? meanings of eigenspan and SVD


? meanings of eigenspan and SVD

Suppose A is a self-adjoint operator in R^N.

Then by spectral thm, A = V*D*V', and the soln

of A*x = b can be written as V*D*V'*x = b. This

can be understood as

b = sum( landa(n)*( v(n)'*x )*v(n), n = 1...N)

= sum( landa(n)*proj(x, v(n)), n = 1...N)

Think in another direction, we know there are 4

subspaces assocaited with this linear map A*x = b:

A maps vector x in domain(A) ( = Range(adjoint(A))+null(A) )

to Range(A)+null(adjoint(A)) ( = what space? )

A is self-adjoint means Range(A) = Range(adjoin(A)) and

null(A) = null(adjoint(A)) and if we express vector x in

domain(A) by the eigenvectors: x = sum( ( v(n)'*x )*v(n), n = 1...N)

= sum( proj(x, v(n), n = 1...N)

The map A plays the role of re-scaling each linearly independent

components, and thus

A*x = sum( A*( v(n)'*x )*v(n), n = 1...N)

= sum( landa(n)*( v(n)'*x )*v(n), n = 1...N)

= sum( landa(n)*proj(x, v(n)), n = 1...N)

In more detailed, we see that those eigenvectors with non-zeros

eigenvalues together spann Range(A) = Range(adjoint(A)) and those

with zero eigenvelus together span null(A) = null(adjoint(A)).

Now, let's see SVD. Given any square matrix A that represents a

linear map A*x = b. We have A = U*S*V' = sum( s(n)*u(n)*v(n)', n = 1...N).

Similarly it means we first express x as

x = sum( proj(x,v(n)), n = 1...N) = sum( ( v(n)'*x )*v(n), n = 1...N ).

But it becomes strange for A*x, because then we have:

A*x = sum( A*( v(n)'*x )*v(n), n = 1...N )

= sum( ( v(n)'*x )*A*v(n), n = 1...N )

= b = sum( s(n)*u(n)*v(n)'*x, n = 1...N)

= sum( ( v(n)'*x )*s(n)*u(n), n = 1...N)

That is,

sum( A*v(n), n = 1...N )

= sum( s(n)*u(n), n = 1...N)

How to interprete the eqn above?

Also in this case we still have 4 subspaces:

domain(A) = Range(adjoint(A))+null(A)

and Range(A)+null(adjoint(A))

How do we assign u(n) and v(n) to span them?

Thanks,
by Cheng Cosine
Mar/03/2k6 NC


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