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Diagonalizable matrix
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<h2 id="definition">Definition</h2> <p>A square n × n {\displaystyle n\times n} matrix A {\displaystyle A} with entries in a <a href="/facts/Field_(mathematics)/xAjAS4ko">field</a> F {\displaystyle F} is called diagonalizable or nondefective if there exists an n × n {\displaystyle n\times n} invertible matrix (i.e. an element of the <a href="/facts/General_linear_group/B54gNILS">general linear group</a> GL<i>n</i>(<i>F</i>)), P {\displaystyle P} , such that P − 1 A P {\displaystyle P^{-1}AP} is a diagonal matrix. </p> <h2 id="characterization">Characterization</h2> <p>The fundamental fact about diagonalizable maps and matrices is expressed by the following: </p> <ul><li>An n × n {\displaystyle n\times n} matrix A {\displaystyle A} over a field F {\displaystyle F} is diagonalizable <a href="/facts/If_and_only_if/bYSxGJ66">if and only if</a> the sum of the <a href="/facts/Dimension_(linear_algebra)/z8m02A3W">dimensions</a> of its eigenspaces is equal to n {\displaystyle n} , which is the case if and only if there exists a <a href="/facts/Basis_(linear_algebra)/89IPoN6c">basis</a> of F n {\displaystyle F^{n}} consisting of eigenvectors of A {\displaystyle A} . If such a basis has been found, one can form the matrix P {\displaystyle P} having these <a href="/facts/Basis_vectors/89IPoN6c">basis vectors</a> as columns, and P − 1 A P {\displaystyle P^{-1}AP} will be a diagonal matrix whose diagonal entries are the eigenvalues of A {\displaystyle A} . The matrix P {\displaystyle P} is known as a <a href="/facts/Modal_matrix/QWWqranv">modal matrix</a> for A {\displaystyle A} .</li> <li>A linear map T : V → V {\displaystyle T:V\to V} is diagonalizable if and only if the sum of the <a href="/facts/Dimension_(linear_algebra)/z8m02A3W">dimensions</a> of its eigenspaces is equal to dim ( V ) {\displaystyle \dim(V)} , which is the case if and only if there exists a basis of V {\displaystyle V} consisting of eigenvectors of T {\displaystyle T} . With respect to such a basis, T {\displaystyle T} will be represented by a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of T {\displaystyle T} .</li></ul> <p>The following sufficient (but not necessary) condition is often useful. </p> <ul><li>An n × n {\displaystyle n\times n} matrix A {\displaystyle A} is diagonalizable over the field F {\displaystyle F} if it has n {\displaystyle n} distinct eigenvalues in F {\displaystyle F} , i.e. if its <a href="/facts/Characteristic_polynomial/7m8roDqo">characteristic polynomial</a> has n {\displaystyle n} distinct roots in F {\displaystyle F} ; however, the converse may be false. Consider [ − 1 3 − 1 − 3 5 − 1 − 3 3 1 ] , {\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 (<a href="/facts/Similar_(linear_algebra)/ySevpav1">similar</a> to A {\displaystyle A} ) [ 1 0 0 0 2 0 0 0 2 ] {\displaystyle {\begin{bmatrix}1&0&0\\0&2&0\\0&0&2\end{bmatrix}}} and <a href="/facts/Change_of_basis/Hnu1Spcc">change of basis matrix</a> P {\displaystyle P} : [ 1 1 − 1 1 1 0 1 0 3 ] . {\displaystyle {\begin{bmatrix}1&1&-1\\1&1&0\\1&0&3\end{bmatrix}}.} The converse fails when A {\displaystyle A} has an eigenspace of dimension higher than 1. In this example, the eigenspace of A {\displaystyle A} associated with the eigenvalue 2 has dimension 2.