A Peculiar Lower Bound to the Spectral Radius |7 March. 2020|

As a follow up to the previous post, we discuss a lower bound, given as exercise P7.1.14 in (GvL13). I have never seen this bound before, especially not with the corresponding proof, so here we go.

Given any Ain mathbf{R}^{ntimes n}, we have the following lower bound:

 rho(A)geq (sigma_1cdots sigma_n)^{1/n},

which can be thought of as a sharpening of the bound |lambda_i(A)|geq sigma_{mathrm{min}}(A). In the case of a singular A, the bound is trivial (rho(A)geq 0), but in the general setting it is less obvious. To prove that this holds we need a few ingredients. First, we need that for any X,Yin mathbf{R}^{ntimes n} we have mathrm{det}(XY)=mathrm{det}(X)mathrm{det}(Y). Then, we construct two decompositions of A, (1) the Jordan Normal form A=TJT^{-1}, and (2) the Singular Value Decomposition A=USigma V^{top}. Using the multiplicative property of the determinant, we find that

 |mathrm{det}(A)| = left|prod^n_{i=1}lambda_i(A) right| = left| prod^n_{i=1}sigma_i(A) right|.

Hence, since rho(A)=max_i{|lambda_i(A)|} it follows that rho(A)^n geq |prod^n_{i=1}sigma_i(A)| and thus we have the desired result.


(Update 8 March) It is of course interesting to get insight in how sharp this bound is, in general. To that end we consider random matrices Ain mathbf{R}^{5times 5} with mathrm{vec}(A){sim}mathcal{N}(0,I_{25}). When we compute the relative error (rho(A)-(sigma_1cdots sigma_n)^{1/n})/ rho(A) for 10000 samples we obtain the histogram as shown on the left. Clearly, the relative error is not concentrated near 0. Nevertheless, it is an interesting bound, from a theoretical point of view.

(GvL13) G.H. Golub and C.F. Van Loan: ‘‘Matrix Computations’’, 2013 John Hopkins University Press.