If $\Sigma$ is a diagonal matrix and $\vec{x}$ a vector, is there a simple formula for the singular values of the matrix $[\Sigma, \vec{x}]$?

A hint: first, write out our matrix $[\Sigma,\vec{x}]$ as

$\left(\begin{array}{cccc} \sigma_1 & & & x_1 \\ & \ddots & & \vdots \\ & & \sigma_n & x_n \end{array}\right)$

Then the rightmost column of the right singular vectors is a basis vector for the nullspace of $[\Sigma,\vec{x}]$ and is proportional to

$\left(\begin{array}{c} -x_1/\sigma_1 \\ \vdots \\ -x_n/\sigma_n \\ 1\end{array}\right)$

Perhaps this can be used as a starting point for a solution.

Edit Apparently a solution to a more general problem can be found in the paper "Updating the Singular Value Decomposition" by J. R. Bunch and C. P. Nielsen. The solution for the eigenvalues involves solving a rational equation and it seems a closed form solution is not practical.

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    $\begingroup$ What does the bracket notation mean? Can you define $[ \Sigma, \vec{x} ]$? $\endgroup$ – Sammy Black Mar 21 '13 at 21:22
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    $\begingroup$ I presume it means 'stack' $x$ to the right of the diagonal matrix $\Sigma$? $\endgroup$ – copper.hat Mar 21 '13 at 21:22
  • $\begingroup$ I think copper.hat is right. Isn't it clear from concatenated? $\endgroup$ – Git Gud Mar 21 '13 at 21:23
  • $\begingroup$ I thought I knew what "concatenated" and what "stack" mean, yet I still don't understand what does this mean in this context... $\endgroup$ – DonAntonio Mar 21 '13 at 21:28
  • $\begingroup$ I think he means something like: given $A=\begin{pmatrix} y & 0 \\ 0 & z \end{pmatrix}$ and $\vec{x}=[a\space b]^T$, then $$[A,\vec{x}]=\begin{pmatrix} y & 0 & a \\ 0 & z & b\end{pmatrix}$$ $\endgroup$ – Git Gud Mar 21 '13 at 21:33

If $A$ is your matrix, $A^T A$ is a rank-2 perturbation of a diagonal matrix, so you should be able to apply the Matrix determinant lemma

EDIT: even better, $A A^T = \Sigma^2 + x x^T$ is a rank-1 perturbation of the diagonal matrix $\Sigma^2$. Its characteristic polynomial is then $$ P(\lambda) = \left(1 + \sum_j (\sigma_j^2-\lambda)^{-1} x_j^2\right) \prod_k (\sigma_k^2 - \lambda) = \prod_k (\sigma_k^2 - \lambda) + \sum_j x_j^2 \prod_{k \ne j} (\sigma_k^2 - \lambda)$$ where $\sigma_j$ are the diagonal elements of $\Sigma$. So you need to solve that for $\lambda$ and take square roots to get the singular values.

  • $\begingroup$ Interesting...how would I use that to get the singluar values? $\endgroup$ – David Pfau Mar 21 '13 at 21:56

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