Faddeev–LeVerrier algorithm

 In mathematics (linear algebra), the Faddeev–LeVerrier algorithm is a recursive method to calculate the coefficients of the characteristic polynomial  of a square matrix, A, named after Dmitry Konstantinovich Faddeev and Urbain Le Verrier. Calculation of this polynomial yields the eigenvalues of A as its roots; as a matrix polynomial in the matrix A itself, it vanishes by the fundamental Cayley–Hamilton theorem. Computing determinant from the definition of characteristic polynomial, however, is computationally cumbersome, because  is new symbolic quantity, whereas this algorithm works directly with coefficients of matrix .

Urbain Le Verrier (1811–1877)
The discoverer of Neptune.

The algorithm has been independently rediscovered several times, in some form or another. It was first published in 1840 by Urbain Le Verrier, subsequently redeveloped by P. Horst, Jean-Marie Souriau, in its present form here by Faddeev and Sominsky, and further by J. S. Frame, and others.^{[1][2][3][4][5]} (For historical points, see Householder.^[6] An elegant shortcut to the proof, bypassing Newton polynomials, was introduced by Hou.^[7] The bulk of the presentation here follows Gantmacher, p. 88.^[8])

The AlgorithmEdit

The objective is to calculate the coefficients c_k of the characteristic polynomial of the n×n matrix A,

${\displaystyle p(\lambda )\equiv \det(\lambda I_{n}-A)=\sum _{k=0}^{n}c_{k}\lambda ^{k}~,}$

where, evidently, c_n = 1 and c₀ = (−1)ⁿ det A.

The coefficients are determined recursively from the top down, by dint of the auxiliary matrices M,

${\displaystyle {\begin{aligned}M_{0}&\equiv 0&c_{n}&=1\qquad &(k=0)\\M_{k}&\equiv AM_{k-1}+c_{n-k+1}I\qquad \qquad &c_{n-k}&=-{\frac {1}{k}}\mathrm {tr} (AM_{k})\qquad &k=1,\ldots ,n~.\end{aligned}}}$

Thus,

${\displaystyle M_{1}=I~,\quad c_{n-1}=-\mathrm {tr} A=-c_{n}\mathrm {tr} A;}$

${\displaystyle M_{2}=A-I\mathrm {tr} A,\quad c_{n-2}=-{\frac {1}{2}}{\Bigl (}\mathrm {tr} A^{2}-(\mathrm {tr} A)^{2}{\Bigr )}=-{\frac {1}{2}}(c_{n}\mathrm {tr} A^{2}+c_{n-1}\mathrm {tr} A);}$

${\displaystyle M_{3}=A^{2}-A\mathrm {tr} A-{\frac {1}{2}}{\Bigl (}\mathrm {tr} A^{2}-(\mathrm {tr} A)^{2}{\Bigr )}I,}$

${\displaystyle c_{n-3}=-{\tfrac {1}{6}}{\Bigl (}(\operatorname {tr} A)^{3}-3\operatorname {tr} (A^{2})(\operatorname {tr} A)+2\operatorname {tr} (A^{3}){\Bigr )}=-{\frac {1}{3}}(c_{n}\mathrm {tr} A^{3}+c_{n-1}\mathrm {tr} A^{2}+c_{n-2}\mathrm {tr} A);}$

etc.,^[9][10]   ...;

${\displaystyle M_{m}=\sum _{k=1}^{m}c_{n-m+k}A^{k-1}~,}$

${\displaystyle c_{n-m}=-{\frac {1}{m}}(c_{n}\mathrm {tr} A^{m}+c_{n-1}\mathrm {tr} A^{m-1}+...+c_{n-m+1}\mathrm {tr} A)=-{\frac {1}{m}}\sum _{k=1}^{m}c_{n-m+k}\mathrm {tr} A^{k}~;...}$

Observe A⁻¹ = − M_n /c₀ = (−1)ⁿ⁻¹M_n/detA terminates the recursion at λ. This could be used to obtain the inverse or the determinant of A.

