Diagonalizable : Diagonalizable matrix 

Diagonalizable matrices and maps are of interest because diagonal matrices are especially easy to handle: their eigenvalues and eigenvectors are known and one can raise a diagonal matrix to a power by simply raising the diagonal entries to that same power.

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

• An n-by-n matrix A over the field F is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to n, which is the case if and only if there exists a basis of Fⁿ consisting of eigenvectors of A. If such a basis has been found, one can form the matrix P having these basis vectors as columns, and P ^-1AP will be a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of A.
• A linear map T : V → V is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to dim(V), which is the case if and only if there exists a basis of V consisting of eigenvectors of T. With respect to such a basis, T will be represented by a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of T.

Another characterization: A matrix or linear map is diagonalizable over the field F if and only if its minimal polynomial is a product of distinct linear factors over F.

The following sufficient (but not necessary) condition is often useful.

• An n-by-n matrix A is diagonalizable over the field F if it has n distinct eigenvalues in F, i.e. if its characteristic polynomial has n distinct roots in F.
• A linear map T : V → V with n=dim(V) is diagonalizable if it has n distinct eigenvalues, i.e. if its characteristic polynomial has n distinct roots in F.

Here is an example of a diagonalizable matrix:

<math>A = \begin{bmatrix}

Since the matrix is triangular[?] (specifically upper triangular), the eigenvalues are 5, 0, and -2. Since A is a 3-by-3 matrix with 3 real, distinct eigenvalues, A is diagonalizable over R.

As a rule of thumb, over C almost every matrix is diagonalizable. More precisely: the set of complex n-by-n matrices that are not diagonalizable over C, considered as a subset of C^n×n, is a null set with respect to the Lebesgue measure. The same isn't true over R; as n increases, it becomes less and less likely that a randomly selected real matrix is diagonalizable over R.

An application

Diagonalization can be used to compute the powers of a matrix A efficiently, provided the matrix is diagonalizable. Suppose we have found that

<math>P^{-1}AP = D</math>

is a diagonal matrix. Then

<math>A^k = (PDP^{-1})^k = PD^kP^{-1}</math>

and the latter is easy to calculate since it only involves the powers of a diagonal matrix.

For example, consider the following matrix:

<math>M =\begin{bmatrix}a & b-a \\ 0 &b \end{bmatrix}.</math>

<math>

The above phenomenon can be explained by diagonalizing M. To accomplish this, we need a basis of R² consisting of eigenvectors of M. One such eigenvector basis is given by

<math>\mathbf{u}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}=\mathbf{e}_1,\quad

<math> \mathbf{e}_1 = \mathbf{u},\qquad \mathbf{e}_2 = \mathbf{v}-\mathbf{u}.</math>

Straighforward calculations show that

<math>M\mathbf{u} = a\mathbf{u},\qquad M\mathbf{v}=b\mathbf{v}.</math>

<math> M^n \mathbf{u} = a^n\, \mathbf{u},\qquad M^n \mathbf{v}=b^n\,\mathbf{v}.</math>

<math> M^n \mathbf{e}_1 = M^n \mathbf{u} = a^n \mathbf{e}_1,</math>

<math> M^n \mathbf{e}_2 = M^n (\mathbf{v}-\mathbf{u}) = b^n \mathbf{v} - a^n\mathbf{a} = (b^n-a^n) \mathbf{e}_1+b^n\mathbf{e}_2.</math>

<math>