## Homework Statement

[/B]
Let $V$ be a vector space, and let $U, W$ be subspaces of $V$ such that $V = U \oplus W$. Let $P_U$ be the projection on $U$ in the direction of $W$ and $P_W$ the projection on $W$ in the direction of $U$. Prove:

$P_U + P_W = Id$, $P_U P_W = P_W P_U = 0$

Reciprocally, given $P_1, P_2 : V \rightarrow V$ that verify $P_1 P_2 = P_2 P_1 = 0$, $P_1+P_2 = Id$, prove that $V = Im(P_1) \oplus Im(P_2)$

## Homework Equations

[/B]
$P^2 = P$ is my (abbreviated) definition of a projection.

## The Attempt at a Solution

[/B]
I understand the basic idea I think. I know what the direct sum of subspaces implies, namely that the intersection of the two subspaces is 0 and that any vector $v \in V$ can be broken down into $v = u \in U + w \in W$... I read something about proving that one of these subspaces represents the kernel and the other the image of $V,$ but is that necessarily the case? I suppose the second part of the exercise indicates otherwise.

I'm not sure how to go about the proof. I feel like I understand the idea but I don't know how to write it. A hint to get me on the right path before a full answer would be appreciated to help refine my intuition. Thanks.

Is it perhaps something like $P_U(v) = P_U(u+w) = u = P(P(u)) \rightarrow P(u) \in Im(P_U)$... or along those lines?

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
OK, so I was puzzling over the proof a bit and I think I may have reached something.

Take $\{v_1 ... v_n\}$ to be a basis for $V$ such that $\{v_1...v_k\}$ and $\{v_{k+1} ... v_n\}$ are bases for $U$ and $W$ respectively. Also, given the direct sum, we know the intersection of the two subspaces will be 0 and therefore any vector belonging to the intersection will be 0.

I can then decompose any vector $v \in V$ into $v = (u = α_1v_1 + ... + α_kv_k) \in U + (w = α_{k+1}v_{k+1} + ... + α_nv_n) \in W$. At this point my notation might get a little weak, so anyone who can help would be appreciated.

Therefore, $P_U(v) = P_U(u+w) = P_U(u) \in U + P_U(w) \in U = P_U(u) + 0 = u$
Similar reasoning leads to $P_W(v) = w$.

Then, $P_U(P_W(v)) = P_U(w) = 0 = P_W(u) = P_W(P_U(v))$ and $P_U(v) + P_W(v) = u + w = v$ is what occurs to me for the other part of the proof, but I'm not sure if it's quite right, given that $Id$ is the identity matrix...

I can see where I'm going for the reciprocal statement as well, but I'm having a little more trouble there. Is it as simple as following my reasoning in my first post that $P_U(u) = u \in Im(P_U)$ leads again to $v = u + w$ and thus $V = Im(P_U) \oplus Im(P_W) = U \oplus W$ ?