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How do you find the minimal polynomial, $\mu_{M^{-1}}(x)$ given $\mu_{M}(x)$? My guess is since if $\lambda$ is an eigenvalue of $M$, then $1\over \lambda$ is an eigenvalue of $M^{-1}$, we might have something like $\mu_{M^{-1}}(x)=\mu_{M}({1\over x})$? But then I am not sure that that is the minimal polynomial...

Thanks

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up vote 12 down vote accepted

Yours is not a bad guess, but, to begin with, there is a problem: $\mu_M\left( \frac{1}{x} \right)$ needs not to be a polynomial.

Nevertheless, I think we can improve your idea: for any degree $n$ polynomial $p(x)$, define its conjugate (I'm not sure if this guy has already a name in the literature: please, correct me [EDIT. According to Georges, this is called the reciprocal polynomial ]) as

$$ \overline{p}(x) = x^n p\left( \frac{1}{x} \right) \ . $$

Clearly, the conjugate of a polynomial is still a polynomial, and you can easily verify that:

  1. $\overline{\overline{p}}(x) = p(x)$. [EDIT: if $p(x)$ has non-zero constant term. See Georges' comment.]
  2. $\overline{p(x)}\overline{q(x)} = \overline{p}(x)\cdot\overline{q}(x)$

I claim that the result is the following: if $\mu_M (x)= a_0 + a_1 x + \dots + a_{n-1}x^{n-1} + x^n$, then

$$ \mu_{M^{-1}} (x) = \frac{1}{a_0}\overline{\mu_M} (x) \ . $$

In order to prove this, we'll need the following lemma.

Lemma. Let $M$ be an invertible matrix and $p(x)$ a polynomial such that $p(M) = 0$. Then $\overline{p}(M^{-1}) = 0$.

Proof of the lemma. Indeed, $\overline{p}(M^{-1}) = (M^{-1})^n p(M) = 0$.

Hence, since $\mu_M$ annihilates $M$, so does $\overline{\mu_M}$ with $M^{-1}$.

We have to prove that $\frac{1}{a_0}\overline{\mu_M}$ has the characteristic property of the minimal polynomial of $M^{-1}$. Namely, that it has no proper divisor which also annihilates $M^{-1}$.

So, assume there were two polynomials $p(x), q(x)$ such that

$$ \frac{1}{a_0}\overline{\mu_M} (x) = p(x)q(x) $$

and moreover $p(M^{-1}) = 0$. Then, taking conjugates in this last equality, we would obtain

$$ \frac{1}{a_0}\mu_M (x) = \overline{p}(x)\cdot\overline{q}(x) \ . $$

But, because of the lemma, $\overline{p}(M) = 0$. So, by definition of the minimal polynomial of $M$,

$$ \mu_M (x) = \overline{p}(x) \qquad \text{(normalized)} \ . $$

Taking conjugates again, we would have that, up to a constant,

$$ \overline{\mu_M} (x) = p(x) \ . $$

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Thanks, Agusti! –  Eigen Nov 7 '11 at 18:52
    
My pleasure, Eigen. –  a.r. Nov 7 '11 at 19:06
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Dear Agustí, the guy is called reciprocal polynomial by his buddies. And reciprocating twice will not get you back where you started if the original polynomial has no constant term. (Fortunately this doesn't happen in the context of the question). –  Georges Elencwajg Nov 7 '11 at 20:12
    
You're right. Thank you, Georges. –  a.r. Nov 7 '11 at 21:10
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You're welcome, Agustí. I forgot to mention that this is a very nice answer! –  Georges Elencwajg Nov 7 '11 at 22:23
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