On connection of distance to finite rank operators and singular values. Im trying to understand why the following; 
    $s_n(T) = \inf\big\{\, \|T-L\| : L\text{ is an operator of finite rank }<n \,\big\}$ where $s_{n}$ are nth singular values,
  is a plausible claim to make. Even after "understanding" the proof, it puzzles me. Is there an interpretation of this which makes it a reasonable conjecture?
 A: By the spectral theorem, we know that $\sqrt{T^*T}$ can be written as a "diagonal" operator with respect to some orthonormal Hilbert-basis $\{e_k\}$.  That is, we can define
$$
\sqrt{T^*T} \left( \sum_{k=1}^\infty a_k e_k\right) =  \sum_{k=1}^\infty \lambda_k a_k e_k
$$
Where $\lambda_k \to 0$ as $k \to \infty$ and $\lambda_k \geq 0$ for each $k$.  Let's make the further assumption that $\lambda_k$ is a decreasing sequence over $k$.  In this case, I think it is easy to believe that the best rank $n-1$ approximation of the map $\sqrt{T^*T}$ is given by
$$
S\left( \sum_{k=1}^\infty a_k e_k\right) =  \sum_{k=1}^{n-1} \lambda_k a_ke_k
$$
In this case, it should be easy to verify that the result holds true.
From there, it suffices to note that applying a partial isometry $U$ (since $T = U\sqrt{T^*T}$, for some $U$) "doesn't really change things"; $US$ is the best rank $n-1$ approximation of $U\sqrt{T^*T}$.
A: I think the idea comes from spectral theory of self-adjoint compact operators. $s_n(T)$ are supposed to be a generalization of eigenvalues of operators. Assume that you have an operator which has the spectral decomposition
$$T x = \sum_{n=1}^{\infty} \lambda_{n} \left\langle x, e_n \right\rangle e_n$$
where $\lambda_1 \leq \lambda_2 \leq \ldots$. For compact self-adjoint operators one knows that $\left\Vert T \right\Vert$ or $-\left\Vert T \right\Vert$ is an eigenvalue of $T$. Now if you want to find the eigenvalues of $T$ one usually does the following:
Take $\lambda_1 = \pm \left\Vert T \right\Vert$ and let $Lx := \lambda_1 \left\langle x, e_1 \right\rangle e_1$. Then $T - L$ is again compact and selfadjoint and you can do the same thing for this operator to get $\lambda_2$.
On the other hand $\sum_{n=1}^N \lambda_n \left\langle x, e_n \right\rangle e_n =: L$ is the best approximation for $T$ using only operators of rank $< N+1$. In fact, $\lambda_{N+1}  = s_{N+1}(T)$ for self-adjoint, compact operators.
