Faddeev–LeVerrier algorithm

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.

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

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

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

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

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

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

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

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

• Exterior algebra § Leverrier's algorithm
• Jacobi's formula
• Fredholm determinant