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proof of Cayley-Hamilton theorem by formal substitutions

Type of Math Object: 
Proof
Major Section: 
Reference

Mathematics Subject Classification

15A18 no label found15A15 no label found

Comments

The proof on this page is not quite correct. I found your page by a link from the wikipedia (http://en.wikipedia.org/wiki/Cayley-Hamilton_theorem), where I recently corrected the also erroneous proof that use to be there, and provided ample discussion of how and why, so please look there for details. When you say "we think of the formal symbol $t$ as representing an $n \times n$ matrix" you ignore the fact that already it has been treated as commuting with matrices, for instance by placing them arbitrarily at the left of the terms in the expansion of $B(t)$. So you simply cannot just substitute a matrix for $t$ and expect to get a valid identity, even if (like you do) you care about left and right _after_ the substitution.

A valid argument can be given by doing some non-commutative theory first, and carefully distinguishing left-evaluation from right-evaluation. For instance another proof I found on PlanteMath, http://planetmath.org/encyclopedia/ProofOfCayleyHamiltonTheoremInACommut... is basically correct, provided one supplies proper definitions of left factor and left hand value, and proves the result used (which however is about as hard as proving Cayley-Hamilton in the first place...).

Just a note. Here on PlanetMath we have a correction systems which allows you to request corrections to the author. Once these are fixed, the corrections are closed, meaning removed from view and the author is no longer bugged by the mistake. However, posts, such as this one, are never removed and so it is preferable to file a correction so that in the future, once the error is cleared up, the author's entry returns to normal. (You didn't know this but just info for the future.) Also, the corrections are stored in the system (and can be seen by all) and if the author does not fix them in time eventually the entry is placed on a pile for public editing.

Unlike Wikipedia, PlanetMath tries to give the original authors a chance to make changes in the style they feel matches the rest of their entry. They may also disagree with a correction so from time to time some interaction and communication is needed -- for which we have a messaging system and forums.
For us it works the best and we are usually quite happy to be told of incorrect aspects in our entry. So thank you very much for taking the time to indicate the mistake you found.

If I were you, I would file a correction with your same comments so that the article will begin the correction process as normal. Thanks for joining in. I hope we see more comments/corrections/entries from you in the future.

Hello van Leeuwen,
I think Steve's proof is some incomplete, but I cannot see mistakes there. My opinion is that one may circumvent commutativity issue. On the ring of matrices (an algebra) is valid X^{-1}X=XX^{-1}=I. The definition of matrix adjugate B(\lambda) leads to
(\lambda I-A)B(\lambda)=|\lambda I-A|I, (1)
and
B(\lambda)(\lambda I-A)=|\lambda I-A|I. (2)
In both equations the RHS as well as B(\lambda) on the LHS are polynomial matrices in \lambda. One see in those eq., that |\lambda I-A|I us divisible without remainder. So by applying generalized Bezout theorem,
http://planetmath.org/encyclopedia/GeneralizedBezoutTheoremOnMatrices.html
Cayley-Hamilton may be proved.
perucho

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