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Possibly stupid question

  1. Aug 19, 2008 #1
    How might one prove that for any degree n polynomial p(x) with coefficients lying in a field k, there exists any nxn matrix with entries in k with characteristic polynomial p?
  2. jcsd
  3. Aug 19, 2008 #2


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    If you have a diagonal matrix with entries [itex]\lambda_1, \lambda_2, \cdots, \lambda_n[/itex] then what is its characteristic polynomial?

    [addition]It's not as easy as I thought. For example, the polynomial [itex]x^2 + 1[/itex] has real coefficients, two complex roots, and there is a matrix with real entries - namely [[0 1] [-1 0]] - of which it is the characteristic polynomial. So it's not as easy as just taking diag(first root, second root, ...) in the case where the roots can be outside the field (this has a name, I think it is "algebraically closed"?). Of course changing the basis does not change the characteristic polynomial and in my example that is the trick to get a real matrix but you'd have to prove that it is generally possible). But the proof is easy for algebraically closed fields (if that is indeed the name).
    Last edited: Aug 19, 2008
  4. Aug 19, 2008 #3
    Yes, but of course I was interested in the general case (and particularly when k=reals)
  5. Aug 19, 2008 #4


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    Look into companion matrices.
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