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Two words can be similar if they co-occur "a lot" together. They can also be similar if they have similar vectors. This similarity can be captured using cosine similarity. Let $A$ be a $n \times n$ matrix counting how often $w_i$ occurs with $w_k$ for $i,k = 1, \dots, n$. Since computing the cosine similarity between $w_i$ and $w_k$ might be expensive, we approximate $A$ using truncated SVD with $k$ components as: $$A \approx W_k \Sigma W^{T}_{k} = CD$$

where $$C = W_{k} \Sigma \\ D = W^{T}_{k}$$

Where are the cosine similarities between the words $w_i$ and $w_k$ captured? In the $C$ matrix or the $D$ matrix?

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  • $\begingroup$ Where did you read this formulation? Edit your question to add these details. $\endgroup$ – nbro Mar 15 at 18:03
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You can find some material here and here but the idea (at least in this case) is the following: consider the full SVD decomposition of the symmetric matrix $A = W \Delta W^T$. We want to calculate the cosine similarity between the $i$-th column (aka word) $a_i$ and the $j$-th column $a_j$ of $A$. Then $a_k = A e_k$, where $e_k$ is the $k$-th vector of the canonical basis of $\mathbb{R}$. Let's call $\cos(a_i,a_j)$ the cosine between $a_i,a_j$. Then $$\cos(a_i,a_j) = \cos(Ae_i,Ae_j) = \cos(W \Delta W^T e_i,W \Delta W^T e_j) = \cos(\Delta W^T e_i,\Delta W^T e_j)$$

where the last equality holds because $W$ is an orthogonal matrix (and so $W$ is conformal). So you can calculate the cosine similarity between the columns of $\Delta W^T$. A $k$-truncated SVD gives a well-enough approximation. In general, columns of $W \Delta$ and rows of $W$ have different meanings!

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  • $\begingroup$ What do you mean by "conformal"? $\endgroup$ – nbro Mar 16 at 13:23
  • $\begingroup$ @nbro Conformal means that preserves angles $\endgroup$ – dcolazin Mar 16 at 13:27
  • $\begingroup$ But isn't $W^T$ also orthogonal and therefore conformal? If yes, why can't we simply use the diagonal matrix? $\endgroup$ – nbro Mar 16 at 13:41
  • $\begingroup$ Also $W^T$ is orthogonal and conformal, but you only "simplify" on the left: if $f$ is a conformal map, $g$ any function, then $\cos(f \circ g(x),f \circ g(y)) = \cos(g(x),g(y))$, but in general $\cos(g \circ f(x),g \circ f(y)) \neq \cos(g(x),g(y))$ $\endgroup$ – dcolazin Mar 16 at 13:48
  • $\begingroup$ @nbro for example, consider this matrix: $\cos(Me_1,Me_3) = 1 = \cos(\Sigma V^T e_1, \Sigma V^T e_3)$ but $\cos(\Sigma e_1,\Sigma e_3) = 0$ (cosine between different columns in a diagonal matrix is always $0$). $\endgroup$ – dcolazin Mar 16 at 13:57

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