The commutant of a non-scalar matrix is a thin subspace

Idea. $C(B)$ is a subspace, so by the subspace problem it is closed, and it has empty interior as soon as it is proper. It remains to see that it is proper unless $B$ is scalar.

Step 1: $C(B)$ is a closed subspace. $C(B)=\ker(A\mapsto AB-BA)$, the kernel of a linear (hence continuous) map: a closed linear subspace.

Step 2: if $C(B)=M_n$ then $B$ is scalar. Suppose $B$ commutes with every matrix. Take $A=E_{ij}$. The matrix $BE_{ij}$ has $j$-th column equal to the $i$-th column of $B$ and zeros elsewhere; $E_{ij}B$ has $i$-th row equal to the $j$-th row of $B$ and zeros elsewhere. Equality of the two matrices gives:
- for $k\neq i$, comparing entries $(k,j)$: $b_{ki}=0$ (off-diagonal entries of $B$ vanish);
- comparing entries $(i,j)$: $b_{ii}=b_{jj}$.
So $B=bI$ is scalar. Conversely if $B$ is scalar, $C(B)=M_n$.

Step 3: conclusion. If $B$ is not scalar, $C(B)$ is a *proper* closed subspace, hence has empty interior by the subspace problem. (If $B$ is scalar, $C(B)=M_n$ is open and has full interior, so the hypothesis cannot be dropped.)

Remark.
- Dimension count: $\dim C(B)\ge n$ always, with equality iff $B$ is *non-derogatory* (minimal polynomial = characteristic polynomial), which holds for a generic $B$ (these are classical facts, not needed for the problem; $C(B)$ contains the span of $I,B,B^2,\dots$). Numerically, for a random $3\times3$ matrix the commutant has dimension $3$.
- Hence for generic $B$ the commutant is a thin set of dimension $n$ in a space of dimension $n^2$.