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Proving ideas about projections

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Homework Statement


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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


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##P^2 = P## is my (abbreviated) definition of a projection.


The Attempt at a Solution


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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?
 
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Answers and Replies

  • #2
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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## ?
 

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