Inlinear algebra, asquare matrix${\displaystyle A}$ is calleddiagonalizable ornon-defective if it issimilar to adiagonal matrix. That is, if there exists aninvertible matrix${\displaystyle P}$ and a diagonal matrix${\displaystyle D}$ such that${\displaystyle P^{-1}AP=D}$. This is equivalent to${\displaystyle A=PD{P}^{-1}}$. (Such${\displaystyle P}$,${\displaystyle D}$ are not unique.) This property exists for any linear map: for afinite-dimensionalvector space${\displaystyle V}$, alinear map${\displaystyle T:V\to V}$ is calleddiagonalizable if there exists anordered basis of${\displaystyle V}$ consisting ofeigenvectors of${\displaystyle T}$. These definitions are equivalent: if${\displaystyle T}$ has amatrix representation${\displaystyle A=PD{P}^{-1}}$ as above, then the column vectors of${\displaystyle P}$ form a basis consisting of eigenvectors of${\displaystyle T}$, and the diagonal entries of${\displaystyle D}$ are the correspondingeigenvalues of${\displaystyle T}$; with respect to this eigenvector basis,${\displaystyle T}$ is represented by${\displaystyle D}$.

Diagonalization is the process of finding the above${\displaystyle P}$ and${\displaystyle D}$ and makes many subsequent computations easier. One can raise a diagonal matrix${\displaystyle D}$ to a power by simply raising the diagonal entries to that power. Thedeterminant of a diagonal matrix is simply the product of all diagonal entries. Such computations generalize easily to${\displaystyle A=PD{P}^{-1}}$.

The geometric transformation represented by a diagonalizable matrix is aninhomogeneous dilation (oranisotropic scaling). That is, it canscale the space by a different amount in different directions. The direction of each eigenvector is scaled by a factor given by the corresponding eigenvalue.

A square matrix that is not diagonalizable is calleddefective. It can happen that a matrix${\displaystyle A}$ withreal entries is defective over the real numbers, meaning that${\displaystyle A=PD{P}^{-1}}$ is impossible for any invertible${\displaystyle P}$ and diagonal${\displaystyle D}$ with real entries, but it is possible withcomplex entries, so that${\displaystyle A}$ is diagonalizable over the complex numbers. For example, this is the case for a genericrotation matrix.

Many results for diagonalizable matrices hold only over analgebraically closed field (such as the complex numbers). In this case, diagonalizable matrices aredense in the space of all matrices, which means any defective matrix can be deformed into a diagonalizable matrix by a smallperturbation; and theJordan–Chevalley decomposition states that any matrix is uniquely the sum of a diagonalizable matrix and anilpotent matrix. Over an algebraically closed field, diagonalizable matrices are equivalent tosemi-simple matrices.

A square${\displaystyle n\times n}$ matrix${\displaystyle A}$ with entries in afield${\displaystyle F}$ is calleddiagonalizable ornondefective if there exists an${\displaystyle n\times n}$ invertible matrix (i.e. an element of thegeneral linear group GL_n(F)),${\displaystyle P}$, such that${\displaystyle P^{-1}AP}$ is a diagonal matrix.

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 diagonalizableif and only if the sum of thedimensions of its eigenspaces is equal to${\displaystyle n}$, which is the case if and only if there exists abasis 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 thesebasis 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 amodal matrix for${\displaystyle A}$.
• A linear map${\displaystyle T:V\to V}$ is diagonalizable if and only if the sum of thedimensions 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 itscharacteristic polynomial has${\displaystyle n}$ distinct roots in${\displaystyle F}$; however, the converse may be false. Consider which has eigenvalues 1, 2, 2 (not all distinct) and is diagonalizable with diagonal form (similar to${\displaystyle A}$) andchange of basis matrix${\displaystyle P}$: 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${\displaystyle {\lambda }_j\ne 0}$) 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 arenot diagonalizable over${\displaystyle \mathbb{C}}$, considered as asubset of${\displaystyle {\mathbb{C}}^{n\times n}}$, hasLebesgue measure zero. One can also say that the diagonalizable matrices form a dense subset with respect to theZariski topology: the non-diagonalizable matrices lie inside thevanishing set of thediscriminant of the characteristic polynomial, which is ahypersurface. From that follows also density in the usual (strong) topology given by anorm. The same is not true over${\displaystyle \mathbb{R}}$.

