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The Cayley-Hamilton Theorem states that if we take the characteristic polynomial $p$ of an $n \times n$ matrix $A$

$$p(\lambda) = \det(A - \lambda I_n),$$

then

$$p(A) = 0.$$

What's wrong with my proof? I'm sure it's something simple, but I don't see it:

If we have $p$ as defined above, then simple replace $\lambda$ with $A$. This gives us

$$p(A) = \det(A - A I_n) = \det(A - A) = \det(0) = 0,$$

thereby proving the theorem.

feralin
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    The wikipedia page does explain why this isn't a real proof. In fact, it has a whole section on it. I suggest reading through that and then asking any questions about its explanation. – Hayden Nov 25 '15 at 05:11
  • Why do you think you can get: $P(A)=\det(A-AI_n)$? That's the fundamental error. The substitution is allowed form real $x$, but not obviously for any matrix $B$ that $P(B)=\det(A-B)$. – Thomas Andrews Nov 25 '15 at 05:15
  • @Hayden you're absolutely right. I'm not sure how I missed that when skimming the article. Thanks for pointing that out! – feralin Nov 25 '15 at 05:15
  • @ThomasAndrews can you explain why that is an error? It may be fundamental, but I don't fully understand why it is wrong. – feralin Nov 25 '15 at 05:16
  • I don't need to - you need t o explain why you can make that substitution. That's the way math works. :) But read the article. $\det(A-A)$ is a scalar, $P(A)$ is a matrix. They can't be equal. – Thomas Andrews Nov 25 '15 at 05:17
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    Try $f(\lambda) = trace(A-\lambda I_n)$ instead. By the same logic, one would conclude $f(A) = 0$, but this is not true. – Ted Nov 25 '15 at 05:33

1 Answers1

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In which ring do you work, in $K$ or in $K[A] \subset M_n(K)$ the commutative subring generated by the matrix $A$. If you work in $K$ you certainly cannot consider $A$ as an element of the ring. If you work in $K[A]$, what is the meaning of $A$ in $det(A-\lambda I_n)$, is it a scalar ?

There is however, a trick which can make the proof correct. We know (Cramer Formula)that for every matrix $M$, on has $M.\tilde M= det(M). Id$, where $\tilde M$ is the transpose of the cofactor matrix of $M$. Applying this formula to $M=A-\lambda Id$ in the ring $K[\lambda]$ we get : $(A-\lambda. I_n)(A_0+\lambda A_1+\lambda ^2 A_2++\lambda^{n_1}A_{n-1})= p(\lambda) Id$. This identity is written in $M_n(K[\lambda])$.

If $p(\lambda)=a_0+a_1\lambda+a_2\lambda ^2+..$, then: $AA_0= a_0 Id$, $AA_1-A_0=a_1 Id$, $AA_2-A_1...=a_2 Id$ etc.

Therefore $p(A)=a_0Id+a_1 A+a_2A^2+...=AA_0-AA_0+A^2A_1-A^2A_1+..=0$ Note that we use the fact that the matrices $A.A_{i+1}-A_i$ are proportional to $Id$ and commute with $A$.

Thomas
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