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Let $A,B\in M_2(\mathbb R)$. Prove that there's $C \in M_2(\mathbb R)$ that is not sum of $f(A)+g(B)$ for any polynomials $f,g \in \mathbb R[X]$.

I know that if $\lambda $ is eigenvalue of $A$ then $f(\lambda)$ of $f(A)$ and also for $B$ and $g$.

What else should I do?

Thank you.

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    Thanks to the equality $A^2 -\mathrm{Tr}(A)A+(\det A)I_2 = 0$ you can show that for all integer $k$: $A^k\in\mathrm{Span}\left\{I,A\right\}$. We get that $V:=\left\{f(A)+g(B),f,g\in\mathbb R[X]\right\} = \mathrm{Span}\left\{I,A,B\right\}$, hence the dimension of $V$ is $\leq 3$. You can conclude since the dimension of $M_2(\mathbb R)$ is $4$.2011-08-18
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    @Davide: Thanks for the comment- I'm looking for an answer which does not involve this equality that you brought.2011-08-18
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    @Davide, you've given a man a fish.2011-08-18
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    @Nir, any other restrictions in the kind of answer you want? So we don't all waste our time coming up with solutions you don't like?2011-08-18
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    @Gerry Yes, it was better to learn this man to fish, and it's what your answer does.2011-08-18
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    @gerry: I won't be able to know until you'll write it, but for now- no.2011-08-18
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    @Gerry: Do you really believe that I *don't* like answers? for real? There are some professors that teach some theorems in some universities around the world, and in order to use some theorem you need to prove it, so let's say that my luck will shine in my test and I'll have this question- I won't be able to use this theorem without proving it, and I don't want or know to prove it.2011-08-18
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    @Davide: your comment is spot on. It took me some time to understand why $V$ equals $\text{Span}\{I,A,B\}$ (and not only, is included in it). But I am slow... :-)2011-08-18
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    @Nir Do you realize that the equality (I guess you meant this one: $A^2 -\mathrm{Tr}(A)A+(\det A)I_2 = 0$) is Cayley-Hamilton for 2-dim. matrices? From your answer below it seems you know it.2011-08-18
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    @Plop: where do I use that?2011-08-18

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