Questions: Orthogonality and Orthogonal Projections

Question 1

A vector x ∈ H is projected onto a closed subspace M, yielding a candidate point m₀ ∈ M. A student checks whether m₀ is the orthogonal projection by computing x − m₀ and testing ⟨x − m₀, m⟩ = 0 for all m ∈ M. The test passes. What can we conclude?

• A. Nothing — this test only works in finite-dimensional spaces
• B. m₀ is the orthogonal projection P_M x, and it is the unique nearest point in M to x
• C. m₀ is a projection but may not be the nearest point in M
• D. x must already be in M, since the error is orthogonal to everything

Answer: (B)

The orthogonality condition ⟨x − m₀, m⟩ = 0 for all m ∈ M uniquely characterizes the orthogonal projection in any Hilbert space, finite- or infinite-dimensional. This condition is equivalent to minimizing ‖x − m‖ over M. The two characterizations — nearest point and orthogonal error — are identical: m₀ is both. Option C is wrong because no two distinct points in a closed convex set can both be nearest.

Question 2

An operator P on a Hilbert space satisfies P² = P and ⟨Px, y⟩ = ⟨x, Py⟩ for all x, y. A student claims P must be the zero operator because 'applying it twice with no change means it collapsed everything to zero.' What is wrong with this reasoning?

• A. Nothing — P² = P does imply P = 0 in infinite dimensions
• B. The student forgot that P² = P is satisfied by the identity operator I as well as by 0
• C. P² = P (idempotence) means that any vector already in the range of P is fixed by P — not sent to zero
• D. The self-adjoint condition overrides the idempotence condition

Answer: (C)

Idempotence P² = P means that once you project onto M, re-applying P leaves you in M unchanged — P fixes every vector in its own range. Both P = 0 and P = I satisfy P² = P, but so does every orthogonal projection onto any subspace. The geometric meaning is: a vector already in M is its own nearest point in M, so projecting it again does nothing. This is the opposite of 'collapsing to zero.'

Question 3

If P_M x = x for some nonzero vector x, then x should not be in the subspace M.

• T. True
• F. False

Answer: False

P_M x = x means x is fixed by the projection — its nearest point in M is itself, which happens precisely when x ∈ M. The statement has it backwards: P_M x = x if and only if x is already in M. Conversely, P_M x = 0 would mean the nearest point of x in M is the origin, which occurs when x ∈ M⊥.

Question 4

The orthogonal projection P_M is idempotent (P_M² = P_M) because projecting a vector that is already in M gives back that same vector.

• T. True
• F. False

Answer: True

This is exactly the geometric content of idempotence. If y = P_M x ∈ M, then the nearest point in M to y is y itself, so P_M y = y. Therefore P_M(P_M x) = P_M x, which is P_M² = P_M. The self-adjoint condition ⟨P_M x, y⟩ = ⟨x, P_M y⟩ is an independent property expressing the symmetry of perpendicularity.

Question 5

Why does the orthogonality of the error vector x − P_M x to every vector in M uniquely characterize the orthogonal projection? What goes wrong if the error is not fully orthogonal to M?

Model answer: If x − m₀ is not orthogonal to some m₁ ∈ M, we can find a better approximation to x by moving m₀ toward m₁, reducing the distance. Only when the error is orthogonal to the entire subspace is there no direction within M that improves the approximation. The Pythagorean theorem in Hilbert space formalizes this: ‖x − m‖² = ‖x − P_M x‖² + ‖P_M x − m‖² for any m ∈ M, showing the projection error is the minimum possible.

The orthogonality condition eliminates all 'first-order improvements' — it says you're at a critical point of the distance function restricted to M. Because M is a convex set and the distance is a strictly convex function, the critical point is a global minimum. The uniqueness follows from strict convexity: two distinct minimizers would average to a point with even smaller distance, contradiction.