Chapter 13: Problem 20

Show that the sum of two projection operators \(P_{u}+P_{M}\) is a projection operator iff \(P_{M} P_{N}=0\). Show that this condition is equavalent to \(M \perp N\).

Short Answer

Expert verified

In conclusion, the sum of two projection operators \(P_M\) and \(P_N\) is a projection operator if and only if \(P_M\) and \(P_N\) are orthogonal.

Step by step solution

Understand the Properties of Projection

By definition, a projection operator \(P\) in linear algebra must satisfy \(P^2 = P\), i.e., it is idempotent. Also, if \(P_M\) and \(P_N\) represent the projection operators on subspaces \(M\) and \(N\), respectively, then \(P_M P_N = P_N P_M = 0\) only if \(M\) and \(N\) are orthogonal.

Prove that if \(P_M\) and \(P_N\) are Orthogonal then their Sum is a Projection Operator

Assuming that \(P_M P_N = P_N P_M = 0\), we now need to prove that the sum \(P = P_M + P_N\) is itself a projection. This can be done by checking the idempotent property. So, calculate \(P^2 = (P_M + P_N)^2 = P_M^2 + P_N^2 + 2P_MP_N = P_M + P_N + 0 = P\). We see that \(P^2 = P\), so it follows that \(P\) is a projection operator.

Prove that if the Sum is a Projection Operator, the Projections must be Orthogonal

Assume that \(P_M + P_N = P\) is a projection operator. Hence, \(P^2 = P\). Using the expansion, \(P^2 = (P_M + P_N)^2 = P_M^2 + P_N^2 + 2P_MP_N = P_M + P_N + 2P_MP_N\). Comparing, we find \(2P_MP_N = 0\). Since projection operators are always non-negative, it must be the case that \(P_MP_N = 0\), which means that \(P_M\) and \(P_N\) are orthogonal.