TheJordan–Chevalley decomposition expresses an operator as the sum of its semisimple (i.e., diagonalizable) part and itsnilpotent 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 itsJordan form has no nilpotent part; i.e., each "block" is a one-by-one matrix.

Consider the two following arbitrary bases${\displaystyle E=\{{\mathit{e}}_i|\mathrm{\forall }i\in [n]\}}$ and${\displaystyle F=\{{\mathit{\alpha }}_i|\mathrm{\forall }i\in [n]\}}$. Suppose that there exists a linear transformation represented by a matrix${\displaystyle A_E}$ which is written with respect to basis E. Suppose also that there exists the following eigen-equation:

${\displaystyle A_E{\mathit{\alpha }}_{E,i}={\lambda }_i{\mathit{\alpha }}_{E,i}}$

The alpha eigenvectors are written also with respect to the E basis. Since the set F is both a set of eigenvectors for matrix A and it spans some arbitrary vector space, then we say that there exists a matrix${\displaystyle D_F}$ which is a diagonal matrix that is similar to${\displaystyle A_E}$. In other words,${\displaystyle A_E}$ is a diagonalizable matrix if the matrix is written in the basis F. We perform the change of basis calculation using the transition matrix${\displaystyle S}$, which changes basis from E to F as follows:

${\displaystyle D_F=S_E^F{A}_E{S}_E^{-1F}}$,

where${\displaystyle S_E^F}$ is the transition matrix from E-basis to F-basis. The inverse can then be equated to a new transition matrix${\displaystyle P}$ which changes basis from F to E instead and so we have the following relationship :

${\displaystyle S_E^{-1F}=P_F^E}$

Both${\displaystyle S}$ and${\displaystyle P}$ transition matrices are invertible. Thus we can manipulate the matrices in the following fashion:$${\displaystyle \begin{array}{r}D=S{A}_E{S}^{-1}\\ D=P^{-1}{A}_{E}P\end{array}}$$The matrix${\displaystyle A_E}$ will be denoted as${\displaystyle A}$, which is still in the E-basis. Similarly, the diagonal matrix is in the F-basis.

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

then:

The transition matrix S has the E-basis vectors as columns written in the basis F. Inversely, the inverse transition matrix P has F-basis vectors${\displaystyle {\mathit{\alpha }}_i}$ written in the basis of E so that we can represent P in block matrix form in the following manner:

as a result we can write:

In block matrix form, we can consider the A-matrix to be a matrix of 1x1 dimensions whilst P is a 1xn dimensional matrix. The D-matrix can be written in full form with all the diagonal elements as an nxn dimensional matrix:

Performing the above matrix multiplication we end up with the following result:Taking each component of the block matrix individually on both sides, we end up with the following:

${\displaystyle A{\mathit{\alpha }}_i={\lambda }_i{\mathit{\alpha }}_i(i=1,2,\dots ,n).}$

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

When a complex matrix${\displaystyle A\in {\mathbb{C}}^{n\times n}}$ is aHermitian matrix (or more generally anormal matrix), eigenvectors of${\displaystyle A}$ can be chosen to form anorthonormal basis of${\displaystyle {\mathbb{C}}^n}$, and${\displaystyle P}$ can be chosen to be aunitary matrix. If in addition,${\displaystyle A\in {\mathbb{R}}^{n\times n}}$ is a realsymmetric 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 anorthogonal 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 besimultaneously 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 diagonalizablematrices 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

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

A set consists of commutingnormal matrices if and only if it is simultaneously diagonalizable by aunitary matrix; that is, there exists a unitary matrix${\displaystyle U}$ such that${\displaystyle U^{\ast }AU}$ is diagonal for every${\displaystyle A}$ in the set.

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

• Involutions are diagonalizable over the reals (and indeed any field of characteristic not 2), with ±1 on the diagonal.
• Finite orderendomorphisms are diagonalizable over${\displaystyle \mathbb{C}}$ (or any algebraically closed field where the characteristic of the field does not divide the order of the endomorphism) withroots of unity on the diagonal. This follows since the minimal polynomial isseparable, because the roots of unity are distinct.
• Projections are diagonalizable, with 0s and 1s on the diagonal.
• Realsymmetric matrices are diagonalizable byorthogonal matrices; i.e., given a real symmetric matrix${\displaystyle A}$,${\displaystyle Q^{\mathrm{T}}AQ}$ is diagonal for some orthogonal matrix${\displaystyle Q}$. More generally, matrices are diagonalizable byunitary matrices if and only if they arenormal. In the case of the real symmetric matrix, we see that${\displaystyle A=A^{\mathrm{T}}}$, so clearly${\displaystyle A{A}^{\mathrm{T}}=A^{\mathrm{T}}A}$ holds. Examples of normal matrices are real symmetric (orskew-symmetric) matrices (e.g. covariance matrices) andHermitian matrices (or skew-Hermitian matrices). Seespectral theorems for generalizations to infinite-dimensional vector spaces.

