Smallest eigenvalue of a random matrix

Question

Let $A \in \mathbb R^{n\times n}$ be a positive semi-definite matrix, and let $b \in \mathbb R^n$. For a random vector $x \sim \mathcal N(0, I_{n\times n})$, consider the random matrices $$ B_1 = A + xx^T, \qquad B_2 = A + (x + b)(x + b)^T. $$ Intuitively, it seems likely to me that the following inequality should be satisfied for all $r \geq 0$: $$ \mathbb P[\lambda_{\min}(B_1) ≥ r] \leq \mathbb P[\lambda_{\min}(B_2) ≥ r] \,. $$ However, I have searched the literature (particularly the literature on non-central Wishart distributions) and have not found an answer to this question. Does this follow from classical results, or is it an open question?

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EDIT

In fact, the inequality does not hold. Indeed, in the example suggested by @Ben Deitmar, it is the other way around. Consider the case $$ A = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}, \qquad b = \begin{pmatrix} x \\ 0 \end{pmatrix}. $$

• For $x = 2$:

• For $x = 5$, the difference is even more pronounced:

It looks as if, in the limit $x \to \infty$, the random varibale $\lambda_{\min}(B_2)$ tends to 1 in probability. The Julia code, for reproducibility purposes:

using LinearAlgebra
using LaTeXStrings
using Plots

A = [2. 0.; 0. 1.]
b = [5., 0.]

n = 10^5
λ₁, λ₂ = zeros(n), zeros(n)

for i in 1:n
    x = randn(2)
    B₁ = A + x*x'
    B₂ = A + (b + x)*(b + x)'
    λ₁[i] = eigvals(B₁)[1]
    λ₂[i] = eigvals(B₂)[1]
end

λ₁ = sort(λ₁)
λ₂ = sort(λ₂)

plot(λ₁, (1:n)/n, label=L"CDF of $\lambda_{\min}(B_1)$")
plot!(λ₂, (1:n)/n, label=L"CDF of $\lambda_{\min}(B_2)$")
savefig("CDF.png")

Answer 1

Let $x \sim \mathcal N(0, I_2)$, and for $z \in \mathbb R$, consider the random matrix $$ B_z = A + (x + b_z)(x + b_z)^T. $$ where $$ A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad b_z = \begin{pmatrix} z \\ 0 \end{pmatrix}. $$ Let us prove that $\lambda_{\min}(B_z)$ tends to 0 in probability as $z \to \infty$, which will prove rigorously that the inequality in the question does not hold.

To this end, let $y_z(x)$ be a unit vector orthogonal to $x + b_z$. Note that since $x$ is a random vector, so is $y_z$. Then $$ \lambda_{\min}(B_z) \leq y_z(x)^T B_z y_z(x) = y_z(x)^TAy_z(x) = \langle y_z(x), e_1 \rangle^2. $$ Clearly, the right-hand side converges to 0 in probability as $z \to \infty$, because in this limit $x + b_z$ aligns with $e_1$. The conclusion the follows.