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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?

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3 Answers 3

up vote 3 down vote accepted

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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So in second case $det(A)=(-1)^n*c_0$ where $c_0$ is the constant term of $P(x)$? –  Robert S. Barnes Oct 15 '12 at 10:15
Here I've used the notation det(A) = |A|. Notice that the matrix on the LHS is $(A - \lambda I)$ and the matrix on the RHS is $(\lambda I - A)$. One property of det(A) is that if we multiply a row of A by a scalar c, the determinant becomes c*det(A). Now notice that $(\lambda I - A)$ is the same matrix we get by multiplying each row of $(A - \lambda I)$ by $(-1)$. So, we have $det(A - \lambda I) = (-1)^n det(\lambda I - A)$. –  Sean O'Brien Oct 15 '12 at 10:20
Yeah, according to this it looks like what I said is correct: mathworld.wolfram.com/CharacteristicPolynomial.html –  Robert S. Barnes Oct 15 '12 at 10:28

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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Why the downvote ? –  Belgi Oct 15 '12 at 9:43
I didn't vote you down, but you should take a look at: en.wikipedia.org/wiki/… –  Robert S. Barnes Oct 15 '12 at 10:16
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 $\det(A - \lambda I) = -\det(\lambda I - A)$. –  TMM Oct 15 '12 at 11:55

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