</li> <li>A linear map T : V → V {\displaystyle T:V\to V} with n = dim ( V ) {\displaystyle n=\dim(V)} is diagonalizable if it has n {\displaystyle n} distinct eigenvalues, i.e. if its characteristic polynomial has n {\displaystyle n} distinct roots in F {\displaystyle F} .</li></ul> <p>Let A {\displaystyle A} be a matrix over F {\displaystyle F} . If A {\displaystyle A} is diagonalizable, then so is any power of it. Conversely, if A {\displaystyle A} is invertible, F {\displaystyle F} is algebraically closed, and A n {\displaystyle A^{n}} is diagonalizable for some n {\displaystyle n} that is not an integer multiple of the characteristic of F {\displaystyle F} , then A {\displaystyle A} is diagonalizable. Proof: If A n {\displaystyle A^{n}} is diagonalizable, then A {\displaystyle A} is annihilated by some polynomial ( x n − λ 1 ) ⋯ ( x n − λ k ) {\displaystyle \left(x^{n}-\lambda _{1}\right)\cdots \left(x^{n}-\lambda _{k}\right)} , which has no multiple root (since λ j ≠ 0 {\displaystyle \lambda _{j}\neq 0} ) and is divided by the minimal polynomial of A {\displaystyle A} . </p><p>Over the complex numbers C {\displaystyle \mathbb {C} } , almost every matrix is diagonalizable. More precisely: the set of complex n × n {\displaystyle n\times n} matrices that are <i>not</i> diagonalizable over C {\displaystyle \mathbb {C} } , considered as a <a href="/facts/Subset/SJGERmbA">subset</a> of C n × n {\displaystyle \mathbb {C} ^{n\times n}} , has <a href="/facts/Lebesgue_measure/8us9cGoW">Lebesgue measure</a> zero. One can also say that the diagonalizable matrices form a dense subset with respect to the <a href="/facts/Zariski_topology/uolkKBpM">Zariski topology</a>: the non-diagonalizable matrices lie inside the <a href="/facts/Algebraic_variety/Vk7f97vn">vanishing set</a> of the <a href="/facts/Discriminant/0SmQtrvU">discriminant</a> of the characteristic polynomial, which is a <a href="/facts/Hypersurface/cTvtLVsV">hypersurface</a>. From that follows also density in the usual (<i>strong</i>) topology given by a <a href="/facts/Norm_(mathematics)/xIbR4uE1">norm</a>. The same is not true over R {\displaystyle \mathbb {R} } . </p><p>The <a href="/facts/Jordan%25E2%2580%2593Chevalley_decomposition/w4PN88Xo">Jordan–Chevalley decomposition</a> expresses an operator as the sum of its semisimple (i.e., diagonalizable) part and its <a href="/facts/Nilpotent/rERm2Pp7">nilpotent</a> 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 <a href="/facts/Jordan_form/M7ezYbnl">Jordan form</a> has no nilpotent part; i.e., each "block" is a one-by-one matrix. </p> <h2 id="diagonalization">Diagonalization</h2> <p class="note">See also: <a href="/facts/Eigendecomposition_of_a_matrix/a2nfF7hJ">Eigendecomposition of a matrix</a></p><p>Consider the two following arbitrary bases E = { e i | ∀ i ∈ [ n ] } {\displaystyle E=\{{{\boldsymbol {e}}_{i}|\forall i\in [n]}\}} and F = { α i | ∀ i ∈ [ n ] } {\displaystyle F=\{{{\boldsymbol {\alpha }}_{i}|\forall i\in [n]}\}} . Suppose that there exists a linear transformation represented by a matrix A E {\displaystyle A_{E}} which is written with respect to basis E. Suppose also that there exists the following eigen-equation: </p><p> A E α E , i = λ i α E , i {\displaystyle A_{E}{\boldsymbol {\alpha }}_{E,i}=\lambda _{i}{\boldsymbol {\alpha }}_{E,i}} </p><p>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 D F {\displaystyle D_{F}} which is a diagonal matrix that is similar to A E {\displaystyle A_{E}} . In other words, A E {\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 S {\displaystyle S} , which changes basis from E to F as follows: </p><p> D F = S E F A E S E − 1 F {\displaystyle D_{F}=S_{E}^{F}\ A_{E}\ S_{E}^{-1F}} , </p><p>where S E F {\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 P {\displaystyle P} which changes basis from F to E instead and so we have the following relationship : </p><p> S E − 1 F = P F E {\displaystyle S_{E}^{-1F}=P_{F}^{E}} </p><p>Both S {\displaystyle S} and P {\displaystyle P} transition matrices are invertible. Thus we can manipulate the matrices in the following fashion: D = S A E S − 1 D = P − 1 A E P {\displaystyle {\begin{aligned}D=S\ A_{E}\ S^{-1}\\D=P^{-1}\ A_{E}\ P\end{aligned}}} The matrix A E {\displaystyle A_{E}} will be denoted as A {\displaystyle A} , which is still in the E-basis. Similarly, the diagonal matrix is in the F-basis. </p> <p>If a matrix A {\displaystyle A} can be diagonalized, that is, </p> P − 1 A P = [ λ 1 0 ⋯ 0 0 λ 2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ λ n ] = D , {\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}}=D,} <p>then: </p> A P = P [ λ 1 0 ⋯ 0 0 λ 2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ λ n ] . {\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}}.} <p>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 α i {\displaystyle {\boldsymbol {\alpha }}_{i}} written in the basis of E so that we can represent P in block matrix form in the following manner: </p> P = [ α E , 1 α E , 2 ⋯ α E , n ] , {\displaystyle P={\begin{bmatrix}{\boldsymbol {\alpha }}_{E,1}&{\boldsymbol {\alpha }}_{E,2}&\cdots &{\boldsymbol {\alpha }}_{E,n}\end{bmatrix}},} <p>as a result we can write: A [ α E , 1 α E , 2 ⋯ α E , n ] = [ α E , 1 α E , 2 ⋯ α E , n ] D . {\displaystyle {\begin{aligned}A{\begin{bmatrix}{\boldsymbol {\alpha }}_{E,1}&{\boldsymbol {\alpha }}_{E,2}&\cdots &{\boldsymbol {\alpha }}_{E,n}\end{bmatrix}}={\begin{bmatrix}{\boldsymbol {\alpha }}_{E,1}&{\boldsymbol {\alpha }}_{E,2}&\cdots &{\boldsymbol {\alpha }}_{E,n}\end{bmatrix}}D.\end{aligned}}} </p><p>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: </p><p> A [ α E , 1 α E , 2 ⋯ α E , n ] = [ α E , 1 α E , 2 ⋯ α E , n ] [ λ 1 0 ⋯ 0 0 λ 2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ λ n ] . {\displaystyle A{\begin{bmatrix}{\boldsymbol {\alpha }}_{E,1}&{\boldsymbol {\alpha }}_{E,2}&\cdots &{\boldsymbol {\alpha }}_{E,n}\end{bmatrix}}={\begin{bmatrix}{\boldsymbol {\alpha }}_{E,1}&{\boldsymbol {\alpha }}_{E,2}&\cdots &{\boldsymbol {\alpha }}_{E,n}\end{bmatrix}}{\begin{bmatrix}\lambda _{1}&0&\cdots &0\\0&\lambda _{2}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &\lambda _{n}\end{bmatrix}}.} </p><p>Performing the above matrix multiplication we end up with the following result: A [ α 1 α 2 ⋯ α n ] = [ λ 1 α 1 λ 2 α 2 ⋯ λ n α n ] {\displaystyle {\begin{aligned}A{\begin{bmatrix}{\boldsymbol {\alpha }}_{1}&{\boldsymbol {\alpha }}_{2}&\cdots &{\boldsymbol {\alpha }}_{n}\end{bmatrix}}={\begin{bmatrix}\lambda _{1}{\boldsymbol {\alpha }}_{1}&\lambda _{2}{\boldsymbol {\alpha }}_{2}&\cdots &\lambda _{n}{\boldsymbol {\alpha }}_{n}\end{bmatrix}}\end{aligned}}} Taking each component of the block matrix individually on both sides, we end up with the following: </p> A α i = λ i α i ( i = 1 , 2 , … , n ) . {\displaystyle A{\boldsymbol {\alpha }}_{i}=\lambda _{i}{\boldsymbol {\alpha }}_{i}\qquad (i=1,2,\dots ,n).