In linear algebra, a square matrix A is called diagonalizable if it is similar to a diagonal matrix, i.e. if there exists an invertible matrix P such that P ^-1AP is a diagonal matrix. If V is a finite-dimensional vector space, then a linear map T : V → V is called diagonalizable if there exists a basis of V with respect to which T is represented by a diagonal matrix. Diagonalization is the process of finding a corresponding diagonal matrix for a diagonalizable matrix or linear map.

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.

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. One can also say that the diagonalizable matrices form a dense subset with respect the Zariski topology: the complement lies inside the set where the discriminant of the characteristic polynomial vanishes, which is a hypersurface. From that follows also density in the usual (strong) topology given by a norm.

The same is not true over R. As n increases, it becomes (in some sense) less and less likely that a randomly selected real matrix is diagonalizable over R.

Example

Consider a matrix

<math>A=\begin{pmatrix}

1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix}<math>

This matrix has eigenvalues

<math>\lambda_1 = 3, \quad \lambda_2 = 2, \quad \lambda_3= 1<math>

Writing the eigenspaces E_λi = ker (λ_iI - A), we find that the dim E_λ1=1, dim E_λ2=1, and dim E_λ3=1. So A is diagonalizable.

By directly calculating the kernels above, we find

E_λ1 = span (−1, −1, 2)^T

E_λ2 = span (0,0,1)^T

E_λ3 = span (−1, 0, 2)^T

Now, (−1, −1, 2)^T, (0,0,1)^T, and (−1, 0, 2)^T are eigenvectors of A, so we can form a matrix with its columns as these eigenvectors, call it P:

<math>P=

\begin{pmatrix} -1 & 0 & -1 \\ -1 & 0 & 0 \\ 2 & 1 & 2 \end{pmatrix}<math>

P diagonalizes A - observe

<math>P^{-1}AP=

\begin{pmatrix} -1 & 0 & -1 \\ -1 & 0 & 0 \\ 2 & 1 & 2 \end{pmatrix}^{-1} \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix} \begin{pmatrix} -1 & 0 & -1 \\ -1 & 0 & 0 \\ 2 & 1 & 2 \end{pmatrix}<math>

<math>=\begin{pmatrix}

0 & -1 & 0 \\
2 & 0  & 1 \\

-1 & -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix} \begin{pmatrix} -1 & -1 & 2 \\

0 & 0  & 1 \\

-1 & 0 & 2 \end{pmatrix}<math>

<math>=\begin{pmatrix}

0 & -1 & 0 \\
2 & 0  & 1 \\

-1 & -1 & 0 \end{pmatrix} \begin{pmatrix}

-3 & 0 & 1 \\
-3 & 0 & 0 \\
 6 & 2 & 2 \end{pmatrix}<math>

<math>=\begin{pmatrix}

3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{pmatrix}<math> as required.

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>

Calculating the various powers of M reveals a surprising pattern:

<math>

M^2 = \begin{bmatrix}a^2 & b^2-a^2 \\ 0 &b^2 \end{bmatrix},\quad M^3 = \begin{bmatrix}a^3 & b^3-a^3 \\ 0 &b^3 \end{bmatrix},\quad M^4 = \begin{bmatrix}a^4 & b^4-a^4 \\ 0 &b^4 \end{bmatrix},\quad \ldots <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

\mathbf{v}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}=\mathbf{e}_1+\mathbf{e}_2,<math> where e_i denotes the standard basis of Rⁿ. The reverse change of basis is given by

<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>

Thus, a and b are the eigenvalues corresponding to u and v, respectively. By linearity of matrix multiplication, we have that

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

Switching back to the standard basis, we have

<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>

The preceding relations, expressed in matrix form, are

<math>

M^n = \begin{bmatrix}a^n & b^n-a^n \\ 0 &b^n \end{bmatrix}, <math> thereby explaining the above phenomenon.

See also

• Jordan form