TITLE: Commutative subalgebra of matrix-algebra
QUESTION [5 upvotes]: Let $M_3$ be the $\mathbb{C}$- vector space of matrices of order 3 with complex entries. What is the dimension of subspace $V$ of $M_3$ satisfying the following both conditions?
(a) $AB=BA$ if $A,B\in V$.
(b) If $W$ is a subspace of $M_3$ satisfying $W\supset V$ and $W \neq V$, then there exist $A,B\in W$ such that $AB\neq BA$.

REPLY [3 votes]: Let $V$ be a space with the given properties.

If the dimension of $V$ is zero, we have immediately a contradiction, let $W$ be the linear space generated by $I$, say.
In fact, without loss of generality, we can already assume (and do this) that $V$ contains as a subspace the vector space $\Bbb C I$, and we always chose a basis starting with $I$. 
If the dimension of $V$ is one, assumed with base $I$,
then pass to $W=\Bbb D$, where $\Bbb D$ is here (and in the sequel) the space of diagonal matrices. Any two matrices in $\Bbb D$ commute. Contradiction.
If the dimension of $V$ is two, assumed with base $I,A$,
then $A$ has a Jordan normal form, using the base change matrix, we can assume that $A$ is already in Jordan block form. If $A$ is diagonal, then take $W=\Bbb D$ again, contradiction. Else consider the space generated by $I$, $A$, and $A^2$. It is clear that any two matrices in this space commute, its vector space dimension is three, because $A^2$ has a $1$ in the "next diagonal" (or corner) of the non-trivial block of $A$.
Contradiction.
If the dimension of $V$ is three we declare ourselves happy with the situation. (Such a space exists, for instance $\Bbb D$.)
If the dimension of $V$ is four, we split in cases. If any matrix in $V$ is diagonalizable, we chose four of them, a basis, then there exists a simultaneous base change (which is a linear from $M_3$ to $M_3$) that makes them diagonal, so $V$ lands isomorphically in $\Bbb D$, a space of dimension three, contradiction. So $V$ also contains a non-diagonalizable matrix, $A$, say. 
After a base change, we can and do assume $A$ is already in Jordan block format. 
We have some cases, depending on its block formation.

First, let us consider the case of a single block,
$$A=
\begin{bmatrix}
\lambda & 1 & \\ & \lambda &1 \\ & & \lambda
\end{bmatrix}
\ .
$$
(Missing entries are zero.) Since $I$ is in $V$, we consider instead of $A$ the matrix $A-\lambda I$, so we can reduce ourselves to the case 
$$A=
\begin{bmatrix}
0 & 1 & \\ & 0 &1 \\ & & 0
\end{bmatrix}
\ .
$$
Then $I,A,A^2$ are already linearly independent, commuting, thus in $V$. (Else extend (with $A^2$, and any matrix commuting with $A$ (in $V$) also commutes with $A^2$).) Which are the matrices $X$ commuting with $A$? Let $X$ be such a matrix, we compute:
$$
\begin{aligned}
\begin{bmatrix}
x_{11} & 
x_{12} & 
x_{13} 
\\
x_{21} & 
x_{22} & 
x_{23} 
\\
x_{31} & 
x_{32} & 
x_{33} 
\end{bmatrix}
\begin{bmatrix}
0 & 1 & \\ & 0 &1 \\ & & 0
\end{bmatrix}
&=
\begin{bmatrix}
0 &
x_{11} & 
x_{12} 
\\
0 &
x_{21} & 
x_{22} 
\\
0 &
x_{31} & 
x_{32} 
\end{bmatrix}
\ ,
\\
\begin{bmatrix}
0 & 1 & \\ & 0 &1 \\ & & 0
\end{bmatrix}
\begin{bmatrix}
x_{11} & 
x_{12} & 
x_{13} 
\\
x_{21} & 
x_{22} & 
x_{23} 
\\
x_{31} & 
x_{32} & 
x_{33} 
\end{bmatrix}
&=
\begin{bmatrix}
x_{21} & 
x_{22} & 
x_{23} 
\\
x_{31} & 
x_{32} & 
x_{33} 
\\
0 &
0 & 
0 
\end{bmatrix}
\ ,
\end{aligned}
$$
(we have shifted columns to the right, respectively rows downwards,)
so the lower diagonal entries are zero, $x_{21}=x_{31}=x_{32}=0$,
and the diagonal entries coincide, $x_{11}=x_{22}=x_{33}$, and also on the "next diagonal" $x_{12}=x_{23}$. So any such $X$ is already in the space generated by $I, A, A^2$, contradiction with the fourth dimension supposed to be in $V$.
