# Homework Help: Spectral Decomposition of a Matrix

1. Aug 2, 2011

### syj

1. The problem statement, all variables and given/known data

Find the projectors of matrix A
(I am not sure how to write it in matrix form here so I have written out each row of the matrix)
Row 1 : [ 2 1 1]
Row 2: [ 2 3 2]
Row 3: [ 1 1 2]
2. Relevant equations

A theorem in my text book says the Lagrange interpolation polynomial hi(A) = the projectors Ei

3. The attempt at a solution

If my understanding is correct, then all I need is the Lagrange interpolation polynomials for this matrix.

Do I have to first express this matrix in terms of a vandermonde matrix to get the fixed points?

I know how to find the Lagrange interpolation polynomials if I am given the fixed points.
I know that the second column of the vandermonde matrix gives you the fixed points.
I am not so sure as to how to express this matrix in terms of a vandermonde matrix.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 5, 2011

### stringy

It's a good thing you told me what book you were using in that other thread, because I'm not familiar with the terminology here. However, I looked your book up on Google Books, so we're good to go now!

I've heard spectral decomposition used sometimes as a synonym for diagaonlization, and that's essentially what we're doing when we look for these projectors.

Let $A$ be an nxn matrix. Suppose it is similar to a diagaonl matrix:

$$P^{-1} AP = \Lambda.$$

The diagonal matrix $\Lambda$ contains the eigenvalues, as usual. Then

$$A = P\Lambda P^{-1} = P(\lambda_1 D_1 + \ldots +\lambda_nD_n)P^{-1} = \lambda_1 PD_1P^{-1} + \ldots + \lambda_n PD_nP^{-1},$$

where the $\lambda$'s are the eigenvalues (not necessarily unique) and the $D_i$ matrices are a zero matrix with a 1 in the $i,i$ position. Then, define

$$E_i = PD_iP^{-1}$$

so that

$$A= \lambda_1 E_1 + \ldots + \lambda_nE_n.$$

The right-hand side of the equation is termed the spectral decomposition of A and the E's are the projectors.

I see the Lagrange polynomials that you are referring to. They are in the proof of Theorem 7.11, right? I don't think we need to use those, we can just do this directly.

We first need to diagonalize our matrix A. You will NOT get 3 unique eigenvalues, but that's OK. A full set of unique eigenvalues is only a sufficient condition for diagonalization, not necessary. Once you diagonalize A, follow the formula I wrote above, $E_i=PD_iP^{-1}$, to get your ith projector. At the end, write down the decomposition using these matrices and see if you recover A.

EDIT: BTW, they are called projectors because they are idempotent, or $E^2=E$. Any matrix that fixes its range is a projection/projector.

Last edited: Aug 5, 2011
3. Aug 5, 2011

### syj

Thanks so much.
:)