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From Hoffman's Linear Algebra (section 6.3, exercise 9):

Let $A$ be a $n \times n$ matrix with characteristic polynomial $$f=(x-c_1)^{d_1} \cdots (x-c_k)^{d_k}.$$ Prove that $$\mbox{trace} (A) = c_1 d_1 + \cdots + c_k d_k$$

This is an exercise in the minimal polynomial section. Thus, in principle, it can be solved using methods covered in that section. Is there a "minimal polynomial"-method to solve this exercise?

Rodrigo de Azevedo
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HeMan
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2 Answers2

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No you cannot do this with minimal polynomials, because the minimal polynomial does not determine the trace. With respect to the characteristic polynomial mentioned in the question, the minimal polynomial may have lower exponents (but not $0$) instead of the $d_i$, but this is not reflected in the trace.

Marc van Leeuwen
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  • Yes. But then, how can I prove the statement with the methods in that section. I mean, yes, not directly with the minimal polynomial, but, there isn't any method related to minimal polynomials for solving the exercise? – HeMan Sep 01 '17 at 04:28
  • if not, then why that exercise is located in that section? – HeMan Sep 01 '17 at 04:28
  • I really don't know, you should ask the author. – Marc van Leeuwen Sep 01 '17 at 13:36
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If A is as you define it, then you can find its Jordan Canonical Form J which is an upper triangular matrix. Then A ~ J therefore they have the same determinant and by definition of the Jordan Canonical Form you can show the result you want.

Here I am using multiple theorems from Hoffman's Linear Algebra book:

  1. Similar matrices have the same determinant
  2. Let J be the Jordan Canonical Form of a matrix A. Then J is an upper triangular matrix that is similar to A and tr(J) are eigenvalues of A (each eigenvalue appears [its multiplicity]-times on the diagonal of J)
student509
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