Next case, let us consider the case of a $2\times 2$ block, an 
$1\times 1$ block, with different eigenvalues,
$$A=
\begin{bmatrix}
\lambda & 1 & \\ & \lambda &1 \\ & & \lambda'
\end{bmatrix}
\ ,\ \lambda\ne \lambda'\ .
$$
Since $I$ is in $V$, we consider instead of $A$ the matrix $A-\lambda I$, so we can reduce ourselves to the case 
$$A=
\begin{bmatrix}
0 & 1 & \\ & 0 & \\ & & \mu
\end{bmatrix}
\ ,\ \mu=\lambda'-\lambda\ne 0\ .
$$
Then $I,A,A^2$ are already linearly independent, commuting, thus 
in $V$. (Else extend...) The matrix $A^2$ has only one non-zero entry at position $(3,3)$, using it, we can linearly clean the corresponding entry in $A$, so $V$ contains the matrices
$$
I\ ,\ 
B=
\begin{bmatrix}
0 & 1 & \\ & 0 & \\ & & 0
\end{bmatrix}
\ ,
C=
\begin{bmatrix}
0 &  & \\ & 0 & \\ & & 1
\end{bmatrix}
\ .
$$
A matrix $X$ commuting with $C$ is of the shape
$$
\begin{bmatrix}
* & * & \\ * & * & \\ & & *
\end{bmatrix}
$$
and the condition that the left upper $2\times 2$ block commutes with $B$ shows (in a similar computation as above) that $X$ is in fact a linear combination of $, B, C$. We miss again the chance to go in the fourth direction.
Final (longer) case, where we have less than in the last case, let us consider the case of a $2\times 2$ block, an 
$1\times 1$ block, with equal eigenvalues,
$$A=
\begin{bmatrix}
\lambda & 1 & \\ & \lambda & \\ & & \lambda
\end{bmatrix}
\ .
$$
Since $I$ is in $V$, we consider instead of $A$ the matrix 
$$
B=
\begin{bmatrix}
0 & 1 & \\ & 0 & \\ & & 0
\end{bmatrix}
\ .
$$
From
$$
\begin{aligned}
\begin{bmatrix}
x_{11} & 
x_{12} & 
x_{13} 
\\
x_{21} & 
x_{22} & 
x_{23} 
\\
x_{31} & 
x_{32} & 
x_{33} 
\end{bmatrix}
\begin{bmatrix}
0 & 1 & \\ & 0 & \\ & & 0
\end{bmatrix}
&=
\begin{bmatrix}
0 &
x_{11} & 
0
\\
0 &
x_{21} & 
0 
\\
0 &
x_{31} & 
0 
\end{bmatrix}
\ ,
\\
\begin{bmatrix}
0 & 1 & \\ & 0 & \\ & & 0
\end{bmatrix}
\begin{bmatrix}
x_{11} & 
x_{12} & 
x_{13} 
\\
x_{21} & 
x_{22} & 
x_{23} 
\\
x_{31} & 
x_{32} & 
x_{33} 
\end{bmatrix}
&=
\begin{bmatrix}
x_{21} & 
x_{22} & 
x_{23} 
\\
0 &
0 & 
0 
\\
0 &
0 & 
0 
\end{bmatrix}
\ ,
\end{aligned}
$$
we get - after replacing $X$ by $X-x_{11}I$ that $X$ is in the space of matrices of the shape
$$
\begin{bmatrix}
0 & * & *
\\
0 &
0 & 
0 
\\
0 &
* & 
* 
\end{bmatrix}
\ .
