Spectral Decomposition of a Matrix

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SUMMARY

The discussion focuses on the spectral decomposition of a matrix A, specifically finding its projectors using Lagrange interpolation polynomials and diagonalization. The matrix A is represented as a 3x3 matrix with specific entries. The participants clarify that the projectors E_i can be derived from the diagonalization of A, where A can be expressed as A = λ_1 E_1 + ... + λ_n E_n. It is emphasized that projectors are idempotent matrices, satisfying E^2 = E.

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  • Understanding of spectral decomposition and diagonalization of matrices
  • Familiarity with Lagrange interpolation polynomials
  • Knowledge of Vandermonde matrices and their properties
  • Basic linear algebra concepts, including eigenvalues and eigenvectors
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  • Study the process of diagonalizing matrices, specifically using eigenvalues and eigenvectors
  • Learn about Lagrange interpolation polynomials and their applications in matrix theory
  • Explore the properties and applications of Vandermonde matrices in linear algebra
  • Investigate the concept of idempotent matrices and their role in spectral decomposition
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Students and professionals in mathematics, particularly those studying linear algebra, matrix theory, and numerical methods. This discussion is beneficial for anyone looking to deepen their understanding of spectral decomposition and its applications.

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 textbook 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.
 
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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 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:
Thanks so much.
:)
 

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