In my professor's notes today he talked about some sort of rank-1 decomposition that had something to do with singular value decomposition. He briefly mentioned for us to go over it, but I have no idea what exactly he is referring to. I did some google searches and found something like this but I'm not entirely sure this is what I'm looking for: http://www.uwlax.edu/faculty/will/svd/norm/index.html

Can anyone give me some idea of what exactly he might be referring to?

If so, what exactly is the purpose of this operation, how is it related to SVD, and do you know of anywhere I can read to learn how to do this (as I don't think its in my textbook Introduction to Linear Algebra by Gilbert Strang)?.


  • $\begingroup$ Your question is vague. It would be of help if you could phrase your question to let us know what exactly you want. Once you have the singular value decomposition as $A = U \Sigma V^*$, the "best" rank approximation is given by $\displaystyle A_r = \sum_{k=1}^{r} \sigma_k u_k v_k^*$, where $u_k,v_k$ are the $k^{th}$ columns of $U$ and $V$ respectively. By "best", of all rank $r$ matrices, $||A-A_r||_2$ (also the frobenius norm of $A-A_r$) is the minimum. $\endgroup$
    – user17762
    Dec 22, 2011 at 1:24
  • 1
    $\begingroup$ I gave a good discussion of the technique and an application here. $\endgroup$ Dec 22, 2011 at 1:25
  • $\begingroup$ @J.M. Thanks for the link. I think that is what I'm looking for. Can you give, or do you know where I can find a simple numerical example of how it works? $\endgroup$
    – Matt
    Dec 22, 2011 at 1:56

1 Answer 1


The link you gave answers your question. You have three $5\times 1$ unit column vectors $h_i$, orthogonal to each other, making up a $5\times3$ matrix, then a $3\times3$ diagonal matrix, then three $1\times4$ unit row vectors $a_j$, orthogonal to each other, making up a $3\times4$ matrix, and that's your singular-value decomposition.

Then $h_1 a_1$, $h_2 a_2$, and $h_3 a_3$ are $5\times4$ matrices, each with rank $1$. If you multiply them by the corresponding singular values, you still have three $5\times4$ matrices, each with rank $1$. Their sum is $A$, the matrix you started with. So you've decomposed $A$ as a sum of three matrices, each with rank $1$. That's the rank-$1$ decomposition.


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