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 October 12th, 2017, 02:39 AM #1 Member   Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1 Direct sum and projection Given a decomposition V= direct sum of V$_i$ where i=1,...,n: find p$_i$:V->V, $(pi)^2$=p$_i$, p$_i$*p$_j$=0 such that V$_i$=im(p$_i$) where p$_i$=projection and im=image. October 12th, 2017, 05:18 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics What do you mean find the projections? Do you mean you want a formula? If so this is just the Euclidean inner product. If not, explain what you mean. October 12th, 2017, 08:50 PM   #3
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Quote:
 Originally Posted by ZMD Given a decomposition V= direct sum of V$_i$ where i=1,...,n: find p$_i$:V->V, $(pi)^2$=p$_i$, p$_i$*p$_j$=0 such that V$_i$=im(p$_i$) where p$_i$=projection and im=image.
If I'm understanding you correctly this is very trivial. It's just the standard projections embedded back into the product.

For example take $V_i = \mathbb Z$, and $V = \mathbb Z^3$ say. So $V$ is the set of all $3$-tuples of integers $(n, m, p)$. The standard projections $\pi_i : V \to \mathbb Z$ select the first, second, and third coordinates respectively. So for example $\pi_2(4, 9, 47) = 9$.

Now we can inject $\mathbb Z$ back into $\mathbb Z^3$ by padding with $0$ coordinates. So we have injection maps $\iota_i : \mathbb Z \to V$ where for example $\iota_2(9) = (0, 9, 0)$.

Then your $p_i = \iota_i \circ \pi_i$.

In other words (composing right to left) given an n-tuple $v$, first we project onto the $i$-th coordinate, then we pad with $0$'s to get back another element of $V$ that has the $i$-th coordinate of $v$ in the $i$-th place and $0$'s elsewhere.

It's clear that $p_i \circ p_i = p_i$ and that if * is multiplication, $p_i * p_j = 0$ for $i \neq j$. However $p_i* p_i = p_i^2$.

However your notation is a little off since it's im($\pi_i$) that's equal to $V_i$. Since you say that $p_i : V \to V$, it must be the case that im($p_i$) $\subset V$, not $V_i$.

Another little problem is that your $p_i$'s are not projections, since they go from $V$ to $V$ rather than from $V$ to $V_i$. They're like "padded projections" to coin a phrase. In one place you called the $p_i$'s projections and notated them correctly, and in another place they are padded projections but not technically projections.

If I'm misunderstanding your question please clarify.

Last edited by Maschke; October 12th, 2017 at 09:06 PM. Tags direct, projection, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ciaran Linear Algebra 0 September 28th, 2014 11:09 AM xzardaz Linear Algebra 0 February 3rd, 2012 03:47 AM Tannz0rz Computer Science 1 May 16th, 2010 10:48 PM redox Algebra 0 April 26th, 2009 10:37 AM gmcsmith@gmail.com Applied Math 0 May 31st, 2007 10:27 AM

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