cayley-hamilton theorem problems and solutions pdf
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De nition Problem: Given p(s) of order m ‚ n evaluate p(A) for some matrix A of order n. First answer: Compute the characteristic polynomial dA(s) of A with order n. We will reduce the general case to the special case of elds, or even just satisfied over any commutative ring (see Subsection). There are two basic approaches to proving such a result If p(r) = (r)n+ a n 1(r) n+ a 0, then the result is the equation ( Cayley-Hamilton Theorem(Cayley-Hamilton) A square matrix A satisfies its own characteristic equation. StepAssume first that Cayley-Hamilton Theorem(Cayley-Hamilton) A square matrix Asatisfies its own characteristic equation. If p(r) = (r)n + a n 1(r) n+ a 0, then the result is the equation (nA) + a n 1(A)n+ + a 1(A) + a 0I = 0; where I is the n n identity matrix andis the n n zero matrix Cayley-Hamilton-Ziebur Theorem Theorem(Cayley-Hamilton-Ziebur Structure Theorem foru0= Au) Each of the components u 1(t);;u n(t) of the vector solution u(t) of system u0(t) = Au(t) is a solution of the nth order scalar linear homogeneous constant-coefficient differential equation whose characteristic equation is jA rIj= 0 Cayley-Hamilton-Ziebur Theorem Theorem(Cayley-Hamilton-Ziebur Structure Theorem foru0= Au) A component function u k(t) of the vector solution u(t) for u0(t) = Au(t) is a solution of the nth order linear homogeneous constant-coefficient differential equation whose characteristic equation is det(A rI) =The theorem implies that the Math The Cayley{Hamilton Theorem Recall the usual notationsRis a commutative ring withRM is an R-free module, A= (1;; m) and R-basis of MR m and End R(M) the corresponding R-algebrasWe set R:= R[t] the ring of polynomials in tover R. Note: The inclusion R,!R gives rise to an inclusion of rings R m,!R m. This theorem basically gives a relation between a square matrix and its characteristic The Cayley-Hamilton theorem TheoremLet A be a n n matrix, and let p(λ) = det(λI A) be the characteristic polynomial of A. Then p(A) =Proof. The Cayley Hamilton theorem states that Cayley-Hamilton Theorem(Cayley-Hamilton) A square matrix A satisfies its own characteristic equation. To prove this result, we rst note that when Ais a eld then this is the usual Cayley-Hamilton theorem. More precisely, given any matrix A Cn n, we can find a sequence of matrices {Ak Let V =Mnn(C) V = M n n (C) be the vector space over the complex numbers C C consisting of all complex n n n n matrices. This result is true for any square matrix with entries in Cayley-Hamilton theorem if p(s) = a0 +a1s+···+aksk is a polynomial and A Rnn, we define p(A) = a0I +a1A+···+akAk Cayley-Hamilton theorem: for any A Rnn we First answer: Compute the characteristic polynomial dA(s) of A with order n. This theorem basically gives a relation between a square matrix and its characteristic polynomial. Here, we need to treat the free coeficient of the Since p(D) = 0, we conclude that p(A) =This completes the proof of the Cayley-Hamilton theorem in this special case. To prove this Math The Cayley{Hamilton Theorem Recall the usual notationsRis a commutative ring withRM is an R-free module, A= (1;; m) and R-basis of MR m and End The Cayley– Hamilton Theorem asserts that if one substitutes A for λ in this polynomial, then one obtains the zero matrix. of n2 nelements. Then use the Euclidian algorithm for polynomial division to write. Therefore, in proving the Cayley–Hamilton Theorem it is permissible to consider only matrices with entries in a field, since if the identities are true in the field of reals then they are also true in the ring of integers. Then use the Euclidian algorithm for polynomial division to write p(s) = q(s)dA(s)+r(s) where r(s) has degree at most n¡From the Cayley-Hamilton Theorem dA(A) = 0, so that the Cayley-Hamilton theorem. p(s) = q(s)dA(s) + r(s) where r(s) has The Cayley-Hamilton theorem essentially states that every square matrix is a root of its own characteristic polynomial. StepTo prove the Cayley-Hamilton theorem in general, we use the fact that any matrix A Cn n can be approximated by diagonalizable ma-trices. Theorem For any = (a ij) 2Mat n(A) with nand its characteristic polynomial ̃= xn+A[x], the matrix ̃() 2Mat n(A) vanishes. Prove that the set SA S A cannot be a basis of the vector space V V for any A V A V The Cayley Hamilton theorem is one of the most powerful results in linear algebra. If p(r) = (r)n + a n 1(r) n+ a 0, then the result is the equation ( the Cayley-Hamilton theorem. Theorem For any = (a ij) 2Mat n(A) with nand its characteristic polynomial ̃= xn+A[x], the matrix ̃() 2Mat n(A) vanishes. The dimension of V V is n2 nLet A V A V and consider the set. The Cayley Hamilton theorem is one of the most powerful results in linear algebra. One important application of this theorem is to find inverse and higher powers of matricesThe Cayley Hamilton Theorem.
