Interpretation of an SVD for recommender systems The idea is to motivate the SVD for use in a recommender system.

Consider a matrix $A\in \mathbb{R}^{f\times u}$ where $A_{ij}$ caputures how user $j$ rates film $i$ (on a scale from 1-10, some entries may be missing).
Considering matrices $K=AA^T$, $L = A^TA$ what do $K_{ij}$ and $L_{ij}$ tell us?
What interpretations can you give for matrices $U$ and $V$ in the SVD $A=U\Sigma V^T$?

Riiight ... so
K = $AA^T$ = $U\Sigma^2U^T$, L = $A^TA = V\Sigma^2V^T$
With $K_{ij} = \langle A[i,:], A[j,:] \rangle$ being the dot product of all ratings for film $i$ with all ratings for film $j$.
Similarly, $L_{ij} = \langle A[:, i], A[:,j] \rangle$ being the dot product of all ratings from user $i$ with all ratings from user $j$.
That tells us ... what exactly? I expect it to amount to some kind of similarity measure, but beyond that, I have no idea.
As for $U$ and $V$ ...  I know the first $r$ of them to be Eigenvectors of $K$ and $L$ for some $r$. I also know
$u_1,..,u_r$ is an orthonormal basis for the column space
$v_1,..v_r$ is an orthonormal basis for the row space
but that doesn't really tell me anything.
Yet have no idea if the notion of &quot;row space&quot; and &quot;column space&quot; even makes sense. What would the &quot;span&quot; of $A$'s columns / rows even mean?
I also have no idea what the nullspace of $A$ or $A^T$ would represent - if it represents anything at all - nor what
$u_{r+1},..,u_{m}$ being an orthonormal basis for $ker(A)$
or $v_{r+1},..,v_{n}$ being an orthonormal basis for $ker(A^T)$
implies - again, if anything at all.
First of all, note that the dot product between two movies/users is by definition the correlation between them. So your intuition for treating it as a similarity measure is not wrong.

Now, applying SVD to $A$ is simply a re-representation of it in some other basis. Let's call it feature space. If we'll take it one step further and keep only $k$ leading vectors of $U$ and $V$, we'll get two matrices: $U^{(k)} \in \mathbb{R}^{f,k}$ and $V^{(k)} \in \mathbb{R}^{k,u}$ (where $f$ and $u$ are the dimensions you supplied in your question). where $U^{(k)}\cdot V^{(k)} \approx A^{(k)}$ is a $k$-rank approximation for $A$ ** (note it has the same dimensions).

So you can of $U^{(k)}$ as a matrix which rows movies, and each movie has some $k$ latent features. Similarly, $V^{(k)}$ will be a "list" (columns) of users and each user is described by some $k$ values (latent features). 

Note that you can "split" the scaling matrix $\Sigma^{(k)}$ between $U^{(k)}$ and $V^{(k)}$: $U'^{(k)} = U^{(k)}\Sigma^{(k)0.5}$ and $V'^{(k)} = V^{(k)}\Sigma^{(k)0.5}$, to make up for the lack of scale of $U$ and $V$.

This factorization can be elaborated into more interpretable methods such as LDA.

** The variational approach to singular values tell us that this approximation minimizes the difference between $A$ and any other $k$ rank matrix in terms of Frobenius-norm and operator-norm (and possibly other norms I'm not aware of).