Spectral Decomposition of a Matrix

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syj
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Homework Statement



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]

Homework Equations



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




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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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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 [itex]A[/itex] be an nxn matrix. Suppose it is similar to a diagaonl matrix:

[tex] P^{-1} AP = \Lambda.[/tex]

The diagonal matrix [itex]\Lambda[/itex] contains the eigenvalues, as usual. Then

[tex] 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},[/tex]

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

[tex] E_i = PD_iP^{-1}[/tex]

so that

[tex]A= \lambda_1 E_1 + \ldots + \lambda_nE_n.[/tex]

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, [itex]E_i=PD_iP^{-1}[/itex], 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 [itex]E^2=E[/itex]. Any matrix that fixes its range is a projection/projector.
 
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  • #3
syj
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Thanks so much.
:)
 

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