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Wouldn't the two formulations give you completely different eigenvalues?

Seems dangerous and callous to me.

dustin
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Fraïssé
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    No, they will give identical solutions. In fact the only difference between them is a sign. – Vim Feb 26 '15 at 06:10
  • @Vim, then it is better to say that the do not give identical solutions, as they don't! – Mariano Suárez-Álvarez Feb 26 '15 at 06:32
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    @MarianoSuárez-Alvarez why? The only difference between the two polynomials is just a $+/-$, their roots are certainly identical – Vim Feb 26 '15 at 06:34
  • Simply because they are not identical! If there is difference, no matter how small, they simply are not the same polynomial. The two have the same roots and probably share other properties, but they are in general not the same polynomial, so saying that they are identical is simply wrong. – Mariano Suárez-Álvarez Feb 26 '15 at 06:36
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    @MarianoSuárez-Alvarez well I think there may be a misunderstanding here. What I mean is that their roots are identical, but not the polynomials themselves.. – Vim Feb 26 '15 at 07:03

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Remember that for any $n\times n$ matrix $X$ and any $r\in K$, with $K=\mathbb{R}$, $\mathbb{C}$,or any commutative ring, as @MarcvanLeeuwen points out. Then we have that: $$ \det (rX) = r^n \det(X). $$ In particular, if $r=-1$, you get $$ \det(-X)=(-1)^n\det(X). $$ So the roots of both of them are the same and their respective characteristic polynomials will only differ by a sign.

hjhjhj57
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    The restriction $K=\Bbb R$ or $\Bbb C$ is superfluous; any commutative ring will do. Indeed I think the coefficient ring of $A$ (which is what matters) is $K[\lambda]$ here. Also, since your matrix is going to be $\lambda I-A$ in the application to the question, maybe choosing another name than $A$ would be convenient here (it avoids having to say "apply this with $A$ equal to $\lambda I-A$"). – Marc van Leeuwen Feb 26 '15 at 06:15
  • @MarcvanLeeuwen Thank you for the useful comment. I wasn't sure about that and I didn't want to go through any problem with the entries of the matrices. I'll incorporate it into the answer. – hjhjhj57 Feb 26 '15 at 06:21
  • Now I need to know what a commutative ring is... O_O – Fraïssé Feb 26 '15 at 06:29
  • @IllegalImmigrant Need ? http://en.wikipedia.org/wiki/Ring_%28mathematics%29#Definition – hjhjhj57 Feb 26 '15 at 06:30
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    @hjhjhj57 Need to know but do not need to use... what a dilemma – Fraïssé Feb 26 '15 at 06:31
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For an $n\times n$ matrix $A$, those two determinants will be the same, except for a change of sign if $n$ is odd. Since you are finding the eigenvalues by setting the determinant equal to zero, the change of sign does not matter.

David
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  • So this is only helpful if you are computing the characteristic polynomial and det(A) =/= det(-A) in general – Fraïssé Feb 26 '15 at 06:05
  • No, it holds for any square matrix: if $M$ is $n\times n$ then $\det(-M)=(-1)^n\det(M)$. – David Feb 26 '15 at 06:09
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No the two determinants differ by a sign $(-1)^n$ where $n$ is the size of the matrix, and (for odd $n$) only one of them is the characteristic polynomial. Which one depends on the book you are using, but I think there is preference to defining it to be always monic, in other words taking $\det(\lambda I-A)$ as the definition of the characteristic polynomial.

This does not prevent either one or the other to be used to have a polynomial whose roots are precisely the eigenvalues; alternatively one could choose to use neither and find eigenvalues as roots of the minimal polynomial of$~A$. In fact there is no absolute link between the notions of eigenvalues and the characteristic polynomial: both are defined independently. So the chosen definition for characteristic polynomial does not affect the set of eigenvalues.

If all one does with the characteristic polynomial is to find the eigenvalues as its roots, then either definition will work equally well. However, there are more advanced applications, for which the characteristic polynomial always being monic is really more convenient (as always, one can make do with inconvenient conventions, like saying "when the minimal polynomial and characteristic polynomial agree up to sign"; only such work-arounds distract attention from what is essential).

Community
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Marc van Leeuwen
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