4

Let $V$ be a $\mathbb{K}$-vector space. Suppose we have a set of projectors $P_1, P_2, ...P_n$. Denote $P = \sum_{i=1}^n P_i$. Suppose $P$ is a projector. Does this fact imply that $P_i P_j = \delta_{ij}P_i$, with characteristic of $\mathbb{K}$ being $0$ or greater than $n$? Is the assumption true at least when the $\dim(V) < \infty$?

For example, the statement holds for $n=2$ and the characteristic of $\mathbb{K} \neq 2$.

Proof

$P_1 + P_2 = (P_1 + P_2)^2 = P_1^2 + P_1P_2 + P_2P_1 + P_2^2 = P_1 + P_1P_2 + P_2P_1 + P_2 \Rightarrow P_1P_2 + P_2P_1 = 0$

Now, multiply on the left and on the right by $P_1$. We then get

$P_1^2P_2 + P_1P_2P_1 = P_1P_2 + P_1P_2P_1 = 0$

$P_1P_2P_1 + P_2P_1^2 = P_1P_2P_1 + P_2P_1 = 0$

Adding both equations we get

$2P_1P_2P_1 + (P_1P_2 + P_2P_1) = 2P_1P_2P_1 = 0 \Rightarrow P_1P_2P_1 = 0 \Rightarrow P_1P_2 = P_2P_1 =0$

I found this post solving the case when the characteristic of the field is 0 and the dimension is finite.

principal-ideal-domain
  • 6,324
Jorge
  • 1,210

1 Answers1

1

The statement is not true in the generality wanted from the inquirer. Let $V$ be a finite dimensional $K$ vector space with $p=\operatorname{char}(K)>0$ and $P_1:V \to V$ any non-zero projection. Now let $$n \equiv 0 \mod p$$ or $$n \equiv 1 \mod p.$$ We definite $P_j := P_1$ for $2 \le j \le n$. The projections are not orthogonal if $n>1$ since $P_iP_j=P_1$ (independent of $i$ and $j$). But $$P := \sum_{j=1}^n P _j = nP_1$$ is always a projections since in the $p \mid n$ case we have $P=0$ and in the other case we have $P=P_1$.

In particular this shows that for $p=2$ it doesn't work for a single $n>1$ since every $n \equiv 0,1 \pmod 2$.

Edit: Actually, a little adaptation turns it into a counter-example for every $n>p$ (this is the case the inquirer was interested in). Simply definie $P_j:=P_1$ for $2 \le j \le p+1$ and $P_j:=0$ for $j > p+1$. Then you get always $P=P_1$ and you always have that $P_1$ and $P_2$ are not orthogonal.

Summary: So all finite characterstics are completely out. For $p=0$ you have already a proof in the finite dimensional case and for infinite dimension even $n=3$ doesn't work (that was pointed out in the comments). If $n=3$ doesn't work, any bigger $n$ doesn't work by the same trick I did above. So the questions seems fully answered.

principal-ideal-domain
  • 6,324
  • 1
    But you proof the case when $char(\mathbb{K}) \leq n$, which was an assumption (I found the same xounterexample as you). The inquirer was interested in the case when $n > char(\mathbb{K})$ and $\dim(V) \geq char(\mathbb{K})$ – Jorge Jun 28 '21 at 08:18
  • @Jorge No, it works for arbitrary big $n$ and arbitrary positive dimension. – principal-ideal-domain Jun 28 '21 at 08:23
  • Yes, but you are assuming $n \geq char(\mathbb{K})$. The inquirer was not interested in this case (it is an assumption that $char(\mathbb{K}) > n$ or $char(\mathbb{K}) = 0$). – Jorge Jun 28 '21 at 08:29
  • Oh, I'm sorry. You're right. I read it the other way. I guess I should delete my answer then? – principal-ideal-domain Jun 28 '21 at 08:31
  • 1
    Delete it and maybe leave the idea as a comment in the original post saying it is not solving the original question but it is a counterexample to the other assumptions (maybe someone finds it interesting) – Jorge Jun 28 '21 at 08:47