The answer to your question is negative in the following two pathological cases:
- When the matrices are $1\times1$, the answer is negative because all matrices commute.
- When $A$ is a $2\times2$ matrix over $GF(2)$ and it is not a scalar matrix, the answer is also negative. In this case, $A$ is similar to either $\operatorname{diag}(1,0)$ or a companion matrix. Therefore $A$ only commutes with polynomials in itself. Now, in order that $B$ commutes with $C=A-B$, we must have $AB=BA$. Hence $B$ must be a polynomial in $A$. So, by Cayley-Hamilton theorem, $B=pA+qI_2$ and $C=A-B=(1-p)A-qI_2$ for some scalars $p,q\in GF(2)=\{0,1\}$. But then one of $B$ or $C$ will be a scalar matrix that commutes with all matrices.
In all other cases, the answer to your question is affirmative.
If $A$ is a scalar matrix whose size is at least $2\times2$, i.e., if $A=aI_n$ for some $n\ge2$, we may take $B=\operatorname{diag}(b,a,\ldots,a)$ and $C=\operatorname{diag}(a-b,0,\ldots,0)$ for any $b\ne a$.
If $A$ is not a scalar matrix, its size is at least $2\times2$ and the field is not $GF(2)$, we may simply take $B=pA$ and $C=(1-p)A$ for any $p\not\in\{0,1\}$.
If $A$ is not a scalar matrix, its size is at least $3\times3$ and the field is $GF(2)$, there are two possibilities:
- $A$ can be diagonalised as $aI_k\oplus(1-a)I_{n-k}$ where $2\le k<n$. Then we may take $B=(1-a)\oplus aI_{k-1}\oplus (1-a)I_{n-k}$ and $C=\operatorname{diag}(1,0,\ldots,0)$.
- $A$ is not diagonalisable. Its minimal polynomial is then at least quadratic. If it is at least cubic, then $A-A^2$ and $A^2$ are not scalar matrices and we may take $B=A-A^2$ and $C=A^2$. If it is quadratic, since $A$ is not diagonalisable, the rational canonical form of $A$ must contain a $2\times2$ companion matrix $M$. Hence up to similarity we may assume that $A$ is in the form of $\pmatrix{M\\ &N}$. Take $B=\pmatrix{M\\ &N-I_{n-2}}$ and $C=\pmatrix{0\\ &I_{n-2}}$ and we are done.
