Sufficient and necessary conditions for matrix to be diagonalize or triangular

Question

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?

Answer 1

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:

1. $T$ is diagonalizable.

2. There is a basis of $V$ consisting of eigenvectors of $T$.

3. The minimal polynomial of $T$ has no repeated roots.

4. There exist 1-dimensional subspaces $U_1, \dots, U_n$ of $V$, each invariant under $T$, such that $V = U_1 \oplus \dots \oplus U_n$.

5. $V = E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$, where $\lambda_1, \dots, \lambda_m$ are the distinct eigenvalues of $T$.

6. $\dim V = \dim E(\lambda_1, T) + \cdots + \dim E(\lambda_m, T)$, where $\lambda_1, \dots, \lambda_m$ are the distinct eigenvalues of $T$.

7. $V = \text{null } (T - \lambda I) \oplus \text{range } (T - \lambda I)$ for every $\lambda \in \mathbf{C}$.

8. $\text{null } (T - \lambda I)^2 = \text{null } (T - \lambda I)$ for every $\lambda \in \mathbf{C}$.

9. $\text{range } (T - \lambda I)^2 = \text{range } (T - \lambda I)$ for every $\lambda \in \mathbf{C}$.