Diagonalizable matrix

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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.

[edit] Characterization

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, has Lebesgue measure zero. One can also say that the diagonalizable matrices form a dense subset with respect to 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.

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., one-by-one matrix.

[edit] Simultaneous diagonalization

A set of matrices are said to be simultaneously diagonalizable if there exists a single invertible matrix P such that $P - 1 A P$ is a diagonal matrix for every A in the set. The following theorem characterizes simultaneously diagonalizable matrices: A set of diagonalizable matrices commutes if and only if the set is simultaneously diagonalizable (Horn & Johnson 1985, pp. 51–53).

The set of all n-by-n diagonalizable matrices (over C) with n > 1 is not simultaneously diagonalizable. For instance, the matrices

$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \quad\text{and}\quad \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$

are diagonalizable but not simultaneously diagonalizable because they do not commute.

A set consists of commuting normal matrices if and only if it is simultaneously diagonalizable by a unitary matrix; that is, there exists a unitary matrix U such that $U * A U$ is diagonal for every A in the set. If the matrices in the set are real, then U can be taken to be an orthogonal matrix.

[edit] Examples

[edit] Diagonalizable matrices

Involutions are diagonalizable over the reals (and indeed any field of characteristic not 2), with 1's and -1's on the diagonal
Finite order endomorphisms (including involutions) are diagonalizable over the complex numbers, or any algebraically closed field where the characteristic of the field doesn't divide the order of the endomorphism (as the roots of unity are distinct), with roots of unity on the diagonal. This is obvious since its minimal polynomial is separable.
Projections are diagonalizable, with 0's and 1's on the diagonal.
Real symmetric matrices are diagonalizable by orthogonal matrices; i.e., given a real symmetric matrix $A$ , $Q T A Q$ is diagonal for some orthogonal matrix $Q$ . More generally, matrices are diagonalizable by unitary matrices if and only if they are normal. Examples of normal matrices are symmetric (or skew-symmetric) matrices and Hermitian matrices (or skew-Hermitian matrices).

[edit] Matrices that are not diagonalizable

Some matrices are not diagonalizable over any field, most notably nonzero nilpotent matrices. This happens more generally if the algebraic and geometric multiplicities of an eigenvalue do not coincide. For instance, consider

$C = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$

This matrix is not diagonalizable: there is no matrix U such that $U - 1 C U$ is a diagonal matrix. Indeed, C has one eigenvalue (namely zero) and this eigenvalue has algebraic multiplicity 2 and geometric multiplicity 1.

Some real matrices are not diagonalizable over the reals. Consider for instance the matrix

$B = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.$

The matrix B does not have any real eigenvalues, so there is no real matrix Q such that $Q - 1 B Q$ is a diagonal matrix. However, we can diagonalize B if we allow complex numbers. Indeed, if we take

$Q = \begin{bmatrix} 1 & \textrm{i} \\ \textrm{i} & 1 \end{bmatrix},$

then $Q - 1 B Q$ is diagonal.

Note that the above examples show that the sum of diagonalizable matrices need not be diagonalizable.

[edit] How to diagonalize a matrix

Consider a matrix

$A=\begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{bmatrix}.$

This matrix has eigenvalues

$\lambda_1 = 3, \quad \lambda_2 = 2, \quad \lambda_3= 1.$

So A is a 3-by-3 matrix with 3 different eigenvalues, therefore it is diagonalizable.

If we want to diagonalize A, we need to compute the corresponding eigenvectors. They are

$v_1 = \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix}, \quad v_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad v_3 = \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}.$

One can easily check that $A v k = λ k v k .$

Now, let P be the matrix with these eigenvectors as its columns:

$P= \begin{bmatrix} -1 & 0 & -1 \\ -1 & 0 & 0 \\ 2 & 1 & 2 \end{bmatrix}.$

Then P diagonalizes A, as a simple computation confirms:

$P^{-1}AP = \begin{bmatrix} 0 & -1 & 0 \\ 2 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 & -1 \\ -1 & 0 & 0 \\ 2 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{bmatrix}.$

Note that the eigenvalues $λ k$ appear in the diagonal matrix.

[edit] 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

$P^{-1}AP = D \,$

is a diagonal matrix. Then, as the matrix product is associative,

$\begin{align} A^k &= (PDP^{-1})^k = (PDP^{-1}) \cdot (PDP^{-1}) \cdots (PDP^{-1}) \\ &= PD(P^{-1}P) D (P^{-1}P) \cdots (P^{-1}P) D P^{-1} \\ &= PD^kP^{-1} \end{align}$

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

This is particularly useful in finding closed form expressions for terms of linear recursive sequences, such as the Fibonacci numbers.

[edit] Particular application

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 e_i denotes the standard basis of Rⁿ. The reverse change of basis is given by

$\mathbf{e}_1 = \mathbf{u},\qquad \mathbf{e}_2 = \mathbf{v}-\mathbf{u}.$

Straightforward 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

$M^n \mathbf{e}_1 = M^n \mathbf{u} = a^n \mathbf{e}_1,$

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

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.

[edit] Quantum mechanical application

In quantum mechanical and quantum chemical computations matrix diagonalization is one of the most frequently applied numerical processes. The basic reason is that the time-independent Schrödinger equation is an eigenvalue equation, albeit in most of the physical situations on an infinite dimensional space (a Hilbert space). A very common approximation is to truncate Hilbert space to finite dimension, after which the Schrödinger equation can be formulated as an eigenvalue problem of a real symmetric, or complex Hermitian, matrix. Formally this approximation is founded on the variational principle, valid for Hamiltonians that are bounded from below. But also first-order perturbation theory for degenerate states leads to a matrix eigenvalue problem.

[edit] See also

[edit] External links

Diagonalization on PlanetMath

[edit] References

Horn, Roger A.; Johnson, Charles R. (1985), Matrix Analysis, Cambridge University Press, ISBN 978-0-521-38632-6 .

Diagonalizable matrix

From Wikipedia, the free encyclopedia

Contents

[edit] Characterization

[edit] Simultaneous diagonalization

[edit] Examples

[edit] Diagonalizable matrices

[edit] Matrices that are not diagonalizable

[edit] How to diagonalize a matrix

[edit] An application

[edit] Particular application

[edit] Quantum mechanical application

[edit] See also

[edit] External links

[edit] References

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