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According to the definition of an eigenvalue it satisfies the equation

$Ax=\lambda x$ where $A\in M_{n\times n}^{\mathbb{F}}$.

So that we could have either:

$(A-\lambda I)x=0$ or $(\lambda I-A)x=0$

such that the characteristic equation is either

$det(A-\lambda I)=0$ or $det(\lambda I-A)=0$.

What practical differences result from the two possible definitions?

3 Answers 3

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There will be no difference in the roots of the characteristic equation.

$ \left| \left( \begin{matrix} a_{11} - \lambda & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} - \lambda & & \\ . \\ . \\ . \\ a_{n1} & ... & & a_{nn} - \lambda \end{matrix} \right) \right| = (-1)^n \left| \left( \begin{matrix} \lambda - a_{11} & - a_{12} & ... & - a_{1n} \\ - a_{21} & \lambda - a_{22} & & \\ . \\ . \\ . \\ -a_{n1} & ... & & \lambda - a_{nn} \end{matrix} \right) \right| $

If $P(x)$ has a root $r$, then $-P(x)$ will have the same root.

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    Yeah, according to this it looks like what I said is correct: http://mathworld.wolfram.com/CharacteristicPolynomial.html2012-10-15
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Since $\det(cX) = c^n\det(X) $for an $n×n$ matrix, both are equivalent for even $n$.

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The definition of the characteristic polynomial is$\det(\lambda I-A)$ and not $\det(A-\lambda I)$ because if $A$ is an $n\times n$ matrix then the characteristic polynomial would of had a leading coefficient$-1$ and not $1$ for odd $n$.

In any case, there is not big difference here since if $\det(\lambda I-A)=P(\lambda)$ then $\det(A-\lambda I)=-P(\lambda)$ and in particular they have the same roots.

Also note $(\lambda I-A)x=0\iff(A-\lambda I)x=0$

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    I downvoted for several reasons: (i) stating that the definition $\det(A - \lambda I)$ is "wrong"; (ii) writing "would of had"; and (iii) stating that $\det(A - \lambda I) = -\det(\lambda I - A)$.2012-10-15