Editing Diagonalizable matrix (section)

== Characterization ==
The fundamental fact about diagonalizable maps and matrices is expressed by the following:

* An <math>n \times n</math> matrix <math>A</math> over a field <math>F</math> is diagonalizable [[if and only if]] the sum of the [[dimension (linear algebra)|dimension]]s of its eigenspaces is equal to <math>n</math>, which is the case if and only if there exists a [[basis (linear algebra)|basis]] of <math>F^n</math> consisting of eigenvectors of <math>A</math>. If such a basis has been found, one can form the matrix <math>P</math> having these [[basis vectors]] as columns, and <math>P^{-1}AP</math> will be a diagonal matrix whose diagonal entries are the eigenvalues of <math>A</math>. The matrix <math>P</math> is known as a [[modal matrix]] for <math>A</math>.
* A linear map <math>T : V \to V</math> is diagonalizable if and only if the sum of the [[dimension (linear algebra)|dimension]]s of its eigenspaces is equal to {{nowrap|<math>\dim(V)</math>,}} which is the case if and only if there exists a basis of <math>V</math> consisting of eigenvectors of <math>T</math>. With respect to such a basis, <math>T</math> will be represented by a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of {{nowrap|<math>T</math>.}}

The following sufficient (but not necessary) condition is often useful.
* An <math>n \times n</math> matrix <math>A</math> is diagonalizable over the field <math>F</math> if it has <math>n</math> distinct eigenvalues in {{nowrap|<math>F</math>,}} i.e. if its [[characteristic polynomial]] has <math>n</math> distinct roots in {{nowrap|<math>F</math>;}} however, the converse may be false. Consider <math display="block">\begin{bmatrix} 
-1 & 3 & -1 \\
-3 & 5 & -1 \\
-3 & 3 & 1 
\end{bmatrix},</math> which has eigenvalues 1, 2, 2 (not all distinct) and is diagonalizable with diagonal form ([[similar (linear algebra)|similar]] to {{nowrap|<math>A</math>)}} <math display="block">\begin{bmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{bmatrix}</math> and [[change of basis|change of basis matrix]] <math>P</math>: <math display="block">\begin{bmatrix}
1 & 1 & -1 \\
1 & 1 & 0 \\
1 & 0 & 3
\end{bmatrix}.</math> The converse fails when <math>A</math> has an eigenspace of dimension higher than 1. In this example, the eigenspace of <math>A</math> associated with the eigenvalue 2 has dimension 2.
* A linear map <math>T : V \to V</math> with <math>n = \dim(V)</math> is diagonalizable if it has <math>n</math> distinct eigenvalues, i.e. if its characteristic polynomial has <math>n</math> distinct roots in <math>F</math>.

Let <math>A</math> be a matrix over {{nowrap|<math>F</math>.}} If <math>A</math> is diagonalizable, then so is any power of it. Conversely, if <math>A</math> is invertible, <math>F</math> is algebraically closed, and <math>A^n</math> is diagonalizable for some <math>n</math> that is not an integer multiple of the characteristic of {{nowrap|<math>F</math>,}} then <math>A</math> is diagonalizable. Proof: If <math>A^n</math> is diagonalizable, then <math>A</math> is annihilated by some polynomial {{nowrap|<math>\left(x^n - \lambda_1\right) \cdots \left(x^n - \lambda_k\right)</math>,}} which has no multiple root (since {{nowrap|<math>\lambda_j \ne 0</math>)}} and is divided by the minimal polynomial of {{nowrap|<math>A</math>.}}

Over the complex numbers <math>\Complex</math>, almost every matrix is diagonalizable. More precisely: the set of complex <math>n \times n</math> matrices that are ''not'' diagonalizable over {{nowrap|<math>\Complex</math>,}} considered as a [[subset]] of {{nowrap|<math>\Complex^{n \times n}</math>,}} has [[Lebesgue measure]] zero. One can also say that the diagonalizable matrices form a dense subset with respect to the [[Zariski topology]]: the non-diagonalizable matrices lie inside the [[Algebraic variety|vanishing set]] of the [[discriminant]] of the characteristic polynomial, which is a [[hypersurface]]. From that follows also density in the usual (''strong'') topology given by a [[norm (mathematics)|norm]]. The same is not true over {{nowrap|<math>\R</math>.}}

The [[Jordan–Chevalley decomposition]] expresses an operator as the sum of its semisimple (i.e., diagonalizable) part and its [[nilpotent]] part. Hence, a matrix is diagonalizable if and only if its nilpotent part is zero. Put in another way, a matrix is diagonalizable if each block in its [[Jordan form]] has no nilpotent part; i.e., each "block" is a one-by-one matrix.