$$
Using $B$ we find a two dimensional subspace of $V$ inside
$$
\begin{bmatrix}
0 & 0 & *
\\
0 &
0 & 
0 
\\
0 &
* & 
* 
\end{bmatrix}
\ .
$$
Then, using this two-dimensionality, there exist a matrix $F\ne 0$ in $V$ of the shape
$$
\begin{bmatrix}
0 & 0 & *
\\
0 &
0 & 
0 
\\
0 &
\boxed{0} & 
* 
\end{bmatrix}
\ .
$$
If the upper right corner in $F$ can be taken to be $0$, then $F=C$ from the last case, and we get the contradiction from there. Else, we can take $F$ of the form
$$
\begin{bmatrix}
0 & 0 & \boxed 1
\\
0 &
0 & 
0 
\\
0 &
\boxed{0} & 
a 
\end{bmatrix}
\ .
$$
Again we are searching for matrices commuting with the above $F$, 
$$
\begin{aligned}
\begin{bmatrix}
x_{11} & 
x_{12} & 
x_{13} 
\\
x_{21} & 
x_{22} & 
x_{23} 
\\
x_{31} & 
x_{32} & 
x_{33} 
\end{bmatrix}
\begin{bmatrix}
0 &  & 1\\ & 0 & \\ & & a
\end{bmatrix}
&=
\begin{bmatrix}
0 & 0 & x_{11} + ax_{13} \\
0 & 0 & x_{21} + ax_{23} \\
0 & 0 & x_{31} + ax_{33} \\
\end{bmatrix}
\ ,
\\
\begin{bmatrix}
0 &  & 1\\ & 0 & \\ & & a
\end{bmatrix}
\begin{bmatrix}
x_{11} & 
x_{12} & 
x_{13} 
\\
x_{21} & 
x_{22} & 
x_{23} 
\\
x_{31} & 
x_{32} & 
x_{33} 
\end{bmatrix}
&=
\begin{bmatrix}
x_{31} & 
x_{32} & 
x_{33} 
\\
0 &
0 & 
0
\\
ax_{31} & 
ax_{32} & 
ax_{33} 
\end{bmatrix}
\ ,
\end{aligned}
$$
and to simplify things, using $I$ (already in $V$), we replace $X$ above by $X-x_{11}I$, so assume $x_{11}=0$, then using $B$ (already in $V$), we replace $X$ above by $X-x_{12}B$, so assume $x_{12}=0$, and then using $F$ (already in $V$), we replace $X$ above by $X-x_{13}B$, so assume $x_{13}=0$. (This is better form me while typing.) So we have instead the situation:
$$
\begin{aligned}
\begin{bmatrix}
0 & 0 & 0
\\
x_{21} & 
x_{22} & 
x_{23} 
\\
x_{31} & 
x_{32} & 
x_{33} 
\end{bmatrix}
\begin{bmatrix}
0 &  & 1\\ & 0 & \\ & & a
\end{bmatrix}
&=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & x_{21} + ax_{23} \\
0 & 0 & x_{31} + ax_{33} \\
\end{bmatrix}
\ ,
\\
\begin{bmatrix}
0 &  & 1\\ & 0 & \\ & & a
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 0
\\
x_{21} & 
x_{22} & 
x_{23} 
\\
x_{31} & 
x_{32} & 
x_{33} 
\end{bmatrix}
&=
\begin{bmatrix}
x_{31} & 
x_{32} & 
x_{33} 
\\
0 &
0 & 
0
\\
ax_{31} & 
ax_{32} & 
ax_{33} 
\end{bmatrix}
\ ,
\end{aligned}
$$
This implies that the third row is zero. The second row is zero because of the condition to commute with $B$, see the pattern above. So we  do not have again place for a fourth dimension in $V$.


$\blacksquare$
(Note: I tried to avoid notions like (maximal) torus, parabolic subgroup, etc. - please search the net for them, if the problem was a beginning  of studying algebraic linear groups.)