} <p>So the column vectors of P {\displaystyle P} are <a href="/facts/Right_eigenvector/8TjEoT8u">right eigenvectors</a> of A {\displaystyle A} , and the corresponding diagonal entry is the corresponding <a href="/facts/Eigenvalue/8TjEoT8u">eigenvalue</a>. The invertibility of P {\displaystyle P} also suggests that the eigenvectors are <a href="/facts/Linearly_independent/8JrHfIIa">linearly independent</a> and form a basis of F n {\displaystyle F^{n}} . This is the necessary and sufficient condition for diagonalizability and the canonical approach of diagonalization. The <a href="/facts/Row_vector/pPuRr9Xk">row vectors</a> of P − 1 {\displaystyle P^{-1}} are the <a href="/facts/Left_eigenvector/8TjEoT8u">left eigenvectors</a> of A {\displaystyle A} . </p><p>When a complex matrix A ∈ C n × n {\displaystyle A\in \mathbb {C} ^{n\times n}} is a <a href="/facts/Hermitian_matrix/qArPblYo">Hermitian matrix</a> (or more generally a <a href="/facts/Normal_matrix/WVKxrEK9">normal matrix</a>), eigenvectors of A {\displaystyle A} can be chosen to form an <a href="/facts/Orthonormal_basis/pzfKHlJG">orthonormal basis</a> of C n {\displaystyle \mathbb {C} ^{n}} , and P {\displaystyle P} can be chosen to be a <a href="/facts/Unitary_matrix/8fsxv3KD">unitary matrix</a>. If in addition, A ∈ R n × n {\displaystyle A\in \mathbb {R} ^{n\times n}} is a real <a href="/facts/Symmetric_matrix/08CGwjNl">symmetric matrix</a>, then its eigenvectors can be chosen to be an orthonormal basis of R n {\displaystyle \mathbb {R} ^{n}} and P {\displaystyle P} can be chosen to be an <a href="/facts/Orthogonal_matrix/WDXBXYB5">orthogonal matrix</a>. </p><p>For most practical work matrices are diagonalized numerically using computer software. <a href="/facts/Eigenvalue_algorithm/AYecadB0">Many algorithms</a> exist to accomplish this. </p> <h2 id="simultaneous-diagonalization">Simultaneous diagonalization</h2> <p class="note">See also: <a href="/facts/Triangular_matrix/qjXSFMM3">Simultaneous triangularisability</a>, <a href="/facts/Weight_(representation_theory)/7N8q5sw0">Weight (representation theory)</a>, and <a href="/facts/Positive_definite_matrix/Hbr6AuPS">Positive definite matrix</a></p> <p>A set of matrices is said to be <i>simultaneously diagonalizable</i> if there exists a single invertible matrix P {\displaystyle P} such that P − 1 A P {\displaystyle P^{-1}AP} is a diagonal matrix for every A {\displaystyle A} in the set. The following theorem characterizes simultaneously diagonalizable matrices: A set of diagonalizable <a href="/facts/Commuting_matrices/R1wGesmf">matrices commutes</a> if and only if the set is simultaneously diagonalizable.