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!