Home > Algebra, Higher Algebra > Cayley Hamilton Theorem alternative proof

Cayley Hamilton Theorem alternative proof


If p=\det(A-\lambda I) is the characteristic polynomial of a square matrix (linear operator) A, then p(A)=0.

Proof: If all the eigenvalues of A are distinct, then there is a base in which A has diagonal matrix, which means that there exists a diagonal matrix consisting of eigenvalues of A, D=diag(\lambda_1,...,\lambda_n) and an invertible matrix T such that A=TDT^{-1}. Then if q is a polynomial we have q(A)=T\cdot diag(q(\lambda_1),...,q(\lambda_n)) \cdot T^{-1}. If q=p then obviously p(A)=0.

If A does not have distinct eigenvalues, then, using this property there exists a sequence (A_r)_{r \geq 1} of matrices which converges to A and each A_r has distinct eigenvalues. Therefore p(A_r) \to p(A), since a polynomial is a continuous function. Now take for every matrix A_r its characteristic polynomial p_r. Then p_r(A_r)=0 and furthermore, since the coefficients of p_r are continuous functions of elements of A_r, we have p_r(X)-p(X) \to 0 for every matrix X, when r \to \infty. This means that p_r(A_r)-p(A_r)\to 0 when r \to \infty, from which we conclude that p(A_r) \to 0 and p(A_r) \to p(A) when r \to \infty. Therefore p(A)=0.

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