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Diagonalizable

Redirected from Diagonalizable matrix 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 : VV 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 : VV 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 : VV 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:

$A = \begin{bmatrix} 5 & -8 & 1 \\ 0 & 0 & 7 \\ 0 & 0 & -2 \end{bmatrix}$

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 Cn×n, is a null set with respect to the Lebesgue measure. The same is not true over R; as n increases, it becomes less and less likely that a randomly selected real matrix is diagonalizable over R.

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

$P^{-1}AP = D$

is a diagonal matrix. Then

$A^k = (PDP^{-1})^k = PD^kP^{-1}$

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

For example, consider the following matrix:

$M =\begin{bmatrix}a & b-a \\ 0 &b \end{bmatrix}.$
Calculating the various powers of M reveals a surprising pattern:
$M^2 = \begin{bmatrix}a^2 & b^2-a^2 \\ 0 &b^2 \end{bmatrix},\quad M^3 = \begin{bmatrix}a^3 & b^3-a^3 \\ 0 &b^3 \end{bmatrix},\quad M^4 = \begin{bmatrix}a^4 & b^4-a^4 \\ 0 &b^4 \end{bmatrix},\quad \ldots$

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

$\mathbf{u}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}=\mathbf{e}_1,\quad \mathbf{v}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}=\mathbf{e}_1+\mathbf{e}_2,$ where ei denotes the standard basis of Rn. The reverse change of basis is given by
$\mathbf{e}_1 = \mathbf{u},\qquad \mathbf{e}_2 = \mathbf{v}-\mathbf{u}.$

Straighforward calculations show that

$M\mathbf{u} = a\mathbf{u},\qquad M\mathbf{v}=b\mathbf{v}.$
Thus, a and b are the eigenvalues corresponding to u and v, respectively. By linearity of matrix multiplication, we have that
$M^n \mathbf{u} = a^n\, \mathbf{u},\qquad M^n \mathbf{v}=b^n\,\mathbf{v}.$
Switching back to the standard basis, we have
$M^n \mathbf{e}_1 = M^n \mathbf{u} = a^n \mathbf{e}_1,$
$M^n \mathbf{e}_2 = M^n (\mathbf{v}-\mathbf{u}) = b^n \mathbf{v} - a^n\mathbf{a} = (b^n-a^n) \mathbf{e}_1+b^n\mathbf{e}_2.$
The preceding relations, expressed in matrix form, are
$M^n = \begin{bmatrix}a^n & b^n-a^n \\ 0 &b^n \end{bmatrix},$ thereby explaining the above phenomenon.