Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E_1$ and $E_2$ be projections on $V$, a vector space over $F$. Why is if $\operatorname{char}F\neq2$ then $E_1+E_2$ is a projection iff $E_1E_2=E_2E_1=0$ ?

share|cite|improve this question
You are quite right, I apologize. I have not encountered char before, nor have been taught about it, so the wording of the question confused me. – Freeman Aug 31 '11 at 16:18
@Asaf: the LaTeX improvement may have increased readability; however, in view of mt_'s comment, LHS had turned the question from a completely wrong statement into a correct one while the LaTeX edit re-incorporated the first version. – t.b. Aug 31 '11 at 16:28
I rolled back because I was mid way through correcting it when you edited it, i'm very grateful for the edit, but it was the original incorrect version. – Freeman Aug 31 '11 at 16:29
@Theo, LHS: Was it not possible to correct the LaTeX edit instead? – Asaf Karagila Aug 31 '11 at 16:33
@Asaf: I'm sorry, but I have never used LaTeX before. – Freeman Aug 31 '11 at 16:35
up vote 4 down vote accepted

Think about it this way: $E_1+E_2$ is a projection if it satisfies: $(E_1+E_2)^2=(E_1+E_2)$

(Use $E_iE_j$ to mean the composition)

1)Assume $E_1E_2=0$

We want to show that $(E_1+E_2)(E_1+E_2)=(E_1+E_2)$ This means that $E_2E_1+E_2E_2+.....=(E_1+E_2)$ Can you see the next step?

For the converse, assume $(E_1+E_2)$ is a projection, then it must satisfy $(E_1+E_2)^2=....$

share|cite|improve this answer
Ok, so I see if E1E2=E2E1=0 as this implies E1E1+E1E2+E2E1+E2E2=E1E1+E2E2, implying (E1+E2)^2=E1+E2 But for the converse you have E1+E2 as a projection.. i'm unsure how you can show E1E2=E2E1=0 – Freeman Aug 31 '11 at 16:46
No problem; I have been kind of slow myself; If we assume $E_iE_j=0$, and we have $1+1=0$, then $(E+E)^2$=$E^2+EE+EE+E^2$=$E^2+0+0+E^2=2E^2=0$ – gary Aug 31 '11 at 17:13
Yes, I think that works; let me double-check. – gary Aug 31 '11 at 17:25
@Arcane1729 That's where $\operatorname{char} F \neq 2$ enters. We can decompose $V$ as $R \oplus K$, where $R$ is the image, and $K$ the kernel of the projection $E_2$ - that works for all projections and all vector spaces, regardless of characteristic. Now you have a simple description of $E_2 + \operatorname{Id}$ - on $R$, it's multiplication by $2$, and on $K$ it's the identity. When $\operatorname{char} F \neq 2$, it follows that $E_2 + \operatorname{Id}$ is invertible. And then you can cancel the invertible $E_2 + \operatorname{Id}$ to get $E_2E_1 = 0$. – Daniel Fischer May 1 at 15:28
@Arcane1729 Also look at this question, where the argument is given in what I consider a more elegant way (the question assumes real scalars, but what is needed for the last step is just $2X = 0 \implies X = 0$). – Daniel Fischer May 1 at 15:44

Easy direction: If $E_1E_2=E_2E_1=0$, then $(E_1+E_2)^2=E_1^2+E_2^2 = E_1+E_2$.

Conversely, suppose $(E_1+E_2)^2= E_1+E_2$. Then $$ E_1E_2+E_2E_1 = 0 \tag{1} $$ By multiplying both sides of (1) by $E_1$ on the left, or by $E_1$ on the right, obtain two equalities: $$E_1 E_2+E_1E_2E_1=0,\qquad E_1E_2E_1+E_2E_1=0$$ By subtracting these, $$E_1E_2-E_2E_1=0. \tag{2}$$ Adding or subtracting (1) and (2), we get $$2E_1E_2=0,\qquad 2E_2E_1=0$$ Since the characteristic is not $2$, the factor $2$ can be cancelled.

(This answer is based on this post by Alex).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.