Is every endomorphism of a vector space a linear combination of idempotents? Is every endomorphism of a $K$-vector space a $K$-linear combination of idempotents? 
This question was asked by Jonas Meyer in a comment to this question. 
To make sure I earn no points thanks to a question raised by somebody else, I shall put a 500 point bounty on this question (but I think I must wait two days). 
George Lowther told me in a comment to this answer that he had a positive answer to Jonas's question. I considered suggesting to George that he posts his answer to Jonas's question as an answer to this question, but I thought Jonas's question deserved to be asked separately.
 A: For a finite dimensional space, it is straightforward to express any endomorphism as a linear combination of idempotents. For an infinite dimensional space, we can use the fact that $V$ is isomorphic to $V\oplus V$. This follows from the fact that vector spaces are determined up to isomorphism by their dimension, and that $\kappa+\kappa=\kappa$ for any infinite cardinal $\kappa$. So,
$$
\mathrm{dim}(V\oplus V)=\mathrm{dim}(V)+\mathrm{dim}(V)=\mathrm{dim}(V).
$$
So, the algebra, $\mathrm{End}(V)$, of endomorphisms of $V$ is isomorphic to the endomorphisms of $V\oplus V$, which can be identified with the $2\times2$ matrices over $\mathrm{End}(V)$,
$$
\mathrm{End}(V)\cong\mathrm{End}(V\oplus V)\cong\mathrm{M}_2(\mathrm{End}(V)).
$$
First, the traceless matrices can be directly expressed as a linear combination of idempotents.

(I) Any matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$ in $\mathrm{M}_2(\mathrm{End}(V))$ with $a+d=0$ is a linear combination of idempotents.

Proof: As, $d=-a$, we have
$$
\left(\begin{matrix}a&b\\c&d\end{matrix}\right)=\left(\begin{matrix}a&a\\1-a&1-a\end{matrix}\right)-\left(\begin{matrix}1&a-b\\0&0\end{matrix}\right)+\left(\begin{matrix}1&0\\c+a&0\end{matrix}\right)-\left(\begin{matrix}0&0\\1&1\end{matrix}\right),
$$
and each of the matrices on the rhs are idempotents. QED
It remains to show that diagonal matrices of the form $(a,0;0,0)$ are linear combinations of idempotents. For this, I will look at the infinite direct sum $V^\omega=V\oplus V\oplus V\oplus\cdots$. This can be identified with the infinite sequences $v=(v_0,v_1,v_2,\ldots)$ for $v_k\in V$, with all but finitely many $v_k$ equal to $0$, under component-wise addition and scalar multiplication. Given any sequence $a_0,a_1,a_2,\ldots$ in $\mathrm{End}(V)$, we use $a_0\oplus a_1\oplus a_2\oplus\cdots$ to represent the linear operator on $V^\omega$ taking $v$ to $(a_0v_0,a_1v_1,\ldots)$.

(II) For any $a\in\mathrm{End}(V)$, the operator $a\oplus 0\oplus 0\oplus\cdots$ is a linear combination of idempotents in $\mathrm{End}(V^\omega)$.

Proof: This is an Eilenberg-Mazur swindle.
The operator $a\oplus(-a)$ on $V\oplus V$ is represented by the traceless matrix $(a,0;0,-a)$ so, by (I), is a linear combination of idempotents in $\mathrm{End}(V\oplus V)$. Grouping the factors of $V$ in the definition of $V^\omega$ into pairs,
$$
V^\omega\cong(V\oplus V)\oplus(V\oplus V)\oplus(V\oplus V)\oplus\cdots
$$
and applying the decomposition of $a\oplus(-a)$ into idempotents to each factor $V\oplus V$ shows that
$$
L\equiv a\oplus(-a)\oplus a\oplus(-a)\oplus a\oplus(-a)\oplus\cdots
$$
decomposes into a linear combination of idempotents in $\mathrm{End}(V^\omega)$. Similarly, grouping the factors of $V^\omega$ as
$$
V^\omega\cong V\oplus(V\oplus V)\oplus(V\oplus V)\oplus\cdots,
$$
we can apply the decomposition of $a\oplus(-a)$ into idempotents to each factor of $V\oplus V$ and the zero map to the first factor $V$ to show that
$$
M\equiv0\oplus a\oplus(-a)\oplus a\oplus(-a)\oplus a\oplus(-a)\oplus\cdots
$$
is a linear combination of idempotents. So, $L+M=a\oplus0\oplus0\oplus\cdots$ is a linear combination of idempotents. QED

