Question on Hoffman and Kunze's proof of the Cayley-Hamilton theorem: why is $\det (xI-A) =x^2-\mathrm{Tr}(A)*x+\det(A)$

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?

• Most probably, they're looking at $n=2$. Do you have a reference to look at? (Page number, scan, &c) – Pedro Tamaroff Apr 13 '13 at 1:15
• Second edition page 195, just after the point where he takes the case n>2 – shooting-squirrel Apr 13 '13 at 1:22
• Try reading it carefully. They might be just looking at the case $n=2$ to develop some idea to work out the general case. – Pedro Tamaroff Apr 13 '13 at 1:23
• 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 $det B = f(T)$. – shooting-squirrel Apr 13 '13 at 1:25

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.
• 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..." – DonAntonio Apr 13 '13 at 1:25
• 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! – DonAntonio Apr 13 '13 at 1:32