Projection operator in spectrum theory

[seek]

Homework Statement

Show that if A is a normal operator in an n-dimensional vector space, and if A has r distinct eigenvalues a₁,a₂,...a_r, then the projection operator onto the subspace with eigenvalue a_i can be written as:

P_i=[(A-a₁)...(A-a_a-1)...(A-a_r)]/[(a_i-a₁) ...(a_i-a_i-1)...(a_i-a_i+1)...(a₁-a_r)]

Homework Equations

AP_iz=a_iP_iz

A= (i=1 to r)$$\Sigma$$a_iP_i

The Attempt at a Solution

First, sorry I'm a noob at text formatting.
The equations I put in are basically my attempt at the solution. They weren't included in the problem, but I was looking around and thought that they are appropriate. I really don't know where to start with this, and after staring at it for a good hour or so, I've decided to ask for direction.

 

[mighty2000]

To show that the projection operator onto the subspace with eigenvalue ai can be written as:

Pi=[(A-a1)...(A-aa-1)...(A-ar)]/[(ai-a1) ...(ai-ai-1)...(ai-ai+1)...(a1-ar)]

We can start by using the fact that A is a normal operator, which means that A commutes with its adjoint A*. This also means that A commutes with any polynomial in A. So, we can write the projection operator Pi as a polynomial in A:

Pi = c0 + c1A + c2A^2 + ... + cnA^n

Where the coefficients ci are determined by the condition that Pi is a projection operator, meaning that Pi^2 = Pi. This gives us the following equation:

(c0 + c1A + c2A^2 + ... + cnA^n)^2 = c0 + c1A + c2A^2 + ... + cnA^n

Expanding this equation, we get:

c0^2 + (c0c1+c1c0)A + (c0c2+c1c1+c2c0)A^2 + ... + (c0cn+c1cn-1+c2cn-2+...+cnc0)A^n = c0 + c1A + c2A^2 + ... + cnA^n

Since A commutes with its adjoint A*, we can rewrite the above equation as:

c0^2 + 2c0c1A + 2(c0c2+c1c1)A^2 + ... + 2(c0cn+c1cn-1+c2cn-2+...+cnc0)A^n = c0 + c1A + c2A^2 + ... + cnA^n

Now, we can compare the coefficients of the powers of A on both sides of the equation. This will give us a system of equations that we can solve for the coefficients ci. We know that Pi is a projection operator, so we can set c0 = 1 and cn = 0, which gives us:

c1 = 0
c2 = 0
...
cn-1 = 0

This leaves us with the following equation:

c0^2 + 2c0c1A + 2(c0c