I am working on a proof of the following problem and I am not sure if I am making the appropriate reduction to a smaller space.
Let $F$ be a subspace of the vector space $\mathbb{C}^{4\times4}$ such that for every $A,B \in F$ we have $AB = BA$. If there exists a matrix $A$ in $F$ having at least two distinct characteristic values prove $\dim(F) \leq 4$
Sketch:
It is easy to rule out the case when the characteristic has four distinct roots since every operator that commutes with A in this case is a polynomial in T and so the family of matrices in $F$ can be simultaneously diagonalized so $\dim(F) \leq 4$ in this case.
We have two cases other for the characteristic polynomial $p_A(x) = (x-c_1)(x-c_2)(x-c_3)^2$ or $P_A(x) = (x-c_1)^{r_1}(x-c_2)^{r_2}$ where $r_1 = 1,2,3$ and $r_2 = 3,2,1$ respectively.
I think I know how to handle the first case but the case with only two distinct characteristic values I am not sure about. I will list the proof I know when the characteristic polynomial has three distinct characteristic values in order to give motivation for my question.
let $W_1$ be the nullspace of$A-c_1I$, $W_2$ the nullspace of $A-c_2I$ and $W_3$ the nullspace of $(A-c_3I)^2$. Observe that $W_1 \cap W_2 \cap W_3 = \{ 0\}$ so that $\mathbb{C}^4 = W_1 \oplus W_2 \oplus W_3$.
Notice any B that commutes with $A$ implies that $W_1$, $W_2$, $W_3$ are invariant subspaces for $B$.
It then follows that the dimension of $F$ is the sum of the dimension of the space of matrices acting via left multiplication on $W_1$ plus the sum of the dimension of the space of matrices acting via left multiplication on $W_2$ plus the sum of the dimension of the space of matrices acting via left multiplication on $W_3$.
Notice dimension of the space the dimesion of the space of matrices acting via left multiplication on $W_1$ is just 1 since $dim W_1 =1$ and similarity for $dim W_2= 1$.
Now the dimension of the family of commuting operators on a 2-dimensional vectors space over $C$ is at most 2. Indeed each of the commuting operators on $W_2$ will have a repeated eigenvalue.
Therefore for the case $ p_A(x) = (x-c_1)(x-c_2)(x-c_3)^2$ we see that $\dim(F) \leq 4$.
I am a little unsure about the case $p_A(x) = (x-c_1)^{r_1}(x-c_2)^{r_2}$ where $r_1 = 1,2,3$ and $r_2 = 3,2,1$. In particular I am not sure if the above argument can deal with the case $r_1 =1 , r_2=3$ or $r_1=3, r_2=1$. If I apply the same arguments to the above as above can I make the following claim when i am studying the dimension of the space of matrices restricted to $W_1$ and if $\dim(W_3) =3$ can I still make the following statement as I did above in the two dimension case:
Question: Is the dimension of the family of commuting operators on a 3-dimensional vector space over $\mathbb{C}$ at most 3? Will each of the commuting operators on $W_2$ have a double or triple repeated eigenvalue?