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At one point, in the proof of the Cayley-Hamilton theorem the authors say that $$\det (xI-A) =x^2-\mathrm{Tr}(A)*x+\det(A)$$ for any $n\times n$ matrix that represents a linear operator, $I$ being the identity matrix. Should not that determinant be a polynomial of degree $n$? In that case how can it equal a polynomial of degree 2?

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  • $\begingroup$ Most probably, they're looking at $n=2$. Do you have a reference to look at? (Page number, scan, &c) $\endgroup$ – Pedro Tamaroff Apr 13 '13 at 1:15
  • $\begingroup$ Second edition page 195, just after the point where he takes the case n>2 $\endgroup$ – shooting-squirrel Apr 13 '13 at 1:22
  • $\begingroup$ Try reading it carefully. They might be just looking at the case $n=2$ to develop some idea to work out the general case. $\endgroup$ – Pedro Tamaroff Apr 13 '13 at 1:23
  • $\begingroup$ I'm absolutely sure that it computes the polynomial $ f = x^2−Tr(A)∗x+det(A) $ when he takes the case n=2, then for n>2 he says that $ det B = f(T) $. $\endgroup$ – shooting-squirrel Apr 13 '13 at 1:25
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I've the book ($2$nd edition) in front of me: if you mean at page 203, where it talks of "...where $\,f\,$ is the characteristic polynomial..." , 6 lines up they wrote "When $\,n=2\,$ ...", so this is only for $\,2\times 2\,$ matrices.

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  • $\begingroup$ Second edition page 195, just after the point where he takes the case n>2. $\endgroup$ – shooting-squirrel Apr 13 '13 at 1:21
  • $\begingroup$ Yes, I made a mistake: it is in page 195, not 203, and no: the formula $$f=x^2-(trace A)x+\det A$$ is one line immediately before they write "For the case $\,n>2 \,$, it is also clear..." $\endgroup$ – DonAntonio Apr 13 '13 at 1:25
  • $\begingroup$ But that makes no sense, he defines the symbol f, then when he takes the case n>2 he talks about f as it was already defined. Was he trying to say that "it is also clear that det(B) is a polynomial in T"? $\endgroup$ – shooting-squirrel Apr 13 '13 at 1:28
  • $\begingroup$ This is the whole parraph: When n = 2 Bii = 6ijT - AjJ. and B = T - Ad -&uI [ - A121 T - AnzI 1 det B = (T - Ad) (T - Ad) - AlsAd = T2 - (An + A&” + (An& - Add = f(T) where f is the characteristic polynomial: f = x2 - (trace A)z + det A. For the case n > 2, it is also clear that.... $\endgroup$ – DonAntonio Apr 13 '13 at 1:31
  • $\begingroup$ As you can see, the whole thing except the very last few words is under the assumption $\,n=2\,$ and there they define $\,f\,$ as the char. polynomial. not aterweards, when the $\,n>2\,$ case begins! $\endgroup$ – DonAntonio Apr 13 '13 at 1:32

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