Question

If C(A)⊂ C(B), then prove that Pᵦ -Pₐ is the orthogonal projector onto C(B)∖C(A). Provide a step-by-step proof for this statement.

Answer 1

To prove that Pᵦ - Pₐ is the orthogonal projector onto C(B)\C(A), we consider vectors in C(A), where both projectors act identically, and vectors in C(B) that are orthogonal to C(A), where the difference acts as the identity operator. The difference between the projectors acts differently on these sets of vectors, confirming the desired property of orthogonal projectors.

Step-by-step explanation:

To prove that Pᵦ - Pₐ is the orthogonal projector onto C(B)\C(A), we need to show that this difference maps any vector in C(B) to itself if already orthogonal to C(A) and annihilates any vector in C(A).

1. By definition of the orthogonal projector Pₐ, for any vector x in C(A), Pₐ(x) = x.
2. Similarly, since C(A) is a subset of C(B), for any vector x in C(A), Pᵦ(x) also equals x based on the definition of Pᵦ.
3. Therefore, for any x in C(A), (Pᵦ - Pₐ)(x) = Pᵦ(x) - Pₐ(x) = x - x = 0, which shows that the difference Pᵦ - Pₐ annihilates vectors in C(A).
4. Next, consider any vector y that is in C(B) but orthogonal to all vectors in C(A). Since y is not in C(A), Pₐ(y) = 0.
5. For such a vector y, since y is in C(B), Pᵦ(y) = y. Thus, (Pᵦ - Pₐ)(y) = Pᵦ(y) - Pₐ(y) = y - 0 = y.
6. This demonstrates that for vectors orthogonal to C(A) but within C(B), the operator Pᵦ - Pₐ acts as the identity operator, confirming that Pᵦ - Pₐ preserves vectors in C(B)\C(A).

Thus, we have shown that Pᵦ - Pₐ is indeed the orthogonal projector onto C(B)\C(A).