Matrices question (1 Viewer)

porcupinetree

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Kingofacting has helped me to prove this for real numbers a_n and b_n, but not integers... can someone shed some light?

 

InteGrand

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Kingofacting has helped me to prove this for real numbers a_n and b_n, but not integers... can someone shed some light?



Edit: Sorry, I think what I wrote it wrong (fails for n = 2. But if you can show the n = 2 case, you can certainly induct this way.).

 
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porcupinetree

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What's getting me is that we need to use the n=2 case in the inductive step, but (as far as I see) there's no guarantee that the statement holds true for integers for the n=2 case (like, what if all the entries of M are irrational?)
 

InteGrand

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What's getting me is that we need to use the n=2 case in the inductive step, but (as far as I see) there's no guarantee that the statement holds true for integers for the n=2 case (like, what if all the entries of M are irrational?)
Yeah it won't hold if M can be any real matrix at all (edited in an example).

If they meant for M to have integer entries, then you can show the n = 2 case using Cayley-Hamilton theorem for instance. The characteristic polynomial will be pM(z) = z2 + bz + c, where b and c are integers (you can show b and c are integers, e.g. by writing out a general 2x2 matrix where the entries are integers and computing its characteristic polynomial). Then by Cayley-Hamilton, O = M2 + bM + cI, which can be rearranged to give what we want.
 
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