# proof of Cayley-Hamilton theorem by formal substitutions

## Primary tabs

Type of Math Object:
Proof
Major Section:
Reference

## Mathematics Subject Classification

15A18 Eigenvalues, singular values, and eigenvectors
15A15 Determinants, permanents, other special matrix functions

### proof needs correction

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...).

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### Re: proof needs correction

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