2
$\begingroup$

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 $(-1)^n$

$\endgroup$
5
  • 2
    $\begingroup$ 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. $\endgroup$
    – Dietrich Burde
    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
  • 1
    $\begingroup$ 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. $\endgroup$
    – Arturo Magidin
    Feb 27, 2020 at 22:02
  • 1
    $\begingroup$ (cont) so the former involves less work, and a smaller possibility of transcription errors. $\endgroup$
    – Arturo Magidin
    Feb 27, 2020 at 22:03
  • 1
    $\begingroup$ @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 $\det(A-\lambda I)$. $\endgroup$
    – Arturo Magidin
    Feb 27, 2020 at 22:05

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.