<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a>: p. 64 </p><p>The set of all n × n {\displaystyle n\times n} diagonalizable matrices (over C {\displaystyle \mathbb {C} } ) with n > 1 {\displaystyle n>1} is not simultaneously diagonalizable. For instance, the matrices </p> [ 1 0 0 0 ] and [ 1 1 0 0 ] {\displaystyle {\begin{bmatrix}1&0\\0&0\end{bmatrix}}\quad {\text{and}}\quad {\begin{bmatrix}1&1\\0&0\end{bmatrix}}} <p>are diagonalizable but not simultaneously diagonalizable because they do not commute. </p><p>A set consists of commuting <a href="/facts/Normal_matrix/WVKxrEK9">normal matrices</a> if and only if it is simultaneously diagonalizable by a <a href="/facts/Unitary_matrix/8fsxv3KD">unitary matrix</a>; that is, there exists a unitary matrix U {\displaystyle U} such that U ∗ A U {\displaystyle U^{*}AU} is diagonal for every A {\displaystyle A} in the set. </p><p>In the language of <a href="/facts/Lie_theory/SpBzkqrH">Lie theory</a>, a set of simultaneously diagonalizable matrices generates a <a href="/facts/Toral_Lie_algebra/AHP3rRDy">toral Lie algebra</a>. </p> <h2 id="examples">Examples</h2> <h3>Diagonalizable matrices</h3> <ul><li><a href="/facts/Involution_(mathematics)/kKxYrUVa">Involutions</a> are diagonalizable over the reals (and indeed any field of characteristic not 2), with ±1 on the diagonal.</li> <li>Finite order <a href="/facts/Endomorphism/s7NNlSea">endomorphisms</a> are diagonalizable over C {\displaystyle \mathbb {C} } (or any algebraically closed field where the characteristic of the field does not divide the order of the endomorphism) with <a href="/facts/Roots_of_unity/EqH0ho1Y">roots of unity</a> on the diagonal. This follows since the minimal polynomial is <a href="/facts/Separable_polynomial/5mcxhRGw">separable</a>, because the roots of unity are distinct.</li> <li><a href="/facts/Projection_(linear_algebra)/sElXGkxD">Projections</a> are diagonalizable, with 0s and 1s on the diagonal.</li> <li>Real <a href="/facts/Symmetric_matrices/08CGwjNl">symmetric matrices</a> are diagonalizable by <a href="/facts/Orthogonal_matrix/WDXBXYB5">orthogonal matrices</a>; i.e., given a real symmetric matrix A {\displaystyle A} , Q T A Q {\displaystyle Q^{\mathrm {T} }AQ} is diagonal for some orthogonal matrix Q {\displaystyle Q} . More generally, matrices are diagonalizable by <a href="/facts/Unitary_matrix/8fsxv3KD">unitary matrices</a> if and only if they are <a href="/facts/Normal_matrix/WVKxrEK9">normal</a>. In the case of the real symmetric matrix, we see that A = A T {\displaystyle A=A^{\mathrm {T} }} , so clearly A A T = A T A {\displaystyle AA^{\mathrm {T} }=A^{\mathrm {T} }A} holds. Examples of normal matrices are real symmetric (or <a href="/facts/Skew-symmetric_matrix/nw5CQeLN">skew-symmetric</a>) matrices (e.g. covariance matrices) and <a href="/facts/Hermitian_matrix/qArPblYo">Hermitian matrices</a> (or skew-Hermitian matrices). See <a href="/facts/Spectral_theorem/Vqc04B21">spectral theorems</a> for generalizations to infinite-dimensional vector spaces.</li></ul> <h3>Matrices that are not diagonalizable</h3> <p>In general, a <a href="/facts/Rotation_matrix/c2K2Ftl8">rotation matrix</a> is not diagonalizable over the reals, but all <a href="/facts/Rotation_matrix/c2K2Ftl8">rotation matrices</a> 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 as <a href="/facts/Jordan_Normal_Form/M7ezYbnl">Jordan normal form</a>. </p><p>Some matrices are not diagonalizable over any field, most notably nonzero <a href="/facts/Nilpotent_matrix/qIns3FJJ">nilpotent matrices</a>. This happens more generally if the <a href="/facts/Eigenvalues_and_eigenvectors/8TjEoT8u">algebraic and geometric multiplicities</a> of an eigenvalue do not coincide. For instance, consider </p> C = [ 0 1 0 0 ] . {\displaystyle C={\begin{bmatrix}0&1\\0&0\end{bmatrix}}.