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Cayley Hamilton Theorem 
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#1
May2514, 11:53 AM

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Given a Matrix A = [a,b;c,d] and it's characteristic polynomial, why does the characteristic polynomial enables us to determine the result of the Matrix A raised to the nth power?



#2
May2514, 03:12 PM

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The scalar "[itex]\lambda[/itex]" is an "eigenvalue" for matrix A if and only if there exist a nonzero vector, v, such that [itex]Av= \lambda v[/itex]. It can be shown that [itex]\lambda[/itex] is an eigenvalue for A if and only if it satisfies A's characteristic equation. A vector v, satisfying [itex]Av= \lambda v[/itex] is an eigenvector for A corresponding to eigenvalue [itex]\lambda[/itex] (some people require that v be nonzero to be an "eigenvector" but I prefer to include the 0 vector as an eigenvector for every eigenvalue).
Further, if we can find n independent eigenvectors for A (always true if A has n distinct eigenvalues but often true even if the eigenvalues are not all distinct) then the matrix, P, having those eigenvectors as columns is invertible and [itex]P^{1}AP=D[/itex] where D is the diagonal matrix having the eigenvalues of A on its diagonal. Then it is also true that [itex]PDP^{1}= A[/itex] and [tex]A^n= (PDP^{1})^n= (PDP^{1})(PDP^{1})\cdot\cdot\cdot(PDP^{1})= PD(P^{1}P)(D)(P^{1}P)\cdot\cdot\cdot(P^{1}P)DP= PD^nP^{1}[/tex] Of course, [itex]D^n[/itex] is easy to calculate it is the diagonal matrix having the nth power of the entries in D on its diagonal. Notice that this is "if we can find n independent eigenvectors for A". (Such a matrix is said to be "diagonalizable" matrix.) There exist nondiagonalizable matrices. They can be put in what is called "Jordan normal form" which is slightly more complicated than a diagonal matrix and it is a little more complicated to find powers. 


#3
May2514, 03:53 PM

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What HallsofIvy said is true, but I think it's not quite the point of the question.
The CayleyHamilton theorem says that the matrix ##A## satisfies its own characteristic equation. For an ##n \times n## matrix, the characteristic equation is of order ##n##, so ##A^n## is a linear combination of ##I, A, \dots, A^{n1}##. It follows that every power ##A^k## where ##k > n## is also a linear combination of the first ##n1## powers. For A ##2 \times 2## matrix, that means every power of ##A## is a linear combination of ##A## and ##I##. That is true even if the eigenvectors are not independent, and you can't diagonalize ##A##. 


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