Matrices question (1 Viewer)

porcupinetree · Jun 23, 2016

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

InteGrand · Jun 23, 2016

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

$\noindent Hint: It can be done quite easily by induction (show it starting from $n=1$, since this is an easy base case, and you can use strong induction if needed). :-). Doing it this way will also be able to give you a recursive formula for the $a_n$ and $b_n$, assuming you are able to find $a_2$ and $b_2$. This inductive method would also mean that it's proven for arbitrary-sized square $M$ too.$

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.).

$\noindent I don't know if it's true for integers, if $M$ can be any real matrix. E.g. if $M = \begin{bmatrix}\pi & 0 \\ 0 & \pi\end{bmatrix}$, then $M^2 = \begin{bmatrix}\pi^2 & 0 \\ 0 & \pi^2\end{bmatrix}$. Suppose $M^2 = a M + b I$ for some integers $a,b$. Then comparing first entries, we have $\pi^2 = a\pi +b$, which contradicts the fact that $\pi$ is transcendental.$

porcupinetree · Jun 23, 2016

InteGrand said:
$\noindent Hint: It can be done quite easily by induction (show it starting from $n=1$, since this is an easy base case, and you can use strong induction if needed). :-). Doing it this way will also be able to give you a recursive formula for the $a_n$ and $b_n$, assuming you are able to find $a_2$ and $b_2$. This inductive method would also mean that it's proven for arbitrary-sized square $M$ too.$

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?)

porcupinetree · Jun 23, 2016

InteGrand said:
$\noindent Hint: It can be done quite easily by induction (show it starting from $n=1$, since this is an easy base case, and you can use strong induction if needed). :-). Doing it this way will also be able to give you a recursive formula for the $a_n$ and $b_n$, assuming you are able to find $a_2$ and $b_2$. This inductive method would also mean that it's proven for arbitrary-sized square $M$ too.$

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.).

Ok yeah that makes sense. Thanks

InteGrand · Jun 23, 2016

porcupinetree said:
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 p_M(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.

Matrices question (1 Viewer)

porcupinetree

not actually a porcupine

InteGrand

Well-Known Member

porcupinetree

not actually a porcupine

porcupinetree

not actually a porcupine

InteGrand

Well-Known Member

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