MHB How to Prove the Decomposition Theorem for Projections on a Vector Space?

  • Thread starter Thread starter Euge
  • Start date Start date
  • Tags Tags
    2015
Click For Summary
The discussion revolves around proving the decomposition theorem for projections on a vector space, specifically showing that if a sequence of projections satisfies certain conditions, then the vector space can be expressed as a direct sum of the images of these projections. The key conditions include that the projections are mutually orthogonal and their sum equals the identity operator. A solution is provided in the thread, but no responses or alternative solutions have been posted by other participants. The problem emphasizes the properties of linear operators and their implications for vector space structure. Overall, the thread highlights an important concept in linear algebra concerning projections and direct sums.
Euge
Gold Member
MHB
POTW Director
Messages
2,072
Reaction score
245
Here is this week's POTW:

-----
Let $P_1,\ldots, P_n$ be a sequence of projections on a vector space $V$ such that $P_iP_j = 0$ whenever $i \neq j$ and $P_1 + \cdots + P_n = I$. Prove that

$$V = \operatorname{Im}(P_1) \oplus \cdots \oplus \operatorname{Im}(P_n).$$

-----

Note: By a projection $P$ on a vector space $V$, I mean a linear operator on $V$ such that $P^2 = P$.

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Last edited:
Physics news on Phys.org
No one answered this week's problem. You can find my solution below.
Given $v \in V$, $v = P_1(v)+ P_2(v) + \cdots + P_n(v)$. Furthermore, since the $P_i$ are projections, $P_i(v) \in \operatorname{Im}(P_i)$. Thus $v \in \operatorname{Im}(P_1) + \cdots + \operatorname{Im}(P_n)$ and consequently $V = \operatorname{Im}(P_1) + \cdots + \operatorname{Im}(P_n)$. Now suppose $0 = v_1 + \cdots + v_n$ for some $v_i \in \operatorname{Im}(P_i)$. For each $k\in \{1,2,\ldots, n\}$, $P_k(v_i) = 0$ for all $i \neq k$; this is due to the condition $P_iP_j = 0$ whenever $i \neq j$. So from the equation $0 = v_1 + \cdots + v_n$, we deduce $0 = P_k(v_k) = v_k$. It follows that $v_1 = \cdots = v_n = 0$. Hence, $V = \operatorname{Im}(P_1) \oplus \cdots \oplus \operatorname{Im}(P_n)$.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K