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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.

Last edited by Maschke; October 12th, 2017 at 09:06 PM.

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