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File: Matrix Pdf 174098 | Cayley
determinants and the cayley hamilton theorem tristan giron throughout this note we try to follow the same notations than in the course f is a eld unless otherwise specied f ...

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                  DETERMINANTS AND THE CAYLEY-HAMILTON THEOREM
                                               TRISTAN GIRON
                Throughout this note, we try to follow the same notations than in the course.
                F is a field, unless otherwise specified F = C.  A will be a matrix in M (F)
                                                                                        n
                representing the linear transformation α : V → V in a given basis of the n-
                dimensional F-vector space V. χ = χ shall denote its characteristic polynomial,
                                              α     A
                and m =m its minimal polynomial.
                      α    A
                Afundamental result in the study of linear transformations is the Cayley-Hamilton
                theorem.
                THEOREM 1 (Cayley-Hamilton). The characteristic polynomial is annihilator.
                In other words,
                                                 χ (α) = 0.
                                                  α
                Exercise 2. Prove the Cayley-Hamilton theorem in the case where α is represented
                by a triangular matrix.
                                                                                           B
                Hint: Proceed by induction. If B = (b ,...,b ) is a basis of V such that A = (α)
                                                   1     n                                 B
                is triangular in B, consider V = span(b ,...,b ) for 0 ≤ k ≤ n and study the linear
                                           k        1      k
                transformation α −µkId, where µk = Akk, for 1 ≤ k ≤ n.
                A bit harder: show that this proof in fact implies the general Cayley-Hamilton as
                stated above.
                This is in essence the proof that you have been given in your course for the
                Cayley-Hamilton theorem.
                In this note we would like to give another proof of the Cayley-Hamilton theorem
                using more directly properties of the determinant.
                                             I. Determinants
                Wereview the main elements of the theory of determinants.
                Definition 3. Let v1,...,vk ∈ V be vectors of V, k ≥ 1.
                     • A map φ : V ×···×V → F is k-multilinear if it is linear in every variable,
                       i.e. for λ ∈ F,w ∈ V, and 1 ≤ j ≤ k,
                     φ(v1,...,vj +λw,...,vk) = φ(v1,...,vj,...,vk)+λφ(v1,...,w,...,vk).
                     • A k-multilinear map is alternated if for all permutation σ ∈ S , we have
                                                                                k
                                 φ(vσ(1),vσ(2),...,vσ(k)) = ε(σ)φ(v1,v2,...,vk).
                       Here ε(σ) is the signature of the permutation σ.
                  Date: December 13, 2019.
                                                      1
                   2                                      TRISTAN GIRON
                   Exercise 4. It is possible that in first year you were taught that the signature of a
                   permutation is the determinant of the permutation matrix, which would make this
                   section circular. Let us therefore give an independent definition.
                                                                                           2
                   Let σ ∈ S , and define i(σ) to be the number of pairs (l,m) ∈ N such that 1 ≤ l <
                              k
                   m≤kandσ(l)>σ(m). Define ε(σ):=(−1)i(σ).
                         • Show that for arbitrary real numbers x ,...,x ∈ R, we have
                                                                       1       k
                                          Y (xσ(l)−xσ(m)) = ε(σ)           Y (xl−xm).
                                        1≤l
						
									
										
									
																
													
					
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