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definition: Let T be a linear mapping on an n-dimensional vector space V with ordered basis $\beta$. Define the characteristic polynomial f(t) of T to to be the characteristic polynomial of $A=[T]_\beta$, that is, $f(t)=det(A-t I_n)$.

Here is another definition:If A is an $n \times n$ matrix, the polynomial $p(\lambda)=(-1)^n det(A- \lambda I)=det(\lambda I -A)$ is called the characteristic polynomial of A.

My question is do definitions collide over the change of positive/negative sign. I notice that they define a little differently in terms of $(-1)^n$

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    $\begingroup$ Yes, this is true. But the sign doesn't matter really. In particular, for eigenvalues, $p(\lambda)=0$ can be multiplied by $(-1)^n$. For polynomials, you would like a monic polynomial, i.e., you normalise the leading coefficient to $1$ anyway. $\endgroup$ Feb 27, 2020 at 20:41
  • $\begingroup$ @DietrichBurde But I was told that if I us$ det(A-\lambda I)$ in $3 \times 3$ matrix, then I would need to change sign. $\endgroup$
    – shine
    Feb 27, 2020 at 20:48
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    $\begingroup$ The first definition gives you a polynomial with leading coefficient $(-1)^n$, hence monic for even $n$, but not monic for odd $n$; the second definition gives you a monic polynomial for all $n$. Normally, we prefer monic polynomials to nonmonic ones (even when all we want is to find the roots). Here, think about the hand computation of $\det(A-\lambda I_n)$ vs. $\det(\lambda I_n-A)$. The former would involve taking the matrix $A$ and subtracting $-\lambda$ on the diagonal items before taking the determinant. The latter would be changing the sign of every entry, adding $\lambda$ in diag. $\endgroup$ Feb 27, 2020 at 22:02
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    $\begingroup$ (cont) so the former involves less work, and a smaller possibility of transcription errors. $\endgroup$ Feb 27, 2020 at 22:03
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    $\begingroup$ @shine: If you define the characteristic polynomial to be monic, or you use the second definition, then you would indeed get minus the characteristic polynomial in the $3\times 3$ case if you use $\det(A-\lambda I)$. $\endgroup$ Feb 27, 2020 at 22:05

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