Spectral Decomposition of a Matrix

In summary, the Homework statement is that you need to find the projectors of matrix A. The Attempt at a Solution says that if you diagonalize the matrix, you will get the projectors.
  • #1
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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  • #2
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.
 
Last edited:
  • #3
Thanks so much.
:)
 

What is spectral decomposition of a matrix?

Spectral decomposition, also known as eigendecomposition, is a mathematical process in which a matrix is decomposed into a set of eigenvectors and eigenvalues. This allows for simplification of complex matrix operations and can provide insights into the behavior of the matrix.

How is spectral decomposition used in science?

Spectral decomposition has various applications in science, including in fields such as physics, engineering, and computer science. It can be used to analyze the behavior of systems, solve differential equations, and perform data compression and dimensionality reduction in machine learning.

What is the significance of eigenvalues and eigenvectors in spectral decomposition?

Eigenvalues represent the scaling factor of the corresponding eigenvector in the matrix transformation. They can provide insights into the behavior of the system and can also be used to determine stability and convergence. Eigenvectors, on the other hand, represent the directions in which the matrix transformation has no effect, and can help identify key patterns and relationships in the data.

Can any matrix be decomposed using spectral decomposition?

Not all matrices can be decomposed using spectral decomposition. A matrix can only be decomposed if it is square and has a full set of linearly independent eigenvectors. Matrices that do not meet these criteria may require other methods of decomposition.

How is spectral decomposition different from other methods of matrix decomposition?

Unlike other methods of matrix decomposition, such as LU decomposition or QR decomposition, spectral decomposition is unique to square matrices. It also provides a direct relation between the eigenvectors and eigenvalues of the matrix, making it useful for certain applications such as principal component analysis.

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