Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

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.

share|improve this answer
    
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
1  
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
add comment

Since $\det(cX) = c^n\det(X) $for an $n×n$ matrix, both are equivalent for even $n$.

share|improve this answer
add comment

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

share|improve this answer
    
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.