What is a "thin" svd?

Question

I've seen lots of mention of this in the implementation of SVDs in various programming environments. What does it actually mean?

Is the definition given in Wikipedia sufficient for your purposes? — mhum, Feb 23 '13 at 00:57

score 15 · Accepted Answer · answered Feb 23 '13 at 04:19

Let the SVD of an $m \times n$ matrix be $A=U \Sigma V^T$. Further, suppose it has rank $r$. Then, $A$ can be written as

$$ A = \sum_{i=1}^r \sigma_i u_i v_i^T + \sum_{i=r+1}^{\min(n,m)}0 \cdot u_i v_i^T$$.

The "thin" SVD is just the first part where the "fat" (?) SVD is the entire expression. In other words, the remaining parts can be discarded. Therefore, if we know the matrix is of rank $r$, we only need to find those $r$ terms.

What is a "thin" svd?

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