Suppose characteristic polynomial is defined as as $\operatorname{Char}(T)(\lambda)=\det(T-\lambda I)$ where determinant is defined using a map on the one-dimensional space of alternating $n$-forms. Then $\lambda$ is an eigenvalue iff it is a root of the characteristic polynomial. Then I want to prove that $p_T(T)=0$.

Proof: Using upper triangular form, we know that :

$(T-\lambda_1 I) (T-\lambda_2 I) \dots (T-\lambda_n I) =0$

where $\lambda_1, \lambda_2, \dots, \lambda_n$ are diagonal entries of the upper triangular matrix and hence eigenvalues. Now, using the definition of the characteristic polynomial as above, we can't say that $\operatorname{Char}(T)(\lambda) = (\lambda - \lambda_1) (\lambda - \lambda_2) \dots (\lambda - \lambda_n)$. We can only say that $\operatorname{Char}(T)(\lambda) = (\lambda - \beta_1)^{r_1} (\lambda - \beta_2)^{r_2} \dots (\lambda - \beta_k)^{r_k}$ where $r_k$'s sum to $n$ and $\beta_k$s are distinct eigenvalues. Is this observation correct ? Then how to prove $p_T(T)=0$?

  • $\begingroup$ I can't understand why you can't say that Char$(T)(\lambda\;$ is that product of linear factors....you're working on a field, right? So then that pol. is that product of linear factors. Perhaps some of them are repeated, but who cares? Anyway, I can't see how you think this can help you to prove C-H... $\endgroup$ – DonAntonio Nov 15 '13 at 16:20

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