pA(λ)=\det(λIn-A)
λ
A
The algorithm has been independently rediscovered several times in different forms. 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 objective is to calculate the coefficients of the characteristic polynomial of the matrix,
pA(λ)\equiv\det(λIn-A)=\sum
n | |
k=0 |
ckλk~,
The coefficients are determined by induction on, using an auxiliary sequence of matrices
\begin{align} M0&\equiv0&cn&=1 &(k=0)\\ Mk&\equivAMk-1+cn-k+1I &cn-k&=-
1 | |
k |
tr(AMk) &k=1,\ldots,n~. \end{align}
Thus,
M1=I~, cn-1=-trA=-cntrA;
M2=A-ItrA, cn-2=-
1 | |
2 |
l(trA2-(trA)2r)= -
1 | |
2 |
(cntrA2+cn-1trA);
M3=A2-AtrA-
1 | |
2 |
l(trA2-(trA)2r)I,
cn-3=-\tfrac{1}{6}l((\operatorname{tr}A)3-3\operatorname{tr}(A2)(\operatorname{tr}A)+2\operatorname{tr}(A
| ||||
(cntr
3+c | |
A | |
n-1 |
trA2+cn-2trA);
Mm=
m | |
\sum | |
k=1 |
cn-m+kAk-1~,
cn-m=-
1 | |
m |
(cntr
m+c | |
A | |
n-1 |
trAm-1+...+cn-m+1trA)=-
1 | |
m |
m | |
\sum | |
k=1 |
cn-m+ktrAk~;...
Observe terminates the recursion at . This could be used to obtain the inverse or the determinant of .
The proof relies on the modes of the adjugate matrix,, the auxiliary matrices encountered. This matrix is defined by
B=(λI-A)-1I~pA(λ)~.
It is evidently a matrix polynomial in of degree . Thus,
B\equiv
n-1 | |
\sum | |
k=0 |
λk~Bk=
n | |
\sum | |
k=0 |
λk~Mn-k,
Inserting the explicit polynomial forms into the defining equation for the adjugate, above,
n | |
\sum | |
k=0 |
λk+1Mn-k-λk(AMn-k+ckI)=0~.
Now, at the highest order, the first term vanishes by =0; whereas at the bottom order (constant in, from the defining equation of the adjugate, above),
MnA=B0A=c0~,
n | |
\sum | |
k=1 |
λk (M1+n-k-AMn-k+ckI )=0~,
\therefore Mm=AMm-1+cn-m+1I~,
This can be easiest achieved through the following auxiliary equation (Hou, 1998),
λ
\partialpA(λ) | |
\partialλ |
-np=\operatorname{tr}AB~.
\partialpA(λ) | |
\partialλ |
=pA(λ)
infty | |
\sum | |
m=0 |
λ-(m+1)\operatorname{tr}Am=pA(λ)~\operatorname{tr}
I | |
λI-A |
\equiv\operatorname{tr}B~.
Inserting the polynomial mode forms in this auxiliary equation yields
n | |
\sum | |
k=1 |
λk (kck-nck-\operatorname{tr}AMn-k )=0~,
n-1 | |
\sum | |
m=1 |
λn-m (mcn-m+\operatorname{tr}AMm )=0~,
\therefore cn-m=-
1 | |
m |
\operatorname{tr}AMm~.
Further note in the algorithm that, more directly,
Mm=AMm-1-
1 | |
m-1 |
(\operatorname{tr}AMm-1)I~,
\operatorname{adj}(A)=(-1)n-1Mn=(-1)n-1(An-1+cn-1An-2+...+c2A+c1I)=(-1)n-1
n | |
\sum | |
k=1 |
ckAk-1~.
The final solution might be more conveniently expressed in terms of complete exponential Bell polynomials as
cn-k=
(-1)n-k | |
k! |
{lB}kl(\operatorname{tr}A,-1!~\operatorname{tr}A2,2!~\operatorname{tr}A3,\ldots,(-1)k-1(k-1)!~\operatorname{tr}Akr).
{\displaystyleA=\left[{\begin{array}{rrr}3&1&5\\3&3&1\\4&6&4\end{array}}\right]}
{\displaystyle{\begin{aligned}M0&=\left[{\begin{array}{rrr}0&0&0\\0&0&0\\0&0&0\end{array}}\right] &&&c3&&&&&=&1\\M\color{blue1}&=\left[{\begin{array}{rrr}1&0&0\\0&1&0\\0&0&1\end{array}}\right]&A~M1&=\left[{\begin{array}{rrr}\color{red3}&1&5\\3&\color{red3}&1\\4&6&\color{red4}\end{array}}\right]&c2&&&=-{
1 | |
\color{blue1 |
}}\color{red10}&&=&-10\\M\color{blue2}&=\left[{\begin{array}{rrr}-7&1&5\\3&-7&1\\4&6&-6\end{array}}\right] &A~M2&=\left[{\begin{array}{rrr}\color{red2}&26&-14\\-8&\color{red-12}&12\\6&-14&\color{red2}\end{array}}\right] &c1&&&=-{
1 | |
\color{blue2 |
}}\color{red(-8)}&&=&4\\M\color{blue3}&=\left[{\begin{array}{rrr}6&26&-14\\-8&-8&12\\6&-14&6\end{array}}\right] &A~M3&=\left[{\begin{array}{rrr}\color{red40}&0&0\\0&\color{red40}&0\\0&0&\color{red40}\end{array}}\right] &c0&&&=-{
1 | |
\color{blue3 |
}}\color{red120}&&=&-40\end{aligned}}}
Furthermore,
{\displaystyleM4=A~M3+c0~I=0}
The characteristic polynomial of matrix is thus
{\displaystylepA(λ)=λ3-10λ2+4λ-40}
{\displaystyle\det(A)=(-1)3c0=40}
{\displaystyleA-1=-{
1 | |
c0 |
A compact determinant of an ×-matrix solution for the above Jacobi's formula may alternatively determine the coefficients,[11] [12]
cn-m=
(-1)m | |
m! |
\begin{vmatrix}\operatorname{tr}A&m-1&0& … &0\\ \operatorname{tr}A2&\operatorname{tr}A&m-2& … &0\\ \vdots&\vdots&&&\vdots\\ \operatorname{tr}Am-1&\operatorname{tr}Am-2& … & … &1\\ \operatorname{tr}Am&\operatorname{tr}Am-1& … & … &\operatorname{tr}A\end{vmatrix}~.
Barbaresco F. (2019) Souriau Exponential Map Algorithm for Machine Learning on Matrix Lie Groups. In: Nielsen F., Barbaresco F. (eds) Geometric Science of Information. GSI 2019. Lecture Notes in Computer Science, vol 11712. Springer, Cham. https://doi.org/10.1007/978-3-030-26980-7_10