Proving V= N(P) * R(P); Projection onto W2 along W1

[special-g]

Note: * = direct sum
Let P : V--> V me a linear map such that P^2 = P . Prove that
V = N(P ) * R(P ). (Here V is not assumed to be finite dimensional.) Conversely,
prove that if V = W1 * W2 then there exists a linear map P : V --> V with P^2 = P
and N(P ) = W1; R(P ) = W2. Such P is called the projection onto W2 along W1

Here’s what I solved:

(-->) V= N(P) * R(P)
Take x is in both N(P) and R(P). Since x is in N(P), P(x) =0.
Now, let v be in V, so P(v) = x. But x is in R(P), so there is a v in V
such that P(v)=x.
Since P^2=P, P^2(v)=P(v), and also P^2(v)=P(P(v))=P(x).
Thus, x=P(v)=P^2(v)=P(x)=0. Hence, N(P) intersect R(P) = {0}.

Let v be in V. Then v=P(v) + (v-P(v)).
Now P(v) is in R(P) (by definition), and P(v-P(v))= P(v)-P^2(v)=0, since
P^2=P, means v-P(v) is in N(P).
Thus, every v in V can be written as a sum of vectors in N(P) and R(P).

Hence, V=N(P)*R(P).

To prove the converse, we must construct P with P^2= P so that N(P)=W_1, R(P)= W_2.
P:V -->V. Let x=W_1 + W_2 for some x_1 in W_1 and x_2 in W_2
i) N(P)= W_1
For all x_1 in W_1, P(x_1)= 0, since x_1 is in N(P). Therefore, W_1 is a subset of N(P).
(<--)For all x in N(P), P(x)=x_2 = 0. Therefore, x= x_1 + x_2 = x_1+ 0= x_1 is in W_1.
(-->)Therefore, N(P) is a subset of W_1.
To conclude, N(P)= W_1
ii) R(P)=W_2
(-->)For all P(x) in R(P), P(x_2) is in W_2. Therefore, R(P) is a subset of W_2.
(<--)For all x_2 in W_2, x_2= P(x_1) is in R(P). Hence, W_2 is a subset of R(P).
To conclude, R(P)= W_2

However, I still need help, given x in V, in describing what P(x) is in terms of V=W_1 * W_2. Could someone help?

 

[lippyka]

Thank you. For x in V, we can write x = x_1 + x_2 where x_1 is in W_1 and x_2 is in W_2. Then the linear map P : V --> V will be defined as P(x) = x_2. This means that P(x) projects the vector x onto the subspace W_2 along W_1. Hence, P^2(x) = P(P(x)) = P(x_2) = x_2 = P(x).