DerivationEdit

The proof relies on the modes of the adjugate matrix, B_k ≡ M_n−k, the auxiliary matrices encountered.   This matrix is defined by

${\displaystyle (\lambda I-A)B=I~p(\lambda )}$

and is thus proportional to the resolvent

${\displaystyle B=(\lambda I-A)^{-1}I~p(\lambda )~.}$

It is evidently a matrix polynomial in λ of degree n−1. Thus,

${\displaystyle B\equiv \sum _{k=0}^{n-1}\lambda ^{k}~B_{k}=\sum _{k=0}^{n}\lambda ^{k}~M_{n-k},}$

where one may define the harmless M₀≡0.

Inserting the explicit polynomial forms into the defining equation for the adjugate, above,

${\displaystyle \sum _{k=0}^{n}\lambda ^{k+1}M_{n-k}-\lambda ^{k}(AM_{n-k}+c_{k}I)=0~.}$

Now, at the highest order, the first term vanishes by M₀=0; whereas at the bottom order (constant in λ, from the defining equation of the adjugate, above),

${\displaystyle M_{n}A=B_{0}A=c_{0}~,}$

so that shifting the dummy indices of the first term yields

${\displaystyle \sum _{k=1}^{n}\lambda ^{k}{\Big (}M_{1+n-k}-AM_{n-k}+c_{k}I{\Big )}=0~,}$

which thus dictates the recursion

${\displaystyle \therefore \qquad M_{m}=AM_{m-1}+c_{n-m+1}I~,}$

for m=1,...,n. Note that ascending index amounts to descending in powers of λ, but the polynomial coefficients c are yet to be determined in terms of the Ms and A.

This can be easiest achieved through the following auxiliary equation (Hou, 1998),

${\displaystyle \lambda {\frac {\partial p(\lambda )}{\partial \lambda }}-np=\operatorname {tr} AB~.}$

This is but the trace of the defining equation for B by dint of Jacobi's formula,

${\displaystyle {\frac {\partial p(\lambda )}{\partial \lambda }}=p(\lambda )\sum _{m=0}^{\infty }\lambda ^{-(m+1)}\operatorname {tr} A^{m}=p(\lambda )~\operatorname {tr} {\frac {I}{\lambda I-A}}\equiv \operatorname {tr} B~.}$

Inserting the polynomial mode forms in this auxiliary equation yields

${\displaystyle \sum _{k=1}^{n}\lambda ^{k}{\Big (}kc_{k}-nc_{k}-\operatorname {tr} AM_{n-k}{\Big )}=0~,}$

so that

${\displaystyle \sum _{m=1}^{n-1}\lambda ^{n-m}{\Big (}mc_{n-m}+\operatorname {tr} AM_{m}{\Big )}=0~,}$

and finally

${\displaystyle \therefore \qquad c_{n-m}=-{\frac {1}{m}}\operatorname {tr} AM_{m}~.}$

This completes the recursion of the previous section, unfolding in descending powers of λ.

Further note in the algorithm that, more directly,

${\displaystyle M_{m}=AM_{m-1}-{\frac {1}{m-1}}(\operatorname {tr} AM_{m-1})I~,}$

and, in comportance with the Cayley–Hamilton theorem,

${\displaystyle \operatorname {adj} (A)=(-)^{n-1}M_{n}=(-)^{n-1}(A^{n-1}+c_{n-1}A^{n-2}+...+c_{2}A+c_{1}I)=(-)^{n-1}\sum _{k=1}^{n}c_{k}A^{k-1}~.}$

The final solution might be more conveniently expressed in terms of complete exponential Bell polynomials as

${\displaystyle c_{n-k}={\frac {(-1)^{n-k}}{k!}}{\mathcal {B}}_{k}{\Bigl (}\operatorname {tr} A,-1!~\operatorname {tr} A^{2},2!~\operatorname {tr} A^{3},\ldots ,(-1)^{k-1}(k-1)!~\operatorname {tr} A^{k}{\Bigr )}.}$

