Thread: Cayley hamilton
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mathwonk
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Jun29-06, 08:56 PM
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If A is an n by n matrix of constants from the commutative ring k, and I is the identity n by n matrix, then Lagrange's expansion formula for determinants implies that

adj[XI-A].[XI-A] = f(X).I where f(X) is the characteristic polynomial of A, and adj denotes the classical adjoint whose entries are + or - the (n-1) by (n-1) minors of A.

Since setting X=A makes the left side equal to zero, we also have f(A) = 0. QED for cayley hamilton. (i.e. any square matrix A satisfies its own characteristic polynomial f(X) = det[XI-A].)


Q: is this a correct argument?

[hint: how could anything so simple and natural not be correct?]
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