2
$\begingroup$

I know that this relation holds in several cases but I'm not sure about the full scope of it. So are there any cases when it doesn't hold?

$\endgroup$

2 Answers 2

6
$\begingroup$

Yes it's always true. It may happen, though, that the eigenvalues do not lie in the field you start with, so you have to enlarge the field to get them, like for the matrix $\left(\begin{array}{cc} 0&{-1}\\ 1&0\end{array}\right)$ which has no eigenvalue in the field of real numbers but does so in $\mathbb C$. Also, you always have to multiply the eigenvalues according to their multiplicity, i.e., one eigenvalue may appear several times. But you find all this in any book on Linear Algebra.

$\endgroup$
3
  • $\begingroup$ thanks, yes, i actually saw this in a linear algebra textbook but it wasn't very clearly put, so that's why i asked $\endgroup$ Jan 17, 2015 at 14:08
  • $\begingroup$ oh and do you mean the algebraic or the geometric multiplicity? i have a gut feeling that it has to be the algebraic one, but i'm not sure. $\endgroup$ Jan 17, 2015 at 14:20
  • $\begingroup$ Yes, the algebraic multiplicity it is. $\endgroup$
    – Echo
    Jan 17, 2015 at 14:53
1
$\begingroup$

the characteristic polynomial, whose roots are the eigenvalues, is the determinant equation: $$ \det(A-\lambda I) = 0 $$ the constant term (which is the product of the roots) is given by the value of the polynomial when $\lambda=0$, which is $\det(A)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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