TITLE: Bounding the spectral gap of a simple symmetric matrix
QUESTION [8 upvotes]: I have a seemingly innocent linear algebra problem that I cannot solve, and which I hope that you would kindly offer some insight into. Here is the description: Let $\mathbf{a} = (a_1, a_2, \dots, a_d)^{T}$ be a positive probability vector, $i.e.$ $\Vert \mathbf{a}\Vert_1=1$ and $a_i > 0$ for all $i$. Let matrix $A$ be defined as follows: $$A = \textrm{diag}(\mathbf{a}) - \mathbf{a}\mathbf{a}^{T}$$ where $\textrm{diag}(\mathbf{a})$ means the diagonal matrix with the $i$th diagonal entry being $a_i$. It is straightforward to show that $\mathbf{1}_d$, the all-one vector of dimension $d$, is an eigenvector of $A$ of eigenvalue $0$. And Gershgorin circle theorem also shows that all $A$'s eigenvalues are greater or equal to $0$. My question is: 
What is the smallest eigenvalue of $A$ that is not zero? 
I carried out the calculation when $d = 3$ and realized that there may not be a simple analytic formula to it and hence a nice lower bound is also greatly appreciated.
Thank you so much!

REPLY [2 votes]: Here is an elementary bound. The second eigenvalue of $A$ satisfies
$$\lambda_2(A)> \max_k\min_{i\ne k}a_ia_k=(\max_ka_k)(\min_ia_i).$$
To prove it, let $A_k$ be the principal submatrix obtained by deleting the $k$th row and column. By interlacing, we have $\lambda_2(A)>\lambda_1(A_k)$. Now apply Gershgorin to $A_k$: there exists an index $i\ne k$ such that 
$$\lambda_1(A_k)\ge a_i(1-a_i)-\sum_{j\ne i,k}a_ia_j=a_ia_k.$$
Improvement.
Actually, one has $\lambda_2(A)\ge\min_ia_i$. To see this, write $A=D+B$ with $D={\rm diag}(a_1,\ldots,a_n)$ and $B=-aa^T$. By Weyl's inequality, we have
$$\lambda_2(A)\ge\lambda_1(D)+\lambda_2(B)=\min_ia_i+0.$$
Remark that this bound is accurate, as if $a_1=\cdots=a_{n-1}=\min_ia_i$ and $a_n=1-(n-1)a_1$, then the spectrum of $A$ is given by $0$, $a_1$ (multiplicity $n-2$) and $na_1a_n$. Hence $\lambda_2=a_1$.