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I have a question regarding matrices and eigen values.

If SVD decomposition was performed on matrix, and the inner matrix of singular values has only one non zero value. Should the left and right singlular matrices be only a matrix of one vector?

The reason I ask is that I have performed SVD on matrix in matlab using function svd, and it turns out that there is one eigen value but the right and left matrices are not one vector

Thanks

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  • $\begingroup$ Yes you can choose to have a compact or a less compact form. What limits how compact you can have is the number of non-zero singular values, just as you say. $\endgroup$ – mathreadler Sep 1 '15 at 14:20
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There are two different conventions for the SVD. First there is the full convention, where you take $A \in \mathbb{K}^{m \times n}$ and you get $U \in \mathbb{K}^{m \times m},\Sigma \in \mathbb{R}^{m \times n},V^* \in \mathbb{K}^{n \times n}$. (Here $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$.) Second there is the reduced convention, where you take $A \in \mathbb{K}^{m \times n}$ with rank $r$ and you get $U \in \mathbb{K}^{m \times r},\Sigma \in \mathbb{R}^{r \times r},V^* \in \mathbb{K}^{r \times n}$. The reduced SVD contains all the information in the full SVD, because in the full SVD, the zero diagonal entries in $\Sigma$ simply destroy whatever extra information is present in $U$ and $V$. The reduced SVD is also faster and more straightforward to compute. So in numerical calculations we tend to use it rather than the full SVD. The full SVD is convenient for theory, however.

This same distinction occurs with the QR decomposition.

In MATLAB you can distinguish between the two using svd(A) (which gives the full SVD) and svd(A,0) (which gives the reduced SVD). Be warned that the rank calculation is not a trivial matter, numerically. For matrices which do not have full rank or are close to such a matrix, the rank will often be computed incorrectly (either nonzero singular values will be treated as zero, or truly zero singular values will have errors that make them be treated as nonzero). If you see very small but positive singular values, or more than $|n-m|$ zero singular values, then you may be in this case.

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  • $\begingroup$ I think there is some problem with the dimension of the matrices in conversion 1. $\endgroup$ – Rajat Sep 1 '15 at 13:50
  • $\begingroup$ @Rajada Typo, thanks. $\endgroup$ – Ian Sep 1 '15 at 13:55
  • $\begingroup$ In my case, the matrix $A$ is 4 by 4 and after I do the SVD in matlab $\Sigma$ has one non singular value but $U$ and $V$ are matrices which are dimension 4 by 4 have entries are all non zero vectors. I would have thought there should be just one vector that should be non zero. Do you think its an error from MATLAB? $\endgroup$ – Henry Sep 1 '15 at 14:57
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    $\begingroup$ @FareedAF In the square case, yes. $\endgroup$ – Ian Jul 15 at 10:11
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    $\begingroup$ @FareedAF It is better to say that the rank of a $n \times n$ matrix is $n$ minus the geometric multiplicity of the eigenvalue $0$. In the diagonalizable case this is the sum of the multiplicities of all the nonzero eigenvalues. $\endgroup$ – Ian Jul 15 at 12:29

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