> The original question (possibly misphrased and not in so many words) asked for a way to project any vector into the space spanned by A.
Yes, it was not completely worded. The interpretation above is backwards. There is a subspace sA. In this subspace there is a set of vectors. These vectors are rows in a matrix A. We condition matrix A to create matrix Ak. All row vectors in A are now in a new subspace sAk. Now, there is a new row vector in subspace sA. What matrix projects this row vector into subspace sAk? One use case is new user recommendations. The new user gives a small set of items. We would like to find other users. For simplicity, take users as rows and items as columns in A. We want a conditioned Ak where we can take a new user with a few preferences, and do a cosine distance search among other users. With the ability to project a new user into a conditioned space, we can find other users who may not have any of the new user's items. Using a conditioned Ak increases recall at the expense of precision. The more conditioned, the more recall. On Sun, Sep 16, 2012 at 4:11 PM, Ted Dunning <ted.dunn...@gmail.com> wrote: > On Sun, Sep 16, 2012 at 1:49 PM, Sean Owen <sro...@gmail.com> wrote: > >> Oh right. It's the columns that are orthogonal. Cancel that. >> >> But how does this let you get a row of Uk out of a new row in Ak? Uk >> != Uk * Uk' * Ak >> > > Well, actually, since a column of A is clearly in the space spanned by A so > if you take any such column, U_k U_k' will send it back to where it came > from. Thus, > > A_k = U_k U_k' A_k > > If you want to talk about rows, then use V_k as with this: > > A_k = A_k V_k V_k' > > > Forgetting Sk, Uk = Ak * Vk, and I miss how these are equivalent so I >> misunderstand the intent of the expression. >> > > Forgetting S_k is a bit dangerous here. It is true that > > U_k S_k = A_k V_k > > But, of course, right multiplying by V_k' gives us the identity above. > > I think that the real confusion is that I am talking about projecting back > into span A and you are talking about expressing things in terms of the > latent variables. > > >> On Sun, Sep 16, 2012 at 8:55 PM, Ted Dunning <ted.dunn...@gmail.com> >> wrote: >> > U_k ' U_k = I >> > >> > U_k U_k ' != I >> -- Lance Norskog goks...@gmail.com
