## Singular Value Decomposition trouble

Next, take:
$$A \, B \, U = A \, U \, \Lambda$$
$$(A \, A^{\dagger}) \, (A \, U) = (A \, U) \, \Lambda$$
$$C \, (A \, U) = (A \, U) \, \Lambda$$
$$V \, K \, (V^{\dagger} \, A \, U) = (A \, U) \, \Lambda$$
$$K \, (V^{\dagger} \, A \, U) = (V^{\dagger} \, A \, U) \, \Lambda$$

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

This is why:
$$V^{\dagger} \, A \, U = \Sigma_{m \times n}$$
where $\Sigma_{m \times n}$ is a diagonal with 1's along the main diagonal (as long as it streches).
Then, we have:
$$A = V_{m \times m} \, \Sigma_{m \times n} \, U^{\dagger}_{n \times n}$$
 I have some mistake in the deduction for the diagonal elements of $\Sigma$.