Prove or disprove: $\forall M \in \mathbb{F}^{n \times n,}: M \cdot M = M \Rightarrow$ $M$ is identity matrix (with columns deleted) I've been wondering about the following:

Let $M$ be a matrix of size $n \times n$ over the field $\mathbb{F}$. 
What does the property $M \cdot M = M$ say about $M$?

Intuition and Example
My intuition tells me that the above property holds for any matrix $M$ that is the identity matrix $I$ (i.e. the matrix with $1$ on the diagonal, and $0$ everywhere else), with an arbitrary set of columns "deleted", i.e. replaced with zero vectors. To demonstrate, let's consider some examples in $\mathbb{R}^3$, thus giving us a $3 \times 3$ matrix:
$$
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
\cdot
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
=
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
$$
(Incomplete) idea for a general proof
We have to show that this property holds for all such "identity matrices with deleted columns", and only for those.
Obviously, it holds for $M = I$. Now let's choose a subset of indices $S_{del} \subseteq \left\{1, \ldots, n\right\}$ which represent the columns (and since the transposed matrix is the same, also the rows) we want to delete from $M = I$, giving us $\tilde{M}$. Then $S_{kep} = \left\{1, \ldots, n \right\} \setminus S_{del}$ are the indices of the columns we decided to keep. 
So now, we need to show that $\tilde{M}$ still maps each of its individual columns onto themselves, because then directly follows $\tilde{M} \cdot \tilde{M} = \tilde{M}$. To do that, we go through each column $\tilde{M}_{\cdot j}$ for $j=1,\ldots,n$.
Is $j \in S_{del}$, then obviously $\tilde{M}_{\cdot j}$ is a zero vector and the resulting matrix product will also be a zero vector. 
Otherwise, $\tilde{M}_{\cdot j}$ will be a column vector with a $1$ in row $j$, and zeros anywhere else. Since a transposition of $\tilde{M}$ will still yield the same matrix, for $\tilde{M}$ still has values only on its diagonal, we also know that $\tilde{M}_{j \cdot}$ is a row vector with a $1$ in column $j$ and zeros anywhere else.
By multiplying this column and row together, we get the $j$-th row of the resulting vector, which turns out to be $1$. Since $\tilde{M}$ is still a diagonal matrix, no row other than the one we just inspected will have a value other than $0$ in its $j$-th column, and since $\tilde{M}_{\cdot j}$ has a value other than $0$ only in its $j$-th column, we can conclude that all other rows of the resulting vector will be zero. So we indeed get our column vector $\tilde{M}_{\cdot j}$ as a result in either case.
What remains to be shown is that no other matrix fulfills the property $M \cdot M = M$; this leaves me stuck.
The questions
Is my above proof correct so far? What's a good way to prove that no other matrix than the ones I considered (identity matrices with deleted columns) fulfill the above property? Is there a shorter, more concise way of proving all of this? 
 A: Your guess is unfortunately not correct, but it's almost correct. Jose Carlos Santos gave a counterexample (which I did not check, but there are many counterexamples so I'm not too worried here), however the matrices which you're interested in are called projector matrices or projectors.
The intuition for that name is that such a matrix "projects" everything onto its image and then doesn't move what's on its image. It's easy to make that precise : $M^2=M$ if and only if the restriction of (the linear operator associated to) $M$ to $\mathrm{im}(M)$ is the identity. 
First of all, this provides tons of counterexamples : if you have a block diagonal matrix with only $0$'s except one of the blocks which is an identity, then it will clearly satisfy this condition and unless the block is the one in the top left hand corner, it won't be of the form you suggest. Reading this, you'll probably feel as though I'm bickering and having one identity block is "almost" the same as what you suggest. The "almost" is what's missing to your suggestion : up to a change of basis, this block is the same as your suggestion. 
Ah ! "up to a change of basis". Note that the condition $M^2=M$ isn't so much a  condition on the matrix as it is a condition on the underlying linear operator : $f\circ f= f$. Therefore it is invariant under base change : if $M^2=M$ then for any invertible matrix $P$, $PMP^{-1}$ satisfies the same property (check it ! there is a computational proof and a proof based on what I explained)
That's what was missing in your suggestion. If $M^2=M$, then there is an invertible $P$ such that $PMP^{-1} $ is of the form you suggest. The best way to prove this is to interpret everything "geometrically", that is : interpret $P(-)P^{-1}$ as a base change and interpret $M^2=M$ as a condition on the image. To get exactly your suggestion, you will also need to prove that $\ker(M)$ is a complement of $\mathrm{im}(M)$
A: You can't prove that, since it is not true. Take$$M=\begin{bmatrix}-3 & 2 \\ -6 & 4\end{bmatrix},$$for instance. It happens that $M^2=M$.
A: Your guess is almost correct (at least over $\mathbb{C}$) in the sense that $M$ is always similar to such a matrix.
$M^2 = M$ implies $M(M-I)=0$ so $M$ is diagonalizable with spectrum $\sigma(M) \subseteq \{0,1\}$. Therefore there exists an invertible matrix $P$ such that
$$P^{-1}MP = \begin{bmatrix} I_k & 0 \\ 0 & 0_{n-k}\end{bmatrix} = \operatorname{diag}(\underbrace{1, \ldots, 1}_k, 0, \ldots, 0)$$
for some $0 \le k \le n$.
