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