} <p>This matrix is not diagonalizable: there is no matrix U {\displaystyle U} such that U − 1 C U {\displaystyle U^{-1}CU} is a diagonal matrix. Indeed, C {\displaystyle C} has one eigenvalue (namely zero) and this eigenvalue has algebraic multiplicity 2 and geometric multiplicity 1. </p><p>Some real matrices are not diagonalizable over the reals. Consider for instance the matrix </p> B = [ 0 1 − 1 0 ] . {\displaystyle B=\left[{\begin{array}{rr}0&1\\\!-1&0\end{array}}\right].} <p>The matrix B {\displaystyle B} does not have any real eigenvalues, so there is no real matrix Q {\displaystyle Q} such that Q − 1 B Q {\displaystyle Q^{-1}BQ} is a diagonal matrix. However, we can diagonalize B {\displaystyle B} if we allow complex numbers. Indeed, if we take </p> Q = [ 1 i i 1 ] , {\displaystyle Q={\begin{bmatrix}1&i\\i&1\end{bmatrix}},} <p>then Q − 1 B Q {\displaystyle Q^{-1}BQ} is diagonal. It is easy to find that B {\displaystyle B} is the rotation matrix which rotates counterclockwise by angle θ = − π 2 {\textstyle \theta =-{\frac {\pi }{2}}} </p><p>Note that the above examples show that the sum of diagonalizable matrices need not be diagonalizable. </p> <h3>How to diagonalize a matrix</h3> <p>Diagonalizing a matrix is the same process as finding its <a href="/facts/Eigenvalues_and_eigenvectors/8TjEoT8u">eigenvalues and eigenvectors</a>, in the case that the eigenvectors form a basis. For example, consider the matrix </p> A = [ 0 1 − 2 0 1 0 1 − 1 3 ] . {\displaystyle A=\left[{\begin{array}{rrr}0&1&\!\!\!-2\\0&1&0\\1&\!\!\!-1&3\end{array}}\right].} <p>The roots of the <a href="/facts/Characteristic_polynomial/7m8roDqo">characteristic polynomial</a> p ( λ ) = det ( λ I − A ) {\displaystyle p(\lambda )=\det(\lambda I-A)} are the eigenvalues λ 1 = 1 , λ 2 = 1 , λ 3 = 2 {\displaystyle \lambda _{1}=1,\lambda _{2}=1,\lambda _{3}=2} . Solving the linear system ( 1 I − A ) v = 0 {\displaystyle \left(1I-A\right)\mathbf {v} =\mathbf {0} } gives the eigenvectors v 1 = ( 1 , 1 , 0 ) {\displaystyle \mathbf {v} _{1}=(1,1,0)} and v 2 = ( 0 , 2 , 1 ) {\displaystyle \mathbf {v} _{2}=(0,2,1)} , while ( 2 I − A ) v = 0 {\displaystyle \left(2I-A\right)\mathbf {v} =\mathbf {0} } gives v 3 = ( 1 , 0 , − 1 ) {\displaystyle \mathbf {v} _{3}=(1,0,-1)} ; that is, A v i = λ i v i {\displaystyle A\mathbf {v} _{i}=\lambda _{i}\mathbf {v} _{i}} for i = 1 , 2 , 3 {\displaystyle i=1,2,3} . These vectors form a basis of V = R 3 {\displaystyle V=\mathbb {R} ^{3}} , so we can assemble them as the column vectors of a <a href="/facts/Change_of_basis/Hnu1Spcc">change-of-basis</a> matrix P {\displaystyle P} to get: P − 1 A P = [ 1 0 1 1 2 0 0 1 − 1 ] − 1 [ 0 1 − 2 0 1 0 1 − 1 3 ] [ 1 0 1 1 2 0 0 1 − 1 ] = [ 1 0 0 0 1 0 0 0 2 ] = D . {\displaystyle P^{-1}AP=\left[{\begin{array}{rrr}1&0&1\\1&2&0\\0&1&\!\!\!\!-1\end{array}}\right]^{-1}\left[{\begin{array}{rrr}0&1&\!\!\!-2\\0&1&0\\1&\!\!\!-1&3\end{array}}\right]\left[{\begin{array}{rrr}1&\,0&1\\1&2&0\\0&1&\!\!\!\!-1\end{array}}\right]={\begin{bmatrix}1&0&0\\0&1&0\\0&0&2\end{bmatrix}}=D.