a question on linear algebra

[joshuashi]

Hi friends,

For the same matrix A, we want to get its eigen value and eigen vector. What is the difference if we use eigenvalue decomposition and singular value decomposition? How to understand the difference of the results?

Thank you very much for your help!


joshuashi

 

[mmatica]

The eigenvalue decomposition for A produces:
A = P^(-1) J P
where J is the so called Jordan form of the matrix (it contains all the eigenvalues on the main diagonal) and P is a nonsingular matrix (its column are the eigenvectors if A is "diagonalizable", in which case J is a diagonal matrix).
[Note that if the eigenvalues are distinct then A is diagonalizable]

The singular value decomposition (SVD) of A produces:
A = U D V'
where D is a diagonal matrix and U,V are orthogonal. (V' denotes the transpose if A has real entries).
D contains the "singular values" of the matrix A; they are in fact the square roots of the eigenvalues of the matrix A A'.

The eigenvalue decomposition is very useful in theoretical problems.
SVD has the advantage that it is much more stable for numerical computations and can be used even if A is rectangular (non square).

M.[/quote]

 

[joshuashi]

Hi mmatica,

Thank you for the reply!

But for a square matrix, how to understand the difference when you use eigenvalue decomposition and SVD decomposition? In particular, when this matrix is the covariance matrix of the signal received at an antenna array? - sorry for this question if you are not work in this field.

Thx,
joshuashi

 

[nebisman]

I'm working on linear systems theory. The SVD decomposition provides useful information about the "directionality" of a system. Suppose th matrix A is a linear operator such that:

Y=A*X


then the decomposition

A=U*sigma *V

Y=U*sigma*V*U


sigma represents the difernte values of amplificacion of this system.
U represents direction of vectro of maximum amplificacion of the vector input. I don't know realy if this is true in your field.