5.6 Hamilton-Cayley Theorem and Consequences 71
5.6 Hamilton-Cayley Theorem and Consequences.
5.6.1 Hamilton-Cayley Theorem
Any symmetric second rank tensorSμν, with principal valuesS(^1 ),S(^2 ),S(^3 )obeys
the equation
(
Sμν−S(^1 )δμν)(
Sνλ−S(^2 )δνλ)(
Sλκ−S(^3 )δλκ)
= 0 , (5.48)
which is essentially theHamilton-Cayley theorem. In linear algebra, this theorem is
applied to square matrices. To prove (5.48), notice firstly, that any vectora,in3D,
can be written as linear combination of the three unit vectorse(^1 ),e(^2 ),e(^3 )which are
parallel to the principal directions:aκ=a(^1 )e(κ^1 )+a(^2 )e(κ^2 )+a(^3 )e(κ^3 ). Multiplication
of the left hand side of (5.48) withaκyields zero. This can be checked, term by
term, using the eigenvalue relationSμνeν(i)=e(μi),fori= 1 , 2 ,3. In particular, one
has immediately(Sλκ−S(^3 )δλκ)eκ(^3 )=0. The multiplication(Sνλ−S(^2 )δνλ)(Sλκ−
S(^3 )δλκ)eκ(^2 )=(Sνλ−S(^2 )δνλ)(S(^2 )−S(^3 ))e(λ^2 )=0 again yields zero. Similarly, one
finds(Sμν−S(^1 )δμν)(Sνλ−S(^2 )δνλ)(Sλκ−S(^3 )δλκ)eκ(^1 )=(Sμν−S(^1 )δμν)(S(^1 )−
S(^2 ))(S(^1 )−S(^3 ))eν(^1 )=0. Since this result is valid for all vectors, the tensorial
relation (5.48) must hold true.
Explicit multiplication of the terms in (5.48) leads to
SμνSνλSλκ−(
S(^1 )+S(^2 )+S(^3 )
)
SμλSλκ+(
S(^1 )S(^2 )+S(^2 )S(^3 )+S(^3 )S(^1 )
)
Sμκ−S(^1 )S(^2 )S(^3 )δμκ= 0.SinceS(^1 )S(^2 )S(^3 ) =det(S), due toS(^1 )+S(^2 )+S(^3 )= SααandS(^1 )S(^2 )+
S(^2 )S(^3 )+S(^3 )S(^1 )=^12 ((S(^1 )+S(^2 )+S(^3 ))^2 −((S(^1 ))^2 +(S(^2 ))^2 +(S(^3 ))^2 ))=
1
2 (SααSββ−SαβSβα), the Hamilton-Cayley theorem is equivalent to
det(S)δμκ=SμνSνλSλκ−SννSμλSλκ+1
2
(SννSλλ−SνλSλν)Sμκ. (5.49)In the case of a symmetric traceless tensorS, this relation reduces to
det(
S
)
δμκ= Sμν SνλSλκ −1
2
SνλSλν Sμκ. (5.50)Consequences of the Hamilton-Cayley relations (5.49) and (5.50) are presented next.
The contraction withμ=κin the Hamilton-Cayley theorem (5.49) leads to the
previously presented equation (5.42) for the calculation of the determinant in terms