Prove that an idempotent $e$ of $R$ is primitive iff $\dim_{\,F}\left(\operatorname{Im} e\right)=1$

Question

Let $V$ be a vector space over the field $F$, $R$ is the ring of linear operators on $V$. Prove that an idempotent $e$ of $R$ is primitive iff $\dim_{\,F}\left(\operatorname{Im} e\right)=1$

Thanks a lot.

Answer 1

I'm going to address this problem under the assumption that $V$ is finite dimensional; my proof, or some proof, may very well be made to fly in the event that $\dim V = \infty$, but it seems to me that such flight may very well require results about infinite dimensional spaces which I just can't remember for sure at the present moment, e.g. the existence of complimentary subspaces etc. (I will try to note the places in my argument which may require special attention in the event that $\dim V = \infty$.) I am working from memory here, don't have any off-line references handy, and don't want to spend the evening googling and wiki'ing around to search for the assertions I might need in the infinite dimensional case. I believe I can, however, make things work when $\dim V < \infty$, so I'll give that a shot, and try to point out as I go along where modifications might be in order if $\dim V = \infty$.

The first thing I want to prove is that if $e^2 = e$ is primitive in $R$, then $\dim (\text{Im} \; e) = 1$. By contraposition. So let us assume that $\dim (\text{Im} \; e) > 1$. For the sake of brevity, let us denote $\text{Im} \; e$ by $W$; then $W$, itself being finite dimensional, may be decomposed into two complimentary subspaces $W_1 \ne \{0\} \ne W_2$, with $W_1 + W_2 = W$ and $W_1 \cap W_2 = \{0\}$. Thus any $w \in W = \text{Im} \; e$ may be uniquely written $w = w_1 + w_2$, with $w_i \in W_i$. Define linear maps $f_1:W \to W_1 \subset W$ and $f_2:W \to W_2\subset W$ by $f_i(w) = w_i$, where the $w_i$ are as just described. The linearity of the $f_i$ is easy to see: for example, if $f_i(w) = w_i$, then since $\alpha w = \alpha w_1 + \alpha w_2$, we have $f_i(\alpha w) = \alpha w_i = \alpha f_i(w)$. Furthermore if $u \in W$, then we have $u = u_1 + u_2$ uniquely with $u_i \in W_i$; thus $f_i(u + w) = f_i(u_1 + u_2 + w_1 + w_2) = u_i + w_i = f_i(u) + f_i(w)$. We also have the $f_i$ idempotent, since $f_i^2(w) = f_i(f_i(w)) = f_i(w_i) = w_i = f_i(w)$, and in addition to $f_i^2 = f_i$, we have $f_1 f_2 = f_2 f_1 = 0$: $f_1 f_2(w) = f_1 f_2(w_1 + w_2) = f_1(w_2) = 0$; $f_2 f_1(w) = f_2 f_1(w_1 + w_2) = f_2(w_1) = 0$. Now set $e_i = f_i e:V \to V$; this is well-defined since for any $v \in V$, $e(v) \in W$, whence $f_i e:V \to W_i \subset W \subset V$. Furthermore, we note that by the idempotence of $e = e^2$ and the fact $\text{Im} \; e = W$, any $w \in W$ satisfies $e(w) = w$ since for each $w \in W$ there is a $v \in V$ with $e(v) = w$; then $e(w) = e(e(v)) = e^2(v) = e(v) = w$. Based upon these observations we have $(f_ie)^2 = f_ie f_ie$ and since $\text{Im} \; f_ie \subset W_i \subset W$, $ef_ie = f_ie$ by what we have just seen and hence $(f_ie)^2 = f_i^2e = f_ie$; thus the $f_ie$ are idempotent elements of $R$. Also, $f_1e f_2e = f_1f_2e = 0 = f_2f_1e = f_2e f_1e$, showing that the $f_ie$ are in fact orthogonal idempotents. Finally, we have $f_1 + f_2 = I_W$, the identity map on $W$; this is evident since $w = w_1 + w_2$ with $w_i = f_i(w)$: $w = w_1 + w_2 = f_1(w) + f_2(w)$. This shows that $e:V \to W$ obeys $e = I_We = f_1e + f_2e$, expressing $e$ as the sum of the non-zero, orthogonal idempotents $f_ie$. Such an $e$ is by definition not primitive. We have proved that $\dim (\text{Im} \; e) > 1 \Rightarrow e \; \text{imprimitive}$; thus $e \;\text{primitive} \Rightarrow \dim (\text{Im} \; e) = 1$.

In the preceding argument the only place I can see that the assumption $\dim V < \infty$ has been explicitly invoked is in the assertion that $W$ may be expressed as the direct sum of $W_1$ and $W_2$; when $\dim V < \infty$ such a decomposition of $W$ may of course be had by explicitly constructing (finite) bases for the $W_i$, which taken together form a (finite) basis for $W$. Now I strongly suspect that in the event that $\dim V = \infty$, so that we may have $\dim W = \dim (\text{Im} \; e) = \infty$ as well, we can still decompose $W$ into the complimentary subspaces $W_i$, but proving such an assertion will, I think, require the invocation of Zorn's lemma or some other result related to the axiom of choice, and as I said I simply can't recall the argument well enough at the present moment to confidently exploit it. If someone else wishes to chime in here with a comment or answer of their own, I would be most interested in what they have to say. Meanwhile, I'm sticking with the assumption $\dim V < \infty$. And I don't see anywhere else it is needed, besides showing that $W$ is the direct sum of the $W_i$. Note: It appears that rschweib's answer may also (implicitly) address this issue.

Going the other way, suppose that $\dim (\text{Im} \; e) = 1$. If $e = e_1 + e_2$ with the $e_i$ non-vanishing, orthogonal idempotents, let $w \in \text{Im} \; e_1 \cap \text{Im} \; e_2$; then since $w = e_1(v)$ for some $v \in V$, $e_1(w) = e_1^2(v) = e_1(v) = w$; likewise $e_2(w) = w$. Thus $w = e_1(w) = e_1 e_2(w) = 0$ by the orthogonality of the $e_i$. This establishes that the $e_i(V)$ are complimentary subspaces of $e(V) = e_1(V) + e_2(V)$. Since we have assumed that the $e_i \ne 0$, we must have $e_1(V) \ne \{0\} \ne e_2(V)$; but then $\dim (\text{Im} \; e) = \dim (\text{Im} \; e_1) + \dim (\text{Im} \; e_2) > 1$: this contradiction forces the primitivity of $e$. QED

Well, I hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 2

Hints: $\dim_F(Im(e))=1$ implies primitive:

If $e$ were to decompose into nonzero $e_1$ and $e_2$, the images of each of these two things would have to be at least 1 dimensional...

In the other direction:

Suppose $Im(e)$ is at least two dimensional. We should look for two mutually orthogonal idempotents adding up to $e$. If $Im(e)=\langle b_1,b_2,\dots\rangle$, extend this to a basis of and $V$ consider the projection $V\to V$ defined by $\pi(b_1)=b_1$ and $\pi(b_j)=0$ for $j\neq 1$. By linearity, this map is $\pi(\sum\alpha_ib_i)=\alpha_1b_1$. Now look at $\pi e$ and $e-\pi e$.