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I was thinking about taking the Cayley-Hamilton Theorem in the opposite direction. Does this hold : If for $F\in$ End$(V)$ $\exists P\in K[t]$ such that $P(F)=0$ then $P$ is the characteristic polynomial of $F$ ?

If not could you give a counterexample?

sigmatau
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  • No - for instance, the zero endomorphism has characteristic polynomial $(-1)^n t^n,$ so for $n>k>1,$ $t^k$ kills this map as well. –  May 15 '13 at 21:31
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    I think the most (only?) sensible statement you can make in this direction is: if $P(F)=0$, then $P$ is a multiple of the minimal polynomial of $F$. –  May 15 '13 at 21:31
  • the question may sound stupid , i just finished reading the theorem and was too tired to think of a proof that the theorem does not hold in the opposite direction. thanks :) – sigmatau May 15 '13 at 21:35

2 Answers2

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No. If $p_F$ is the characteristic polynomial of $F$ and $q$ is any other polynomial, then $$P(F) = p_F(F) q(F) = 0 \cdot q(F) = 0.$$ So $P$ is a polynomial with $P(F) = 0$ that is not the characteristic polynomial of $F$ (just take $q$ to be nontrivial).

We can also have something like the following happen: Let $$F = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$ Then $$p_F(t) = (1 - t)^2.$$ But of course if $$q(t) = 1 - t,$$ we have $$q(F) = 0.$$

Henry T. Horton
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    Even if we additionally demand that $\deg P=\dim V$ and the leading term is $1$, the answer is still no. $P$ might simply be a suitable multiple of the minimal polynomial of $F$. – Hagen von Eitzen May 15 '13 at 21:31
  • @HagenvonEitzen I was editing an example in that direction in as you commented. :) – Henry T. Horton May 15 '13 at 21:33
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For an extreme example in which this fails, consider the zero matrix $\left(0\right) \in M_n(k)$. Here, the characteristic polynomial $\chi_0(t)=t^k$, corresponding to the fact that $0$ occurs as an eigenvalue of $(0)$ with multiplicity $n$. Obviously, though, $(0)$ is annihilated by many other endomorphisms, e.g. the identity map.

Your proposition happens to be true if and only if the characteristic polynomial of $A$ equals the minimal polynomial of $A$ (the polynomial that truly satisfies the statement of your proposition), and if you replace "is the characteristic polynomial" with "is a multiple of the characteristic polynomial".

Edit: In the comments below, it's pointed out that $\chi_A$ need not be irreducible for this to work. Of course, if $\chi_A$ is irreducible, then this is true (and what I meant, originally.)

awwalker
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    The second paragraph is wrong. Over $\Bbb C$ the characteristic polynomial cannot be irreducible unless $n=1$, yet the minimal polynomial is quite often equal to the characteristic polynomial, for instance if the latter has simple roots. For a discussion of when this happens exactly see http://math.stackexchange.com/q/81467/ for instance. – Marc van Leeuwen May 15 '13 at 21:56
  • @MarcvanLeeuwen Thanks for catching that. I realize now that I wanted "if", not "if and only if". – awwalker May 15 '13 at 22:21