Sentence examples for all eigenvalues equal from inspiring English sources

Root loci show that any system designed to have all eigenvalues equal is optimally stable with respect to variation of any design parameter from its nominal value.

Since we are interested in polynomial solutions of Eq. (A.2), it is clear that they can be obtained only if λ = 0. Having all eigenvalues equal to zero is a very restrictive requirement that is not relevant for the problem we are considering here.

Perfect sphericity requires ∊ = 1, which occurs with all eigenvalues equal.

Moreover, a (Qtimes Q) matrix whose cell values are all equal to 1 / Q, which is an environment sometimes described as "complete mobility", has an eigenvalue equal to one and all the other eigenvalues equal to zero so that its measure of intergenerational mobility is equal to 1.

Note in this respect that the identity matrix has all its eigenvalues equal to one and, therefore, its measure of intergenerational mobility is equal to zero.

Conversely, if the j P-values are perfectly correlated (i.e. if the phenotypes m and the SNPs n are perfectly correlated), then the first eigenvalue equals j, all other eigenvalues equal 0, and q ej =  j − (j − 1) = 1 (i.e. testing the association of perfectly correlated phenotypes with perfectly correlated SNPs yields only one unique unit of information).

If all j P-values are uncorrelated (i.e. if the phenotypes m and/or SNPs n involved in the statistical tests giving rise to the j P-values, are uncorrelated), then all j eigenvalues equal 1, and q ej =  j − 0 =  j.

With the constraint that the sum of the eigenvalues remains constant, the maximum of the product is achieved when all eigenvalues are equal.

There are many works on the distribution of eigenvalues in this asymptotic regime, such as information-plus-noise [16] and spiked models where all eigenvalues are equal excluding a small number of fixed eigenvalues (spikes) [17].

When all the c k are zero (i.e., all eigenvalues are equal to zero), then f is identically zero and one can get the eigenfunctions of L by considering the (N − 1 -order operator obtained from L by replacing y by Dy.

The judgment matrix has an eigenvalue equal to n if the comparisons are perfectly consistent.