(a) Let X={(x,0) | x∈R} and D={(x,x) | x∈R}.
Prove that R2 = X ⊕ D.
If P is the projection map onto X, describe P (x, y) for an arbitrary vector (x, y) ∈ R2. What is P (3, −4)?
(b) Let V be a finite dimensional vector space and let A1,A2,...,Ak be subspaces of V such that V = A1 ⊕A2 ⊕···⊕Ak. If (k)Σ(i=1) ai = 0, where each ai is an element of Ai, explain why we can deduce that ai =0 for all i.
For (a), I know that to prove R2 = X ⊕ D it needs to satisfy R2 = X + D and X ∩ D = {0} which is simple to prove, but I do not understand the second part of the question, what am I suppose to describe for P(x,y)? Could I please get some pointers to start this? And for P(3,-4), do I incorporate both X and D to form the (3,-4)? (7X - 4D)?
For (b), I am not sure if I am oversimplifying the question, but it seems like I can prove it by simply its own definition of:
Let V be a vector space with subspaces U1, U2, . . . , Uk.
Then V = U1 ⊕ U2 ⊕···⊕ Uk if and only if the following conditions hold: (i) V = U1 + U2 +···+ Uk; (ii) Ui ∩ (U1+···+Ui−1+Ui+1+···+Uk)={0}for each i.
So that each i is 0? If this is incorrect how should I go about proving this part?
Your argument for (b) is fine. If $\sum_{i=1}^k a_i=0$, then $-a_1 = a_2 + \cdots+a_k$ which implies $a_1 \in A_1 \cap (A_2+\cdots+A_k)$, so $a_1=0$. Similarly, $a_i=0$ for each $i$.
For the second part of (a) you should have some formula for the projection onto a subspace. $(x,y)$ is some point in $R^2 = X \oplus D$, and $P(x,y)$ is the closest point in the subspace $X$ to $(x,y)$.