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When calculating the characteristic polynomial as $$\det \; (A−t E_n)$$ I get the same polynomial as when I calculate the characteristic polynomial as $$\det\;(t E_n−A).$$ Only the signs are changed. Are they still aquivalent?

When heading on to the minimal polynomial we often multiply the whole polynomial by $−1$ to get the leading coefficient $=1$. Why is it that multiplying the minimal polynomial by $−1$ is not a problem?


Example

$$ A=\pmatrix{ 2 & 1 & -3 \\ 1 &2 & -3 \\ 1& 1 &-2 }.$$

Using the first definition of the characteristic polynomial I get

$$\chi_{A_1}(t) = \det \; (A - t E_n) = - t^3 + 2 t^2 -t = -t ( t-1) (t-1).$$

When continuing from there to get the minimal polynomial, which needs to be normed, I would first multiply by $(-1)$ to get

$$t \;\cdot \; ( -t + 1) \;\cdot \; (-t+1) = t \;\cdot \; (-1)\cdot (t-1) \;\cdot \; (-1)\cdot (t-1) = t \cdot ( t-1) \cdot (t-1)$$

Then I finally would get the minimal polynomial $$\mu(t) = t(t-1)$$

On the other hand, if I had defined the characteristic polynomial the other way round, it would result in the following polynomial

$$\chi_{A_2}(t) = \det \; (t E_n - A) = t^3 - 2 t^2 + t = t ( t-1) (t-1)$$

So this is already normed and I could go on to get the minimal polynomial easily.

But obviously $$t \chi_{A_2}(t) = t(t-1)(t-1) \neq -t (t-1)(t-1) =\chi_{A_1}(t).$$

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Note that what you call normed is usually called "monic" (i.e., the leading coefficient is $1$). No norm is involved. Also, you seem to suggest that you can deduce the minimal polynomial from the characteristic polynomial, which is not the case (it does happen to be $t(t-1)$ in this case, but for that you need to check that $A(A-E_3)$ is the null matrix). You don't in fact need the characteristic polynomial to define or to compute the minimal polynomial (though it can be of some help). –  Marc van Leeuwen Feb 5 '12 at 13:47

1 Answer 1

It really depends on the convention. The two polynomials you get are the same in even dimension, and opposite to each other in odd dimension.

In most English-speaking countries, the convention is to have the leading coefficient to be 1, ie. $\chi_A(t) = \det(tI-A)$. But in French-speaking countries (I don't know about other countries' conventions) it's $(-1)^{dim(V)}$, ie. $\chi_A(t) = \det(A-tI)$. It doesn't really matter, as the two polynomials you get are unique up to a unit of the underlying ring.

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The convention of non-monic definition characteristic polynomial is certainly not universal in France, and is tending to disappear. Although one can live with either convention (but not with both!) the monic one has several advantages (especially when comparing to the minimal polynomial, but also for instance in relation to algebraic integers, whose definition asks for a monic polynomial). –  Marc van Leeuwen Feb 5 '12 at 13:54
    
I have to admit, I'm just a student, but I've only ever encountered in France the definition with the constant term equal to $\det(A)$. But what are the advantages you speak of? When you compare polynomials between them you almost always factor them, so a leading coefficient isn't a problem, and for algebraic integers you can have as well a unit in factor of your polynomial, it doesn't change anything since you can multiply by its inverse. –  Najib Idrissi Feb 5 '12 at 14:13
    
Being a student you would maybe not be affected if your salary were multiplied by an invertible factor (just multiply by its inverse), but for me, I would care. Seriously though, it is rather convenient to be able to say "the companion matrix of $P$ has $P$ as its minimal polynomial and also as its characteristic polynomial" or in other cases to be able to provide the characteristic polynomial in a situation where a monic polynomial is required. Or just to ask "when are the minimal and characteristic polynomials equal?" on SE. –  Marc van Leeuwen Feb 5 '12 at 14:37
    
What prevents you from defining the minimal polynomial up to a unit? It seems to me it's an usual thing, for example for the GCD in a commutative ring, etc. Or just rephrasing the question as "when are they associated". (As a matter of fact I'm lucky enough to get a salary as a student :) ) –  Najib Idrissi Feb 5 '12 at 14:52
    
Defining things up to a unit factor is a real pain when writing expressions. You would do so when there is no alternative, as for GCD's in an abstract setting, but not when there is. Note that prime numbers for instance are defined to be positive, not "defined up to a sign", although that would suffice in some contexts. –  Marc van Leeuwen Feb 5 '12 at 14:56

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