Singular Value Decomposition trouble

Dickfore

Next, take:
[tex]
A \, B \, U = A \, U \, \Lambda
[/tex]
[tex]
(A \, A^{\dagger}) \, (A \, U) = (A \, U) \, \Lambda
[/tex]
[tex]
C \, (A \, U) = (A \, U) \, \Lambda
[/tex]
[tex]
V \, K \, (V^{\dagger} \, A \, U) = (A \, U) \, \Lambda
[/tex]
[tex]
K \, (V^{\dagger} \, A \, U) = (V^{\dagger} \, A \, U) \, \Lambda
[/tex]

We didn't show this, but:
[tex]
\kappa_{k} = K_{k k} = (V^{\dagger} \, C \, V)_{k k} = (V^{\dagger} \, A \, A^{\dagger} \, V)_{k k} = ((V^{\dagger} \, A) (A^{\dagger} \, V))_{k k} = \sum_{i = 1}^{n}{|V^{\dagger} \, A|^{2}_{k i}}
[/tex]
so the set of eigenvalues of [itex]\Lambda[/itex] and [itex]K[/itex] are subsets to each other.

This is why:
[tex]
V^{\dagger} \, A \, U = \Sigma_{m \times n}
[/tex]
where [itex]\Sigma_{m \times n}[/itex] is a diagonal with 1's along the main diagonal (as long as it streches).
Then, we have:
[tex]
A = V_{m \times m} \, \Sigma_{m \times n} \, U^{\dagger}_{n \times n}
[/tex]

Dickfore

I have some mistake in the deduction for the diagonal elements of [itex]\Sigma[/itex].