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

I would like to understand part of a proof that is unclear to me.


Let $V$ be a $K$-vectorspace and $q_1, \dots, q_r$ a set of projection operators on $V$ such that $\sum_{i=1}^r{q_i} = 1_V$ and for all $i \neq j$, $q_iq_j = 0$. Let $V_i =$ im $q_i$. Then $V = V_1 \oplus \dots \oplus V_r$.


Since $1_V = \sum_{i=1}^r{q_i}$, then $v = \sum_{i=1}^r{q_i(v)}$. Therefore $V = \sum_{i=1}^r{V_i}$. All that is left to prove is that this decomposition is unique. Since $v = v_1 + \dots + v_r$, $v_i \in V_i$, then we must show that $v_i = q_i(v)$. Because $V_i =$ im $q_i$, then $v_i = q_i(w_i)$ with $w_i \in V$.

This implies $q_j(v) = q_j(\sum_{i=1}^r{q_i(w_i)}) = q_j^2(w_j) = q_j(w_j) = v_j$. Q.E.D.


How does showing that $v_i = q_i(v)$, imply that the decomposition is unique?

share|cite|improve this question
up vote 1 down vote accepted

You made a transcription error, I think, the fourth senctence of the proof should not start with "Since" but with "Assume that". We know at that point that $v = \sum_{i=1}^r{q_i(v)}$ is one way to write $v$ as sum of elements of each of the $V_i$, and we assume that $v = v_1 + \dots + v_r$ is also a way to write $v$ as sum of elements of each of the $V_i$; showing uniqueness amounts to showing that these two ways must be one and the same. But that means precisely showing that $q_i(v)=v_i$ for $i=1,\ldots,r$, which is what is done in the last line of the proof.

share|cite|improve this answer
Thank you @Marc van Leeuwen – xcrypt Dec 22 '12 at 20:47

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.