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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?

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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 $\lambda \in \mathbf{C}$.

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  • $\begingroup$ where can I find a proof to 3? Thanks $\endgroup$ – gbox Sep 8 '15 at 14:59
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    $\begingroup$ Suppose the minimal polynomial of T has no repeated roots. Then every generalized eigenvector of T is an eigenvector of T. Because there is a basis of V consisting of generalized eigenvectors of T, this implies that there is a basis of V consisting of eigenvectors of T (and hence T is diagonalizable). For more on these results, see Chapter 8 of Linear Algebra Done Right, third edition, by Sheldon Axler. $\endgroup$ – Sheldon Axler Sep 8 '15 at 17:34

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