# proof of inner direct sum of projection operators

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

Theorem:

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 $V = V_1 \oplus \dots \oplus V_r$.

Proof:

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.

Question:

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

-

## 1 Answer

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.

-
Thank you @Marc van Leeuwen – xcrypt Dec 22 '12 at 20:47