# $4$ idempotent operators $E_1,E_2,E_3,E_4$ $:V\to V$ such that $E_1+E_2+E_3+E_4=I$ but don't partition the identity

Let $V$ be a vector space over $F$ such that $charF \neq 2$

Can anyone help me think of $4$ idempotent operators $E_1,E_2,E_3,E_4$ $:V\to V$ such that $E_1+E_2+E_3+E_4=I$ but $\{E_1,E_2,E_3,E_4 \}$ doesn't partition the identity on V (e.g. $E_iE_j\neq0$ when $i\neq j$)?

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Constant functions (unless they are constant zero) are not linear operators. –  Henning Makholm Sep 23 '11 at 15:59
@HenningMakholm would you mind explaining why? –  Freeman Sep 23 '11 at 16:05
@LHS: I believe the linear operator no matter where does it act, applied to zero gives zero. –  Ilya Sep 23 '11 at 16:11
A linear operator $f$ must satisfy $f(\vec 0)=f(0\cdot v)=0\cdot f(v)=\vec 0$. –  Henning Makholm Sep 23 '11 at 16:11
What kind of vector space? –  André Nicolas Sep 23 '11 at 16:33

A partial answer: At least in characteristic 0 and finite dimension this is impossible.

The trace of an idempotent matrix equals its rank (as can be seen by choosing a basis that extends a basis for the range of the linear operator). On the other hand, the sum of the four traces must be $\mathop{\rm Tr}I$, which is the dimension of the entire vector space. But since $E_1+E_2+E_3+E_4=I$, this means that the ranges of the four operators must be direct summands of the entire space. Then choose a basis for each of the summands, combine them to a basis for $V$ and write down the matrices in this basis. For each row only one of the matrices is allowed to have nonzero entries, but since the sum of the matrices is given, this determines them completely, and it is seen that indeed they partition the identity.

On the other hand, there's a trivial solution in characteristic $3$: simply let $E_1=E_2=E_3=E_4=I$.

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Thank you for your answer, about the trivial solution, do they not have to be distinct? –  Freeman Sep 23 '11 at 17:24
You didn't ask for distinct operators in your question. –  Henning Makholm Sep 23 '11 at 17:27
Hmm.. I guess they don't have to be as a set of one element could still partition the identity –  Freeman Sep 23 '11 at 17:28
+1 for the "trivial" solution (that is not so trivial to me). –  user1551 Sep 23 '11 at 17:41
But if you want distinct solutions you could take diag(1,1,0,0,0), diag(1,0,1,0,0), diag(1,0,0,1,0), and diag(1,0,0,0,1) in $(\mathbb F_3)^5$. –  Henning Makholm Sep 23 '11 at 19:24