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Today, I learnt that there is a surprisingly easy proof of the Cayley-Hamilton theorem: First we assume $F$ is algebraically closed WLOG, then we treat matrices as points in $F^{n^2}$ where $n$ is the dimension of the vector space $V$. Now, note that the condition that a matrix is annihilated by its characteristic polynomial is a zariski-closed condition, so we only need to verify this theorem for diagonal matrixes with distinct diagonal entries, which is transparent.

I'm deeply impressed by this proof, not only because it's short, but also because this idea seems unbelievably natural and straight-forward once one adopted some of the most basic ideas in Algebraic Geometry.

I'm looking forward to some background explanation or similarly easy-and-natural proof of the Cayley-Hamilton theorem.

penseur_32
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  • There are several proofs listed on the wikipedia page; you might find these to be interesting – Ben Grossmann May 14 '21 at 14:08
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    Can we consider easy a proof that relies on a heavier theory and other possibly heavy theorems ? –  May 14 '21 at 14:11
  • For diagonalizable matrices, this is easy: applying the polynomial to the matrix results in the polynomial being applied to the elements of the diagonal. –  May 14 '21 at 14:14
  • I see what you mean. Well, I think if that is indeed the case, then saying "easy" might be a little unreasonable, but I think a proof that relies on a heavier theory and other possibly heavy theorems might still be more $natural$ than more "elementary" proofs. By the way, the proof I mentioned neither relies on a heavier theory nor relies on other possibly heavy theorems, which is exactly why I find this proof extremely intriguing. – penseur_32 May 14 '21 at 14:17
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    The Cayley-Hamilton theorem holds in any commutative ring, and even in any commutative semiring (in a sligthly modified version). I am afraid your approach would not recover these results. – J.-E. Pin May 14 '21 at 14:25
  • There is a proof in "Undergraduate Commutative Algebra", by M. Reid, for the case of commutative rings: it is a quick and simple application of Cramer's rule, and only requires basic knowledge of determinants and the adjugate matrix. – Johnny El Curvas May 14 '21 at 14:27
  • Is it correct that you need Zorns Lemma (or the likes) to construct the algebraic closure? That is way beyond `easy' in my eyes. – daw May 14 '21 at 14:28
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    Also, when I first studied Linear Algebra we proved it for matrices in $\mathbb{C}$. First, you prove that every square matrix over $\mathbb{C}$ is triangularisable, and then check the Cayley-Hamilton Theorem for triangular matrices. For arbitrary fields, just take the splitting field of the characteristic polynomial and the proof is exactly the same. – Johnny El Curvas May 14 '21 at 14:31
  • @daw Yes, you need Zorn's Lemma. – Johnny El Curvas May 14 '21 at 14:45
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    The conventional proof is already very natural: since $\operatorname{adj}(Ix-A)(Ix-A)=p(x)I$ (where $p$ denotes the characteristic polynomial of $A$), $Ix-A$ factors $p$ on the right. Therefore $p(A)=0$, by the Factor Theorem. – user1551 May 14 '21 at 15:05
  • @user1551 ?? How do you get $;p(A)=0;$ exactly ? Apparently you seem to imply that you substitute $;A;$ for $;x;$ ...which of course cannot be done directly just like that. Am I missing something? – DonAntonio May 14 '21 at 15:54
  • @DonAntonio Let $A$ be an $n\times n$ matrix over a commutative ring $R$. The equality $\operatorname{adj}(xI-A)(xI-A)=p(x)I$ between members of $M_n(R[x])$ is identified with an equality between members of $(M_n(R))[x]$. E.g. when $A=\pmatrix{1&2\ 3&4}$, $$ \pmatrix{x-4&2\ 3&x-1}(xI-A)=(x^2-5x-2)I $$ gives rises to $$ \underbrace{\left[Ix+\pmatrix{-4&2\ 3&-1}\right]}{f(x)}(Ix-A) =\underbrace{Ix^2-5Ix-2I}{p(x)}. $$ The exact value of $f$ is unimportant. The key here is that $x-A$ factors $p$ on the right. By the Factor Theorem for polynomials over rings, $p(A)$ must be zero. – user1551 May 14 '21 at 17:58
  • @user1551 Again, I can't see why. You seem, as already I mentioned it, to imply that you substitute $;A;$ for $;x;$ in that polynomial equality and that's all...and that can't be done, as far as I am aware. And the factor theorem for polynomials works fine if the ring is, at least, a n integral domain, otherwise $;ab=0;$ (or anything equivalent to this) doesn't mean $;a=0;$ or $;b=0;$ . – DonAntonio May 14 '21 at 20:31
  • @DonAntonio I did substitute $A$ for $x$. This is justified by the factor theorem, which states that when $p=fg$ in $R[x]$ where $R$ is a ring, and if $g(a)=0$ for some $a\in R$, then $p(a)=0$. I am not talking about the converse (which is false). $R$ may contain zero divisors. – user1551 May 14 '21 at 21:11
  • @user1551 That's true when both the polynomials' coefficients and $;a;$ belong to $;R;$ . In this case we have a rather different thing, since the char. polynomial's coefficients belong to $;R;$ yet the matrix $;A;$ does not belong to $;R;$ . Only its entries do...! You may want to check one of the best explanations of this in this site, though it is in fact the standard way to rebuke the substitution mistake trying to prove CH theorem. First answer in https://math.stackexchange.com/questions/1141648/to-prove-cayley-hamilton-theorem-why-cant-we-substitute-a-for-lambda-in – DonAntonio May 14 '21 at 21:20
  • @DonAntonio My bad for overloading the symbols. Let $A\in M_n(R)$. The ring on which the factor theorem is involved is not the coefficient ring $R$, but the matrix ring $M_n(R)$ (and $A$ belongs to this ring). The polynomials $p,f$ and $g$ are polynomials in $x$ whose coefficients are matrices over $R$ rather than scalars in $R$. – user1551 May 14 '21 at 21:25
  • maybe helpful: @user1551 's post on this https://math.stackexchange.com/questions/3696442/on-the-cayley-hamilton-theorem/3800748#3800748 . Regarding simplest proofs: my vote is either do it for triangular matrices over $\mathbb C$ or arguably easier: work over $\mathbb Q$ and then consider the splitting field over the characteristic polynomial -- this shows C-H holds everywhere in $\mathbb Q$ (or $\mathbb C$ if preferred), then Principle of Permanence of Identities gives the result for matrices over a commutative ring with identity. – user8675309 May 14 '21 at 22:46

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