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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).2012-02-05

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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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    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.2012-02-05