In general, arotation matrix is not diagonalizable over the reals, but allrotation matrices are diagonalizable over the complex field. Even if a matrix is not diagonalizable, it is always possible to "do the best one can", and find a matrix with the same properties consisting of eigenvalues on the leading diagonal, and either ones or zeroes on the superdiagonal – known asJordan normal form.

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

This matrix is not diagonalizable: there is no matrix${\displaystyle U}$ such that${\displaystyle U^{-1}CU}$ is a diagonal matrix. Indeed,${\displaystyle C}$ 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

The matrix${\displaystyle B}$ does not have any real eigenvalues, so there is noreal matrix${\displaystyle Q}$ such that${\displaystyle Q^{-1}BQ}$ is a diagonal matrix. However, we can diagonalize${\displaystyle B}$ if we allow complex numbers. Indeed, if we take

then${\displaystyle Q^{-1}BQ}$ is diagonal. It is easy to find that${\displaystyle B}$ is the rotation matrix which rotates counterclockwise by angle$\theta =-\frac{\pi }{2}$

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

Diagonalizing a matrix is the same process as finding itseigenvalues and eigenvectors, in the case that the eigenvectors form a basis. For example, consider the matrix

The roots of thecharacteristic polynomial${\displaystyle p(\lambda )=det(\lambda I-A)}$ are the eigenvalues${\displaystyle {\lambda }_1=1,{\lambda }_2=1,{\lambda }_3=2}$. Solving the linear system${\displaystyle \left(1I-A\right)\mathbf{v}=\mathbf{0}}$ gives the eigenvectors${\displaystyle {\mathbf{v}}_1=(1,1,0)}$ and${\displaystyle {\mathbf{v}}_2=(0,2,1)}$, while${\displaystyle \left(2I-A\right)\mathbf{v}=\mathbf{0}}$ gives${\displaystyle {\mathbf{v}}_3=(1,0,-1)}$; that is,${\displaystyle A{\mathbf{v}}_i={\lambda }_i{\mathbf{v}}_i}$ for${\displaystyle i=1,2,3}$. These vectors form a basis of${\displaystyle V={\mathbb{R}}^3}$, so we can assemble them as the column vectors of achange-of-basis matrix${\displaystyle P}$ to get:We may see this equation in terms of transformations:${\displaystyle P}$ takes the standard basis to the eigenbasis,${\displaystyle P{\mathbf{e}}_i={\mathbf{v}}_i}$, so we have:$${\displaystyle P^{-1}AP{\mathbf{e}}_i=P^{-1}A{\mathbf{v}}_i=P^{-1}({\lambda }_i{\mathbf{v}}_i)={\lambda }_i{\mathbf{e}}_i,}$$so that${\displaystyle P^{-1}AP}$ has the standard basis as its eigenvectors, which is the defining property of${\displaystyle D}$.

Note that there is no preferred order of the eigenvectors in${\displaystyle P}$; changing the order of theeigenvectors in${\displaystyle P}$ just changes the order of theeigenvalues in the diagonalized form of${\displaystyle A}$.^[2]

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

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:

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

This is particularly useful in finding closed form expressions for terms oflinear recursive sequences, such as theFibonacci numbers.

For example, consider the following matrix:

Calculating the various powers of${\displaystyle M}$ reveals a surprising pattern:

The above phenomenon can be explained by diagonalizing${\displaystyle M}$. To accomplish this, we need a basis of${\displaystyle {\mathbb{R}}^2}$ consisting of eigenvectors of${\displaystyle M}$. One such eigenvector basis is given by

${\displaystyle \mathbf{u}=\left[\begin{array}{c}1\\ 0\end{array}\right]={\mathbf{e}}_1,\mathbf{v}=\left[\begin{array}{c}1\\ 1\end{array}\right]={\mathbf{e}}_1+{\mathbf{e}}_2,}$

wheree_i denotes the standard basis ofRⁿ. The reverse change of basis is given by

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

Straightforward calculations show that

${\displaystyle M\mathbf{u}=a\mathbf{u},M\mathbf{v}=b\mathbf{v}.}$

Thus,a andb are the eigenvalues corresponding tou andv, respectively. By linearity of matrix multiplication, we have that

${\displaystyle M^n\mathbf{u}=a^n\mathbf{u},M^n\mathbf{v}=b^n\mathbf{v}.}$

Switching back to the standard basis, we have

The preceding relations, expressed in matrix form, are

thereby explaining the above phenomenon.

