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I'm investigating the eigenvalue ratios

$$ \frac{\lambda_1}{\sum_{j=2}^N\lambda_j} \quad\mbox{and}\quad \frac{\sum_{j=1}^N\lambda_j}{\sum_{j=2}^N\lambda_j} $$

of the NxN matrix $B=AA^T$. $\lambda_1$ denotes the largest eigenvalue. The ratios can be thought of as a measure of "rank-1-ness" of $B$.

I haven't found any mention of either ratio in literature, regardless of constraints on $B$. Has any work been done on this before, and could someone point me there?

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  • $\begingroup$ Not heard of this before, but what do you do to make sure the denominator isn't 0? $\endgroup$
    – Anthony Quas
    Commented Jun 18, 2012 at 15:17
  • $\begingroup$ @Anthony: The matrix for my application originates from a noisy sampling process, so the denominator is almost surely positive. $\endgroup$
    – Anna
    Commented Jun 18, 2012 at 15:37
  • $\begingroup$ @Anna: Which data are available? Is your matrix $B$ a sample covariance matrix? $\endgroup$
    – Stanislav
    Commented Jun 18, 2012 at 20:08
  • $\begingroup$ @Stanislav: It is indeed. The only assumption on $A$ so far is that the expected energy of each sample series is equal, i.e. the 2-norm of each column of $A$ is roughly equal. $\endgroup$
    – Anna
    Commented Jun 18, 2012 at 20:35
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    $\begingroup$ @Stanislav: In my case, A is a function of a a lot of parameters, and takes ~0.1s to calculate (Monte Carlo simulation). I'm trying to optimise the ratio with regard to those parameters. But for this question, I'm just asking if some properties of these ratios have been investigated before. $\endgroup$
    – Anna
    Commented Jun 19, 2012 at 12:18

4 Answers 4

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I am not sure if you are allowed to change your objective function, but a natural alternative for measuring "rank-1-ness" is $$ \frac{\lambda_1^2}{\lambda_1^2+\lambda_2^2+\dots+\lambda_n^2}. $$ This ratio is easy to compute: the denominator is the squared Frobenius norm of the matrix (i.e., sum of squares of all entries), and the numerator is the squared spectral radius, which you can easily estimate using a few iterations of the power method.

The norm $\lambda_1+\lambda_2+\dots+\lambda_n$ is called nuclear norm, and as far as I know it is not as easy to compute as the other two.

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  • $\begingroup$ Sorry about the late answer. This helped me, thanks a lot! $\endgroup$
    – Anna
    Commented Aug 8, 2021 at 14:32
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The eigenvalues you are computing are the singular values of $A.$ The largest singular value is the operator norm of the matrix, the sum of all singular values is another norm (I am not sure what it is called, it's analogous to the $L^1$ norm, while the largest s.v. is the analogue of the $L^\infty$ norm). I am not sure what you are interested in, but if you google "largest singular value", you will see a wealth of information.

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Also, if you have a sample covariance matrix, you should look into the Wishart distribution, there is a lot of work on the distribution of its eigenvalues. See Muirhead: "Aspects of Multivariate Statistical Theory".

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Old question - but it appears that the above answers may have overlooked the fact that the denominator is simply $\sum_{j=2}^N \lambda_j = \left(\sum_{j=1}^{N} \lambda_j \right)- \lambda_1 = \text{Tr}(B) - \lambda_1.$

Thus your quantity of interest is $\frac{\lambda_1}{\text{Tr}(B) - \lambda_1}.$ Especially with $B=AA^T$, the trace should be much easier to calculate then

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    $\begingroup$ This answer doesn't add much to the question itself. It just performs some simple combinatorics with the eigenvalues, but this doesn't shed any light on the problem (which, to be honest, is itself very vague). $\endgroup$
    – Alex M.
    Commented Aug 6, 2021 at 19:41

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