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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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    @Plop: where do I use that?2011-08-18

2 Answers 2

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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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    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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    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