Inquantum mechanical andquantum chemical computations matrix diagonalization is one of the most frequently applied numerical processes. The basic reason is that the time-independentSchrödinger equation is an eigenvalue equation, albeit in most of the physical situations on an infinite dimensionalHilbert space.

A very common approximation is to truncate (or project) the 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 thevariational principle, valid for Hamiltonians that are bounded from below.

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

Matrices can be generalized tolinear operators. A diagonal matrix can be generalized to diagonal operators on Hilbert spaces.

Let${\displaystyle H}$ be a Hilbert space. An operator${\displaystyle D:H\to H}$ is a diagonal operator iff there exists an orthonormal basis${\displaystyle (e_n{)}_n}$ of${\displaystyle H}$, such that${\displaystyle D{e}_n={\lambda }_n{e}_n}$ for some${\displaystyle {\lambda }_n\in \mathbb{C}}$.

For any${\displaystyle p\ge 1}$, define thep-Schatten norm as follows. Let${\displaystyle T:H\to H}$ be an operator, then${\displaystyle \Vert T{\Vert }_p:=\mathrm{Tr}(|T{|}^p{)}^{1/p}}$, where${\displaystyle \mathrm{Tr}}$ is thetrace. The p-Schatten class is the set of all operators with finite p-Schatten norm.

Weyl,^[3]von Neumann,^[4] and Kuroda,^[5] showed the following:

For any${\displaystyle p>1}$, any self-adjoint operator${\displaystyle T}$ on a Hilbert space${\displaystyle H}$, and any${\displaystyle ϵ>0}$, there exists a diagonal operator${\displaystyle D}$, such that${\displaystyle \Vert T-D{\Vert }_p\le ϵ}$.

In other words, any self-adjoint operator is an infinitesimal perturbation from a diagonal operator, where "infinitesimal" is in the sense of p-Schatten norm. In particular, since the Hilbert–Schmidt operator class is the 2-Schatten class, this means that any self-adjoint operator is diagonalizable after a perturbation by an infinitesimal Hilbert–Schmidt operator.In fact, the above result could be further generalized:

For anynorm ideal that is not the trace class, with norm${\displaystyle \Vert \cdot {\Vert }_J}$, any self-adjoint operator${\displaystyle T}$ on a Hilbert space${\displaystyle H}$, and any${\displaystyle ϵ>0}$, there exists a diagonal operator${\displaystyle D}$, such that${\displaystyle \Vert T-D{\Vert }_J\le ϵ}$.

The result is false for${\displaystyle p=1}$ (thetrace class). This is a simple corollary of the Kato^[6]–Rosenblum^{[7][8]: Theorem XI.8} theorem, which states that if${\displaystyle T}$ is self-adjoint, and${\displaystyle A}$ is trace class, then${\displaystyle T,T+A}$ have the sameabsolutely continuous part of the spectrum. The result is sharp, however, in the sense that if${\displaystyle T}$ has no absolutely continuous part, then itcan be diagonalized after perturbation by an infinitesimal trace class operator.^[9]

Forsimultaneous diagonalization, it's known that, given a finite list of${\displaystyle T_1,\dots ,T_n}$ self-adjoint operators that commute with each other, for any${\displaystyle ϵ>0}$, there exists a sequence of diagonal operators${\displaystyle D_1,\dots ,D_n}$, such that${\displaystyle \Vert T_1-D_1{\Vert }_n\le ϵ,\dots ,\Vert T_n-D_n{\Vert }_n\le ϵ}$, where${\displaystyle \Vert \cdot {\Vert }_n}$ is the n-Schatten norm. Note that${\displaystyle n\ge 2}$^[10]

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

• Matrices (mathematics)

• CS1 German-language sources (de)
• Articles with short description
• Short description is different from Wikidata
• Use American English from April 2019
• All Wikipedia articles written in American English
• Pages that use a deprecated format of the math tags