Subring of $\text{End}(V)$ generated by $\text{GL}(V)$ Let $V$ be a $\mathbb{C}$-vector space of arbitrary dimension and let $R$ be its endomorphism ring. Can we describe the subring $S$ generated by $R^\times = \text{GL}(V)$?
If $\text{dim}(V) < \infty$, it's easy to see $R = S$. But in general?
 A: More generally, we have the following theorem.

Theorem: Let $K$ be a field with more than two elements and let $V$ be a vector space over $K$.  Then every element of $\operatorname{End}(V)$ is a sum of two units.

Proof: First we consider the case that $V$ is finite-dimensional.  We use the following Lemma.

Lemma: Let $K$ be a field with more than two elements and let $V$ be a finite-dimensional vector space over $K$.  Let $W_1,W_2\subset V$ be proper subspaces and $a\in V$.  Then $W_1\cup(W_2+a)\neq V$.
Proof of Lemma: We may assume $V=K^n$ and $W_1=K^{n-1}$, identified as the subspace of $K^n$ where the last coordinate is $0$.  Let $P:V\to W_1$ be the projection that drops the last coordinate.  If $P(W_2)$ is not all of $W_1$, then $P(W_2+a)=P(W_2)+P(a)$ is not all of $W_1$ either so we can pick any $v\in W_1\setminus P(W_2+a)$ and then $(v,1)\not\in W_1\cup(W_2+a)$.  If $P(W_2)$ is all of $W_1$, then $P$ restricts to an isomorphism $W_2\to W_1$ since $\dim W_2\leq n-1=\dim W_1$.  Thus $W_1\cup(W_2+a)$ contains at most two points in $\{v\}\times K$ for each $v\in W_1$ and so cannot be all of $V$ since $K$ has more than two elements.

Now let $A:V\to V$ be an endomorphism and let $e_1,\dots,e_n$ be a basis for $V$.  We define automorphisms $B,C:V\to V$ such that $B+C=A$ by choosing $Be_i$ and $Ce_i$ one by one to be linearly independent.  Having chosen $Be_j$ and $Ce_j$ for all $j<i$, let $W_1$ be the span of the $Be_j$ for $j<i$, let $W_2$ be the span of the $Ce_j$ for $j<i$.  By the Lemma, we can pick $v\in V$ that is not in $W_1\cup(W_2-Ae_i)$.  We can then define $Be_i=v$ and $Ce_i=Ae_i-v$, since these will be linearly independent from the $Be_j$ and the $Ce_j$ for $j<i$.
Now we consider the case that $V$ is infinite-dimensional, say of dimension $\kappa$.  We first show that any endomorphism $A:V\to V$ whose image has dimension $\kappa$ is a sum of two units.
To prove this, fix a basis $(e_\alpha)_{\alpha<\kappa}$ for $V$, indexed by the cardinal $\kappa$.  We will construct units $B,C$ such that $B+C=A$ by defining $Be_\alpha$ and $Ce_\alpha$ for each $\alpha$ by transfinite recursion.  In the transfinite recursion, we loop through the following three tasks:

*

*Defining $Be_\alpha$ and $Ce_\alpha$ for the first $\alpha$ for which they are not yet defined.

*Defining $Be_\alpha$ and $Ce_\alpha$ for some $\alpha$ to guarantee that $e_\beta$ is in the image of $B$, for the first $\beta$ for which we have not yet done so.

*Same as 2, but for the image of $C$ instead of the image of $B$.

In the end, if we have arranged at each step that the values $Be_\alpha$ and $Ce_\alpha$ are all linearly independent, we guarantee that $B$ and $C$ are both invertible.
Task 1 is easy: the span of the values of $B$ and $C$ we have defined so far has dimension less than $\kappa$, so we can find $v\in V$ such that both $v$ and $Ae_\alpha-v$ are not in that span.  We then define $Be_\alpha=v$ and $Ce_\alpha=Ae_\alpha-v$.
For Task 2, the span of the values of $C$ we have defined so far has dimension less than $\kappa$.  Since the image of $A$ has dimension $\kappa$, we can pick some $\alpha$ for which we have not yet defined $Be_\alpha$ and $Ce_\alpha$ and for which $Ae_\alpha-e_\beta$ is not in the span of the values of $C$ we have defined so far.  We can then define $Be_\alpha=e_\beta$ and $Ce_\alpha=Ae_\alpha-e_\beta$.
Task 3 of course is the same as Task 2 but with the roles of $B$ and $C$ swapped.
Finally, it remains to be shown that if $\dim V=\kappa$ is infinite and an endomorphism $A:V\to V$ has image of dimension less than $\kappa$, then $A$ is a sum of two units.  In this case, let $U_0=\ker A$, let $W_0$ be a linear complement of $U_0$, let $W_1=W_0+\operatorname{im}(A)$, and let $U_1$ be a linear complement of $W_1$ that is contained in $U_0$.  Note then that $A$ respects the direct sum decomposition $V=U_1\oplus W_1$, since $A$ is $0$ on $U_1$ and $W_1$ contains the image of $A$.  So, it suffices to show we can write the restrictions of $A$ to each of $U_1\to U_1$ and $W_1\to W_1$ as sums of two automorphisms.  For $U_1$ this is trivial: use the identity and its negative.  For $W_1$, note that either $\dim W_1$ is finite (if $\dim\operatorname{im}(A)$ is finite), or else $\dim W_1=\dim \operatorname{im}(A)$.  Either way, we are in a case we have already proved above.

Note that this theorem is not true when $K=\mathbb{F}_2$: if $V$ is $1$-dimensional, then the identity cannot be written as a sum of two units in $\operatorname{End}(V)\cong \mathbb{F}_2$.  However, it is still true for $K=\mathbb{F}_2$ as long as $\dim V\neq 1$.  The only part of the proof that fails is the case that $V$ is finite-dimensional, so it suffices to check the case that $\dim V$ is finite but different from $1$ (in the last infinite-dimensional case that potentially reduces to the finite-dimensional case, you can always assume $\dim W_1>1$ since you can move finitely many basis vectors from $U_1$ to $W_1$ if necessary).
So, suppose $V$ is a vector space over $\mathbb{F}_2$ of finite dimension different from $1$ and $A:V\to V$ is an endomorphism; we wish to write $A$ as a sum of two units.  We can multiply $A$ by a unit to replace it with any other endomorphism with the same kernel.  So, we may assume $A$ is a projection, and we can split $V=\operatorname{im}(A)\oplus\ker(A)$ (with $A$ acting as the identity on the first summand and $0$ on the second).  On $\ker(A)$ we can write $A=I+I$ (where $I$ is the identity).  If $\dim \operatorname{im}(A)\neq 1$, we can multiply by a unit to replace $A$ on $\operatorname{im}(A)$ with an automorphism of $\operatorname{im}(A)$ which does not have $1$ as an eigenvalue (here we use the fact that there are irreducible polynomials over $\mathbb{F}_2$ of arbitrary positive degree).  Then, on $\operatorname{im}(A)$, $A+I$ is invertible and we can write $A=(A+I)+I$.
It remains to consider the case that $\operatorname{im}(A)$ is $1$-dimensional.  In that case, we can write $A$ as an $n\times n$ matrix whose only nonzero entry is the top left corner, and $n>1$ since $\dim V\neq 1$.  Now let $P$ be any permutation matrix which does not have a $1$ in the top left corner (such a $P$ exists since $n>1$), and $A+P$ will be invertible, so we can write $A=(A+P)+P$.