(III) For any $a\in\mathrm{End}(V)$, the matrix $\left(\begin{matrix}a&0\\0&0\end{matrix}\right)$ is a linear combination of idempotents.

Proof: First, using the fact that $\aleph_0\times\kappa=\kappa$ for any infinite cardinal $\kappa$, we have
$$
\mathrm{dim}(V^\omega)=\aleph_0\times\mathrm{dim}(V)=\mathrm{dim}(V),
$$
so $V\cong V^\omega$. Identifying the second factor of $V$ with $V^\omega$ in $V\oplus V$ gives the sequence of isomorphisms
$$
\begin{align}
V\oplus V&\cong V\oplus V^\omega= V\oplus(V\oplus V\oplus V\oplus\cdots)\\
&\cong V\oplus V\oplus V\oplus\cdots=V^\omega.
\end{align}
$$
Under this map, the operator $a\oplus 0$ represented by the matrix $(a,0;0,0)$ gets taken to $a\oplus0\oplus0\oplus\cdots$, which is a linear combination of idempotents by (II). QED
Putting these together gives,

(IV) Every endomorphism of $V$ is a linear combination of idempotents.

Proof: As above, using the fact that $\mathrm{End}(V)\cong \mathrm{M}_2(\mathrm{End}(V))$, we need to show that any $2\times2$ matrix of endomorphisms of $V$ is a linear combination of idempotents. Writing
$$
\left(\begin{matrix}a&b\\c&d\end{matrix}\right)=\left(\begin{matrix}-d&b\\c&d\end{matrix}\right)+\left(\begin{matrix}a+d&0\\0&0\end{matrix}\right),
$$
the first term on the rhs is a linear combination of idempotents by (I) and, similarly for the second term using (III). QED
A: Here is a corollary to George's answer:

(a) If $V$ is an infinite dimensional $K$-vector space, then the set of sub-vector spaces of $V$ has cardinal $\operatorname{card}(K)^{\dim(V)}$. 

Proof. Let $\mathcal S$ be the set in question, and let $\mathcal E$ and $\mathcal I$ be respectively the set of endomorphisms and of idempotent endomorphisms of $V$. We have 
$$
\operatorname{card}(\mathcal E)=\operatorname{card}(K)^{\dim(V)}.
$$
George proved that any endomorphism is a $\mathbb Z$-linear combination of idempotents. This implies 
$$
\operatorname{card}(\mathcal E)=\aleph_0\operatorname{card}(\mathcal I).
$$
The two above displays imply 
$$
\operatorname{card}(\mathcal I)=\operatorname{card}(K)^{\dim(V)}.
$$
The map 
$$
\mathcal I\to\mathcal S,\quad e\mapsto\operatorname{Im}(e)
$$ 
being surjective, and the map 
$$
\mathcal I\to\mathcal S\times\mathcal S,\quad e\mapsto(\operatorname{Im}(e),\operatorname{Ker}(e))
$$ 
being injective, $\mathcal I$ and $\mathcal S$ are equipotent, and the proof is complete.
Note that George's previous answer was already sufficient for the above argument.
Questions: 
(b) Is there a simpler proof of (a)?
(c) Was Statement (a) previously recorded? Where?
Thank you very much in advance, dear reader, if you can tell me what the answers to these questions are.
