Diagonalizable matrix

A square ${\displaystyle n\times n}$ matrix, ${\displaystyle A}$, with entries in a field ${\displaystyle F}$ is called diagonalizable or nondefective if there exists an ${\displaystyle n\times n}$ invertible matrix (i.e. an element of the general linear group GL_n(F)), ${\displaystyle P}$, such that ${\displaystyle P^{-1}AP}$ is a diagonal matrix. Formally,

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

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

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

• An ${\displaystyle n\times n}$ matrix ${\displaystyle A}$ is diagonalizable over the field ${\displaystyle F}$ if it has ${\displaystyle n}$ distinct eigenvalues in ${\displaystyle F}$, i.e. if its characteristic polynomial has ${\displaystyle n}$ distinct roots in ${\displaystyle F}$; however, the converse may be false. Consider
  $${\displaystyle {\begin{bmatrix}-1&3&-1\\-3&5&-1\\-3&3&1\end{bmatrix}},}$$

  
  which has eigenvalues 1, 2, 2 (not all distinct) and is diagonalizable with diagonal form (similar to ${\displaystyle A}$)
  $${\displaystyle {\begin{bmatrix}1&0&0\\0&2&0\\0&0&2\end{bmatrix}}}$$

  
  and change of basis matrix ${\displaystyle P}$:
  $${\displaystyle {\begin{bmatrix}1&1&-1\\1&1&0\\1&0&3\end{bmatrix}}.}$$

  
  The converse fails when ${\displaystyle A}$ has an eigenspace of dimension higher than 1. In this example, the eigenspace of ${\displaystyle A}$ associated with the eigenvalue 2 has dimension 2.
• A linear map ${\displaystyle T:V\to V}$ with ${\displaystyle n=\dim(V)}$ is diagonalizable if it has ${\displaystyle n}$ distinct eigenvalues, i.e. if its characteristic polynomial has ${\displaystyle n}$ distinct roots in ${\displaystyle F}$.

Let ${\displaystyle A}$ be a matrix over ${\displaystyle F}$. If ${\displaystyle A}$ is diagonalizable, then so is any power of it. Conversely, if ${\displaystyle A}$ is invertible, ${\displaystyle F}$ is algebraically closed, and ${\displaystyle A^{n}}$ is diagonalizable for some ${\displaystyle n}$ that is not an integer multiple of the characteristic of ${\displaystyle F}$, then ${\displaystyle A}$ is diagonalizable. Proof: If ${\displaystyle A^{n}}$ is diagonalizable, then ${\displaystyle A}$ is annihilated by some polynomial ${\displaystyle \left(x^{n}-\lambda _{1}\right)\cdots \left(x^{n}-\lambda _{k}\right)}$, which has no multiple root (since ${\displaystyle \lambda _{j}\neq 0}$) and is divided by the minimal polynomial of ${\displaystyle A}$.

Over the complex numbers ${\displaystyle \mathbb {C} }$, almost every matrix is diagonalizable. More precisely: the set of complex ${\displaystyle n\times n}$ matrices that are not diagonalizable over ${\displaystyle \mathbb {C} }$, considered as a subset of ${\displaystyle \mathbb {C} ^{n\times n}}$, 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 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. The same is not true over ${\displaystyle \mathbb {R} }$.

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., each "block" is a one-by-one matrix.

The diagonalization of a symmetric matrix can be interpreted as a rotation of the axes to align them with the eigenvectors.

If a matrix ${\displaystyle A}$ can be diagonalized, that is,

${\displaystyle P^{-1}AP={\begin{bmatrix}\lambda _{1}&0&\cdots &0\\0&\lambda _{2}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &\lambda _{n}\end{bmatrix}},}$

then:

${\displaystyle AP=P{\begin{bmatrix}\lambda _{1}&0&\cdots &0\\0&\lambda _{2}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &\lambda _{n}\end{bmatrix}}.}$

Writing ${\displaystyle P}$ as a block matrix of its column vectors ${\displaystyle {\boldsymbol {\alpha }}_{i}}$

${\displaystyle P={\begin{bmatrix}{\boldsymbol {\alpha }}_{1}&{\boldsymbol {\alpha }}_{2}&\cdots &{\boldsymbol {\alpha }}_{n}\end{bmatrix}},}$

the above equation can be rewritten as

${\displaystyle A{\boldsymbol {\alpha }}_{i}=\lambda _{i}{\boldsymbol {\alpha }}_{i}\qquad (i=1,2,\dots ,n).}$

So the column vectors of ${\displaystyle P}$ are right eigenvectors of ${\displaystyle A}$, and the corresponding diagonal entry is the corresponding eigenvalue. The invertibility of ${\displaystyle P}$ also suggests that the eigenvectors are linearly independent and form a basis of ${\displaystyle F^{n}}$. This is the necessary and sufficient condition for diagonalizability and the canonical approach of diagonalization. The row vectors of ${\displaystyle P^{-1}}$ are the left eigenvectors of ${\displaystyle A}$.