} We may see this equation in terms of transformations: P {\displaystyle P} takes the standard basis to the eigenbasis, P e i = v i {\displaystyle P\mathbf {e} _{i}=\mathbf {v} _{i}} , so we have: P − 1 A P e i = P − 1 A v i = P − 1 ( λ i v i ) = λ i e i , {\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 P − 1 A P {\displaystyle P^{-1}AP} has the standard basis as its eigenvectors, which is the defining property of D {\displaystyle D} . </p><p>Note that there is no preferred order of the eigenvectors in P {\displaystyle P} ; changing the order of the <a href="/facts/Eigenvectors/8TjEoT8u">eigenvectors</a> in P {\displaystyle P} just changes the order of the <a href="/facts/Eigenvalues/8TjEoT8u">eigenvalues</a> in the diagonalized form of A {\displaystyle A} .<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> </p> <h2 id="application-to-matrix-functions">Application to matrix functions</h2> <p>Diagonalization can be used to efficiently compute the powers of a matrix A = P D P − 1 {\displaystyle A=PDP^{-1}} : </p> A k = ( P D P − 1 ) k = ( P D P − 1 ) ( P D P − 1 ) ⋯ ( P D P − 1 ) = P D ( P − 1 P ) D ( P − 1 P ) ⋯ ( P − 1 P ) D P − 1 = P D k P − 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}}} <p>and the latter is easy to calculate since it only involves the powers of a diagonal matrix. For example, for the matrix A {\displaystyle A} with eigenvalues λ = 1 , 1 , 2 {\displaystyle \lambda =1,1,2} in the example above we compute: </p> A k = P D k P − 1 = [ 1 0 1 1 2 0 0 1 − 1 ] [ 1 k 0 0 0 1 k 0 0 0 2 k ] [ 1 0 1 1 2 0 0 1 − 1 ] − 1 = [ 2 − 2 k − 1 + 2 k 2 − 2 k + 1 0 1 0 − 1 + 2 k 1 − 2 k − 1 + 2 k + 1 ] . {\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}}} <p>This approach can be generalized to <a href="/facts/Matrix_exponential/WID5cG2g">matrix exponential</a> and other <a href="/facts/Matrix_function/T2e9sFm9">matrix functions</a> that can be defined as power series. For example, defining exp ( A ) = I + A + 1 2 ! A 2 + 1 3 ! A 3 + ⋯ {\textstyle \exp(A)=I+A+{\frac {1}{2!}}A^{2}+{\frac {1}{3!}}A^{3}+\cdots } , we have: </p> exp ( A ) = P exp ( D ) P − 1 = [ 1 0 1 1 2 0 0 1 − 1 ] [ e 1 0 0 0 e 1 0 0 0 e 2 ] [ 1 0 1 1 2 0 0 1 − 1 ] − 1 = [ 2 e − e 2 − e + e 2 2 e − 2 e 2 0 e 0 − e + e 2 e − e 2 − e + 2 e 2 ] . {\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}}} <p>This is particularly useful in finding closed form expressions for terms of <a href="/facts/Linear_recursive_sequences/JolXC71M">linear recursive sequences</a>, such as the <a href="/facts/Fibonacci_number/mqxpAUQD">Fibonacci numbers</a>. </p> <h3>Particular application</h3> <p>For example, consider the following matrix: </p> M = [ a b − a 0 b ] . {\displaystyle M={\begin{bmatrix}a&b-a\\0&b\end{bmatrix}}.} <p>Calculating the various powers of M {\displaystyle M} reveals a surprising pattern: </p> M 2 = [ a 2 b 2 − a 2 0 b 2 ] , M 3 = [ a 3 b 3 − a 3 0 b 3 ] , M 4 = [ a 4 b 4 − a 4 0 b 4 ] , … {\displaystyle 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 } <p>The above phenomenon can be explained by diagonalizing M {\displaystyle M} . To accomplish this, we need a basis of R 2 {\displaystyle \mathbb {R} ^{2}} consisting of eigenvectors of M {\displaystyle M} . One such eigenvector basis is given by </p> u = [ 1 0 ] = e 1 , v = [ 1 1 ] = e 1 + e 2 , {\displaystyle \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},} <p>where e<i>i</i> denotes the standard basis of R<i>n</i>. The reverse change of basis is given by </p> e 1 = u , e 2 = v − u . {\displaystyle \mathbf {e} _{1}=\mathbf {u} ,\qquad \mathbf {e} _{2}=\mathbf {v} -\mathbf {u} .