ExampleEdit

${\displaystyle {\displaystyle A=\left[{\begin{array}{rrr}3&1&5\\3&3&1\\4&6&4\end{array}}\right]}}$

${\displaystyle {\displaystyle {\begin{aligned}M_{0}&=\left[{\begin{array}{rrr}0&0&0\\0&0&0\\0&0&0\end{array}}\right]\quad &&&c_{3}&&&&&=&1\\M_{\mathbf {\color {blue}1} }&=\left[{\begin{array}{rrr}1&0&0\\0&1&0\\0&0&1\end{array}}\right]&A~M_{1}&=\left[{\begin{array}{rrr}\mathbf {\color {red}3} &1&5\\3&\mathbf {\color {red}3} &1\\4&6&\mathbf {\color {red}4} \end{array}}\right]&c_{2}&&&=-{\frac {1}{\mathbf {\color {blue}1} }}\mathbf {\color {red}10} &&=&-10\\M_{\mathbf {\color {blue}2} }&=\left[{\begin{array}{rrr}-7&1&5\\3&-7&1\\4&6&-6\end{array}}\right]\qquad &A~M_{2}&=\left[{\begin{array}{rrr}\mathbf {\color {red}2} &26&-14\\-8&\mathbf {\color {red}-12} &12\\6&-14&\mathbf {\color {red}2} \end{array}}\right]\qquad &c_{1}&&&=-{\frac {1}{\mathbf {\color {blue}2} }}\mathbf {\color {red}(-8)} &&=&4\\M_{\mathbf {\color {blue}3} }&=\left[{\begin{array}{rrr}6&26&-14\\-8&-8&12\\6&-14&6\end{array}}\right]\qquad &A~M_{3}&=\left[{\begin{array}{rrr}\mathbf {\color {red}40} &0&0\\0&\mathbf {\color {red}40} &0\\0&0&\mathbf {\color {red}40} \end{array}}\right]\qquad &c_{0}&&&=-{\frac {1}{\mathbf {\color {blue}3} }}\mathbf {\color {red}120} &&=&-40\end{aligned}}}}$

Furthermore, ${\displaystyle {\displaystyle M_{4}=A~M_{3}+c_{0}~I=0}}$, which confirms the above calculations.

The characteristic polynomial of matrix A is thus ${\displaystyle {\displaystyle p_{A}(\lambda )=\lambda ^{3}-10\lambda ^{2}+4\lambda -40}}$; the determinant of A is ${\displaystyle {\displaystyle \det(A)=(-1)^{3}c_{0}=40}}$; the trace is 10=−c₂; and the inverse of A is

${\displaystyle {\displaystyle A^{-1}=-{\frac {1}{c_{0}}}~M_{3}={\frac {1}{40}}\left[{\begin{array}{rrr}6&26&-14\\-8&-8&12\\6&-14&6\end{array}}\right]=\left[{\begin{array}{rrr}0{.}15&0{.}65&-0{.}35\\-0{.}20&-0{.}20&0{.}30\\0{.}15&-0{.}35&0{.}15\end{array}}\right]}}$.

An equivalent but distinct expressionEdit

A compact determinant of an m×m-matrix solution for the above Jacobi's formula may alternatively determine the coefficients c,^[11][12]

${\displaystyle c_{n-m}={\frac {(-1)^{m}}{m!}}{\begin{vmatrix}\operatorname {tr} A&m-1&0&\cdots \\\operatorname {tr} A^{2}&\operatorname {tr} A&m-2&\cdots \\\vdots &\vdots &&&\vdots \\\operatorname {tr} A^{m-1}&\operatorname {tr} A^{m-2}&\cdots &\cdots &1\\\operatorname {tr} A^{m}&\operatorname {tr} A^{m-1}&\cdots &\cdots &\operatorname {tr} A\end{vmatrix}}~.}$

This article uses material from the Wikipedia article  Metasyntactic variable, which is released under the  Creative Commons Attribution-ShareAlike 3.0 Unported License.