TITLE: Converse of Schur's Lemma in finite dimensional vector spaces
QUESTION [5 upvotes]: I am trying to prove (or disprove) the converse of Schur's Lemma in finite dimensional vector spaces. I am not sure if it holds in this case, but I have tried to apply the idea that proves it in representation theory (see for example Theorem 4.3 here that uses Maschke's theorem or in questions here and here).
The converse of Schur's Lemma in finite dimensional vector spaces is:

Let $V$ be a finite dimensional vector space over the complex numbers. Let $S$ be a set of endomorphisms of $V$ and assume that every endomorphism $A$ of $V$ such that $$AB=BA\text{ for all }B\in S$$ is of the form $\lambda I, \ \lambda\in\mathbb{C}$. Then $V$ is a simple $S$-space.

(*an endomorphism of $V$ is a linear operator from $V$ to $V$
*$V$ is a simple $S$-space if the only $S$-invariant subspaces of $V$ are $V$ itself and the zero subspace)
What I've tried so far:
Assume to the contrary that $V$ is not a simple $S-$space. Then, there exists a subspace $W$ of $V$ such that $W\not =\{0\}$, $W\not= V$ and $W$ is $S$-invariant. Let $W'=V\setminus W$. Then, since $V$ is finite, we can easily prove that $V=W\oplus W'$ (this is my attempt to translate Maschke's theorem in vector spaces). Consequently, for every $v\in V$, there exists unique $w\in W$ and $w'\in W'$, such that $v=w+w'$. Define the projection $P:V\rightarrow V$ by $Pv=w$ for every $v\in V$. 
What is left to prove is that $PB=BP$ for all $B\in S$. Then $P$ is clearly not a scalar and thus we have the contradiction we are looking for.
There is a difficulty in showing that $PB=BP$ for all $B\in S$, because $W'$ might be $S$-invariant or not. More precisely: 
Let $v\in V=W\oplus W'$ and $v\not =0_V$.
If $v\in W$ then $Bv\in W$ because $W$ is $S$-invariant, therefore: $PBv=Bv$ and $BPv=Bv$.
If $v\in W'$ then $Pv=0$, hence $BPv=B\cdot 0=0$, and
(i) if $Bv\in W'$ then $PBv=0$.
(ii) if $Bv\in W$ then $PBv=Bv$ and this is where the problem occurs.
Any hints, ideas or counterexamples would be very helpful.

REPLY [4 votes]: Here's a counterexample: let $S = \{A \in \Bbb C^{2 \times 2} : Ax = x\}$ where $x = (1,0)$ (the column-vector $(1,0)$).  That is, $S$ is the set of all matrices whose first column is $(1,0)$.
Now, the span of $(1,0)$ is an $S$-invariant subspace.
However, consider any $A$ that is not a multiple of the identity.  If $A$ is diagonalizable, then it has eigenvectors $y_1,y_2$ with $Ay_i = \lambda_i y_i$. One of these eigenvectors is linearly independent to $x$, suppose WLOG that $y_1$ and $x$ are linearly independent.  We can then define $B$ by
$$
Bx = x\\
By_1 = y_1 + y_2
$$
we find that $ABy_1 \neq BAy_1$, so that $AB \neq BA$.
Now, suppose that $A$ is not diagonalizable.  It follows that $A$ is not diagonal.  So, it fails to commute with 
$$
B = \pmatrix{1&0\\0&2}
$$
Thus, $S$ is a counterexample to your claim.