Dimension of a maximal vector subspace of $M_3(\mathbb{C})$ which is commutative under the multiplication 
Let $V$ be a $\mathbb{C}$-vector subspace of $M_3(\mathbb{C})$ satisfying following properties:
  (A) for every $A,B \in V$, $AB = BA$;
  (B) if $W$ is a $\mathbb{C}$-vector subspace of $M_3(\mathbb{C})$ which contains $V$ as a proper subset, then there exists $A,B \in W$ such that $AB \neq BA$.
  Then what is $\dim V$?

The vector space V generated by 
$$\begin{bmatrix}1 & 0 & 0\\0 & 0 & 0 \\ 0 & 0& 0\end{bmatrix},
\begin{bmatrix}0 & 0 & 0\\0 & 1 & 0 \\ 0 & 0& 0\end{bmatrix} \text{ and }
\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0 \\ 0 & 0& 1\end{bmatrix},$$
satisfies these properties.
So I think $\dim V = 3$ in general.
I think that,  if $\dim \lt 3$, then $V$ is too small to be maximal, and if $\dim \gt 3$, then $V$ is too big to be a commutative algebra.
So it seems to be reasonable for me.
Thank you very much.
 A: We prove that $\dim(V)=3$ by the following analysis.
First we prove that it is enough to consider the spaces $V$ with $I\in V$. If $I\notin V$, then $W=\mathrm{span}(V,I)=\{v+cI\mid v\in V, \ c\in \mathbb{C}\}$ contains $V$ properly and has the property (A). 
Case 1: $\dim(V)=0$
This case is not possible since we assumed $I\in V$. 
Case 2: $\dim(V)=1$
We have $V=\mathrm{span}(I)$. 
Consider a matrix $T$ which is not scalar multiple of identity, and
$W=\mathrm{span}(I,T)=\{cI+dT\mid c, d\in\mathbb{C}\}$. This has dimension $2$ it satisfies property (A) and properly contains $V$.
Case 3: $\dim(V)=2$
Let $T$ be an element in $V$ which is not scalar multiple of identity. Such element exists by dimension comparison. Then the minimal polynomial of $T$ is at least $2$. 
Since $\dim(V)=2$, we have $V=\mathrm{span}(I,T)=\{cI+dT\mid c, d\in\mathbb{C}\}$. 
The following theorem will be useful. 
Theorem (A corollary of Cecioni-Frobenius)

Let $T\in M_3(\mathbb{C})$ and define $C_T=\{X\in M_3(\mathbb{C})\mid XT=TX\}$. Then $\dim(C_T)\geq 3$. 

This is true for $3$ replaced by any $n\in\mathbb{N}$. 
Take any element $S\in C_T-V$ which exists by the  dimension comparison. Then consider $W=\mathrm{span}(V, S)=\{v+cS\mid v\in V, \ c\in \mathbb{C}\}=\mathrm{span}(I,T,S)$. 
This space contains $V$ properly and has the property (A). 
This proves that $\dim(V)\geq 3$. 
For the upper bound direction, we appeal to Schur's theorem. The statement is as follows:
Theorem (Schur)

Let $k$ be any field. Let $M_n(k)$ be the set of $n\times n$ matrices over $k$. Then the maximal number $N(n)$ of linearly independent commuting matrices is $\lfloor \frac{n^2}4\rfloor +1$. Moreover, the equality is attained. 

The proof of this theorem is simplified over time. Schur's theorem is first proved in 1905. Then in 1944, it is simplified by Jacobson: https://projecteuclid.org/download/pdf_1/euclid.bams/1183505933
Then the proof is further simplified in 1998 by M. Mirzakhani: https://www.jstor.org/stable/pdf/2589084.pdf?refreqid=excelsior%3A48601ff1deca611937144ea64534f3fb
For our problem, $n=3$ and we have $\mathrm{dim}(V)\leq \lfloor \frac{3^2}{4}\rfloor +1 = 3$. Therefore, we must have $\dim(V)=3$. 