When a complex matrix ${\displaystyle A\in \mathbb {C} ^{n\times n}}$ is a Hermitian matrix (or more generally a normal matrix), eigenvectors of ${\displaystyle A}$ can be chosen to form an orthonormal basis of ${\displaystyle \mathbb {C} ^{n}}$, and ${\displaystyle P}$ can be chosen to be a unitary matrix. If in addition, ${\displaystyle A\in \mathbb {R} ^{n\times n}}$ is a real symmetric matrix, then its eigenvectors can be chosen to be an orthonormal basis of ${\displaystyle \mathbb {R} ^{n}}$ and ${\displaystyle P}$ can be chosen to be an orthogonal matrix.

For most practical work matrices are diagonalized numerically using computer software. Many algorithms exist to accomplish this.

A set of matrices is said to be simultaneously diagonalizable if there exists a single invertible matrix ${\displaystyle P}$ such that ${\displaystyle P^{-1}AP}$ is a diagonal matrix for every ${\displaystyle A}$ 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.[1]^{: p. 64}

The set of all ${\displaystyle n\times n}$ diagonalizable matrices (over ${\displaystyle \mathbb {C} }$) with ${\displaystyle n>1}$ is not simultaneously diagonalizable. For instance, the matrices

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

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 ${\displaystyle U}$ such that ${\displaystyle U^{*}AU}$ is diagonal for every ${\displaystyle A}$ in the set.

In the language of Lie theory, a set of simultaneously diagonalizable matrices generates a toral Lie algebra.

Diagonalization can be used to efficiently compute the powers of a matrix ${\displaystyle A=PDP^{-1}}$:

${\displaystyle {\begin{aligned}A^{k}&=\left(PDP^{-1}\right)^{k}=\left(PDP^{-1}\right)\left(PDP^{-1}\right)\cdots \left(PDP^{-1}\right)\\&=PD\left(P^{-1}P\right)D\left(P^{-1}P\right)\cdots \left(P^{-1}P\right)DP^{-1}=PD^{k}P^{-1},\end{aligned}}}$

and the latter is easy to calculate since it only involves the powers of a diagonal matrix. For example, for the matrix ${\displaystyle A}$ with eigenvalues ${\displaystyle \lambda =1,1,2}$ in the example above we compute:

${\displaystyle {\begin{aligned}A^{k}=PD^{k}P^{-1}&=\left[{\begin{array}{rrr}1&\,0&1\\1&2&0\\0&1&\!\!\!\!-1\end{array}}\right]{\begin{bmatrix}1^{k}&0&0\\0&1^{k}&0\\0&0&2^{k}\end{bmatrix}}\left[{\begin{array}{rrr}1&\,0&1\\1&2&0\\0&1&\!\!\!\!-1\end{array}}\right]^{-1}\\[1em]&={\begin{bmatrix}2-2^{k}&-1+2^{k}&2-2^{k+1}\\0&1&0\\-1+2^{k}&1-2^{k}&-1+2^{k+1}\end{bmatrix}}.\end{aligned}}}$

This approach can be generalized to matrix exponential and other matrix functions that can be defined as power series. For example, defining ${\textstyle \exp(A)=I+A+{\frac {1}{2!}}A^{2}+{\frac {1}{3!}}A^{3}+\cdots }$, we have:

${\displaystyle {\begin{aligned}\exp(A)=P\exp(D)P^{-1}&=\left[{\begin{array}{rrr}1&\,0&1\\1&2&0\\0&1&\!\!\!\!-1\end{array}}\right]{\begin{bmatrix}e^{1}&0&0\\0&e^{1}&0\\0&0&e^{2}\end{bmatrix}}\left[{\begin{array}{rrr}1&\,0&1\\1&2&0\\0&1&\!\!\!\!-1\end{array}}\right]^{-1}\\[1em]&={\begin{bmatrix}2e-e^{2}&-e+e^{2}&2e-2e^{2}\\0&e&0\\-e+e^{2}&e-e^{2}&-e+2e^{2}\end{bmatrix}}.\end{aligned}}}$

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

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 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.

First-order perturbation theory also leads to matrix eigenvalue problem for degenerate states.

• Defective matrix
• Scaling (geometry)
• Triangular matrix
• Semisimple operator
• Diagonalizable group
• Jordan normal form
• Weight module – associative algebra generalization
• Orthogonal diagonalization