} <p>Straightforward calculations show that </p> M u = a u , M v = b v . {\displaystyle M\mathbf {u} =a\mathbf {u} ,\qquad M\mathbf {v} =b\mathbf {v} .} <p>Thus, <i>a</i> and <i>b</i> are the eigenvalues corresponding to u and v, respectively. By linearity of matrix multiplication, we have that </p> M n u = a n u , M n v = b n v . {\displaystyle M^{n}\mathbf {u} =a^{n}\mathbf {u} ,\qquad M^{n}\mathbf {v} =b^{n}\mathbf {v} .} <p>Switching back to the standard basis, we have </p> M n e 1 = M n u = a n e 1 , M n e 2 = M n ( v − u ) = b n v − a n u = ( b n − a n ) e 1 + b n e 2 . {\displaystyle {\begin{aligned}M^{n}\mathbf {e} _{1}&=M^{n}\mathbf {u} =a^{n}\mathbf {e} _{1},\\M^{n}\mathbf {e} _{2}&=M^{n}\left(\mathbf {v} -\mathbf {u} \right)=b^{n}\mathbf {v} -a^{n}\mathbf {u} =\left(b^{n}-a^{n}\right)\mathbf {e} _{1}+b^{n}\mathbf {e} _{2}.\end{aligned}}} <p>The preceding relations, expressed in matrix form, are </p> M n = [ a n b n − a n 0 b n ] , {\displaystyle M^{n}={\begin{bmatrix}a^{n}&b^{n}-a^{n}\\0&b^{n}\end{bmatrix}},} <p>thereby explaining the above phenomenon. </p> <h2 id="quantum-mechanical-application">Quantum mechanical application</h2> <p>In <a href="/facts/Quantum_mechanics/BRc3Mzgr">quantum mechanical</a> and <a href="/facts/Quantum_chemistry/H9AXV0lx">quantum chemical</a> computations matrix diagonalization is one of the most frequently applied numerical processes. The basic reason is that the time-independent <a href="/facts/Schr%25C3%25B6dinger_equation/3csgSIWF">Schrödinger equation</a> is an eigenvalue equation, albeit in most of the physical situations on an infinite dimensional <a href="/facts/Hilbert_space/aDFMh4lV">Hilbert space</a>. </p><p>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 the <a href="/facts/Variational_principle/a5UkKEBj">variational principle</a>, valid for Hamiltonians that are bounded from below. </p><p><a href="/facts/Perturbation_theory_(quantum_mechanics)/lYGgk9Ph">First-order perturbation theory</a> also leads to matrix eigenvalue problem for degenerate states. </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Defective_matrix/I5uysqAs">Defective matrix</a></li> <li><a href="/facts/Scaling_(geometry)/pFm6fvpu">Scaling (geometry)</a></li> <li><a href="/facts/Triangular_matrix/qjXSFMM3">Triangular matrix</a></li> <li><a href="/facts/Semisimple_operator/J3Jdl8Qh">Semisimple operator</a></li> <li><a href="/facts/Diagonalizable_group/Rl2Wq6YQ">Diagonalizable group</a></li> <li><a href="/facts/Jordan_normal_form/M7ezYbnl">Jordan normal form</a></li> <li><a href="/facts/Weight_module/7N8q5sw0">Weight module</a> – associative algebra generalization</li> <li><a href="/facts/Orthogonal_diagonalization/yrJhWcCc">Orthogonal diagonalization</a></li></ul> <h2 id="notes">Notes</h2> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Horn, Roger A.; Johnson, Charles R. (2013). Matrix Analysis, second edition. Cambridge University Press. ISBN 9780521839402. <a href="9780521839402" target="_blank">9780521839402</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Anton, H.; Rorres, C. (22 Feb 2000). Elementary Linear Algebra (Applications Version) (8th ed.). John Wiley & Sons. ISBN 978-0-471-17052-5. <a href="978-0-471-17052-5" target="_blank">978-0-471-17052-5</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> </ol>