# Characteristic and minimal polynomial - leading coefficient and norming

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

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.
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