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What is wrong with this proof of Cayley hamilton?

If $A$ is $n \times n$ matrix and $P$ is its characteristic polynomial then $P(A) = 0.$

Proof: $P(x) = \det(A - xI) \implies P(A) = \det(A - A) = \det(0) = 0.$

coldnumber
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Sara
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    How do you think polynomials in matrices work? Remember that we have not commutativity here, so for example it is not necessarily true that $;(C-A)(C-B)=C^2-(A+B)C+AB;$ , for any three square matrices of order $;n\ge2;$ , but rather $;(C-A)(C-B)=C^2-CB-AC+AB;$ . Check now why this kind of things makes your "proof" wrong. – Timbuc Jul 15 '15 at 08:36
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    Cayley-Hamilton states that $P(A)$ is the zero matrix, not the number $0$. – Empiricist Jul 15 '15 at 08:44
  • It seems that this question pops up more often: To prove Cayley-Hamilton theorem, why can't we substitute $A$ for $\lambda$ in $p(\lambda)=det(\lambda I−A)$? – Josse van Dobben de Bruyn Jul 15 '15 at 08:49
  • refer to matrix mathematics page 243 for a proof. $P(A)$ is a zero matrix but $P(x)$ is zero. This is one reason why your proof can't be write. – Sepideh Abadpour Jul 15 '15 at 09:28
  • If $p(\lambda)$ is the characteristic polynomial, then $p(\lambda)I=(Q_{0}+Q_{1}\lambda + \cdots + Q_{n-1}\lambda^{n-1})(A-\lambda I)$, where $Q_{k}$ are $n\times n$ matrices. (You can derive that factoring from the adjunct matrix.) That IS enough to imply that $p(A)=0$. – Disintegrating By Parts Jul 16 '15 at 03:53
  • TrialAnd Error, you are correct but it may need a little argument, since you cannot just substitute a matrix into a polynomial with (non commutative) matrix coefficients. I.e. as you no doubt know, there is right and left substitution, and you need the lemma that right substitution yields the remainder after right division, (same for left). I.e. even though right evaluation does not respect products, you still get that right evaluation at A is zero iff right division by (A-cI) has zero remainder. This proof is on pages 40, 84, in A.Adrian Albert's Fundamental concepts of higher algebra. – roy smith Sep 19 '16 at 19:09

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