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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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Show that the polynomials in $A$ form a vector space, a subspace of $M_2({\bf R})$; find the dimension, and a basis. Do the same for $B$. Then you can see what matrices you get as $f(A)+g(B)$, and what matrices you don't get.

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    I don't understand what does it prove and the way, sorry.2011-08-18
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    In my opinion, It's a "hint", and not an answer.2011-08-18
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    It's a little more than a hint. It's a roadmap. You still have to do the driving.2011-08-18
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After reading Davide comment couple of times, I'll try to answer it and hope you'll fix me on my way:

Couple of things we should take as a consideration:

$\deg M_A, \deg M_B \leq2$ (Because of the size of the matrices)

From the theorem of dividing polynomials in the Ring of polynomials for every two polynomials $f,g \in \mathbb R[X]$ there are polynomials $q_A,r_A,q_B,r_B$ so that:

$f(x)=q_AM_A(x)+r_A(x)$ ; $g(x)=q_BM_B(x)+r_B(x)$

$\deg r_A<\deg M_A\leq 2$ ; $\deg r_B<\deg M_B\leq 2$

Now if I'll plug in $A$ I'll get $f(A)=q_A(A)M_A(A)+r_A(A)$ and because of Kyley-Hemilton theorem $f(A)=r_A(A)$ ;$f(B)=r_B(B)$ $\to$ $f(A)+g(B)=r_A(A)+r_B(B)$.

Polynomials with degree less with 1 can be written:

$r_A=a_1x+a_0$;
$r_B=b1_x+b_0$

$f(A)+g(B)=a_1A+b_1B+(a_0+b_0)I$

so $f(A)+g(B)\in span$ {$A,B,I$}

But obviously I have some $C$ from $M_2(\mathbb R)$$(=4)$ that $C\neq f(A)+g(B)$ since $f(A)+g(B)$ is from dimension smaller or equal to 3.

Ok, Did I talk non-sense?

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    Hi Nir! I'd state this as follows. We have the following formula for finite dimensional linear subspaces of a vector space: $$\dim(V+W)=\dim(V)+\dim(W)-\dim(V\cap W).$$ Apply this to $$V:=\mathbb R[A]=\{polynomials\ in\ A\},$$ $$W:=\mathbb R[B]=\{polynomials\ in\ B\}.$$ Does it help?2011-08-18
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    There are two possibilities. You can define $M_A$ as the characteristic polynomial or as the minimal polynomial. I suggest the characteristic polynomial. In this case the degree is **exactly** 2. I think you should write explicitly how you define $M_A$ and $M_B$. Apart from this, your proof looks fine to me!2011-08-18
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    Thanks @Pierre! yeah I meant to the minimal polynomial by $M_A$2011-08-18
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    OK! But in this case, where you use Cayley-Hamilton for the first time, that's when you say "$\deg M_A\le2$". And you don't need to invoke it afterward. [The equality $M_A(A)=0$ comes from the definition of $M_A$, not from C-H.] [At least that's my opinion.]2011-08-18