Re: Help with eigenvalue decomposition
- From: "Sumit Mitra" <sumit.mitra@xxxxxxxxxxxxxx>
- Date: Thu, 6 Oct 2005 04:29:41 +0200
Let A be full-ranked, symmetric, K^(n x n).
Let b_1 be eigenvector of A, then c_1*b_1 (c element of K) is also
eigenvector of A.
=> If A can be represented as A = B*D*inv(B), then A can be also represented
as A = B~*D*inv(B~), where B~ = [c_1*b_1 | c_2*b_2 | ... | c_n*b_n ]. As you
can see, you can "use" for every eigenvector an own "c".
=> If A is diagonalizable, the diagonalization is not unique.
But in your case, you have some stricter assumptions.
> B is a unitary matrix ( inv(B)=B^T) and D is the diagonal matrix that
contains the eigen values of A in increasing order. The elements of B are
orthonormal eigenvectors of A.
What I showed you beyond, is now no longer possible.
Let P be unitary. P~ = c*P is no longer unitary, c <> 1, so you cannot use
inv(P~)=(P~)^T.
We could think, now the diagonalization is unique: If we have the
eigenvalues in 1 matrix D, we have 1 Matrix B.
But you think, there could be a second solution with the same matrix D, but
another unitary matrix, E, that should be a permutation of B.
> Is it correct to say that F=D and E=Perm(B) i.e. the columns of E are
permutation of columns of B?
Ok, let me do a "short" 2 x 2 counter-example.
B=[b11, b12; b21, b22]
D=[d1,0; 0, d2]
B^T=[b11, b21; b12, b22]
B*D =
[b11*d1, b12*d2;
b21*d1, b22*d2]
B*D*B^T=
[b11^2 *d1 + b12^2 *d2, b11*b21*d1 + b12*b22*d2;
b21*b11*d1 + b22*b12*d2, b21^2 *d1 + b22^2 * d2]
If you permutate B, you get:
E=[b12, b11; b22, b21]
F=D=[d1,0; 0, d2]
E^T=[b11, b21; b12, b22]
E*F =
[b12*d1, b11*d2;
b22*d1, b21*d2]
E*F*E^T=
[b11^2 *d2 + b12^2 *d1, b11*b21*d2 + b12*b22*d1;
b21*b11*d2 + b22*b12*d1, b21^2 *d2 + b22^2 * d1]
As you can see, it is almost the same. The only difference is, that d1 and
d2 changed their places.
But this is a contradiction to your assumption F=D. It would be only true,
if F is "the right permutation" of rows of D.
Best wishes,
Sumit
.
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