes the eigenspace of $T$ corresponding to $\lambda$, meaning that $E(\lambda, T)$ is the null space of $T - \lambda I$.
Theorem. Suppose $T$ is a linear map from a finite-dimensional complex vector space $V$ to $V$. Then the following are equivalent:

$T$ is diagonalizable.
There is a basis of $V$ consisting of eigenvectors of $T$.
The minimal polynomial of $T$ has no repeated roots.
There exist 1-dimensional subspaces $U_1, \dots, U_n$ of $V$, each
invariant under $T$, such that $V = U_1 \oplus \dots \oplus U_n$.
$V = E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$, where
$\lambda_1, \dots, \lambda_m$ are the distinct eigenvalues of $T$.
$\dim V = \dim E(\lambda_1, T) + \cdots + \dim E(\lambda_m, T)$,
where $\lambda_1, \dots, \lambda_m$ are the distinct eigenvalues of
$T$.
$V = \text{null } (T - \lambda I) \oplus \text{range } (T - \lambda I)$ for every $\lambda \in \mathbf{C}$.
$\text{null } (T - \lambda I)^2 = \text{null } (T - \lambda I)$ for every $\lambda \in \mathbf{C}$.
$\text{range } (T - \lambda I)^2 = \text{range } (T - \lambda I)$ for every $\lambda \in \mathbf{C}$.