Question

Let P= {Pᵢⱼ} be an n×n orthogonal projector. Show that: (i) 0≤Pᵢᵢ ≤1 for each i=1,2,…,n. (ii) −1/2≤Pᵢⱼ≤1/2 for i≠J in the range 1≤i, j≤n. Please provide step-by-step solutions for each part.

Answer 1

An orthogonal projector matrix satisfies 0≤Pᵢᵢ≤1 for each diagonal element and −1/2≤Pᵢⱼ≤1/2 for each off-diagonal element in the matrix.

Step-by-step explanation:

(i) 0≤Pᵢᵢ ≤1 for each i=1,2,…,n:

To show that 0≤Pᵢᵢ≤1 for each i, we can analyze the properties of an orthogonal projector. An orthogonal projector is a matrix P where P²=P and Pᵀ=P. This means that P is idempotent (P²=P) and symmetric (Pᵀ=P).

Now, for each diagonal element Pᵢᵢ, we have Pᵢᵢ² = Pᵢᵢ. Since P is idempotent, Pᵢᵢ must be either 0 or 1.

Therefore, 0≤Pᵢᵢ≤1 for each i.

(ii) −1/2≤Pᵢⱼ≤1/2 for i≠j in the range 1≤i, j≤n:

For each off-diagonal element Pᵢⱼ (where i≠j), we have 2Pᵢⱼ = (Pᵢⱼ + Pⱼᵢ) = (Pⱼᵢ + Pᵢⱼ)ᵀ = 2Pⱼᵢ. Therefore, Pᵢⱼ = Pⱼᵢ.

Since P is symmetric, Pᵢⱼ must be equal to Pⱼᵢ for each off-diagonal element.

Thus, we can conclude that −1/2 ≤ Pᵢⱼ ≤ 1/2 for each i≠j in the range 1≤i, j≤n.