Let $A\in \mathbb{R}^{n\times n}$ be positive definite matrix with least eigenvalue $\lambda$. We create a new matrix $\widetilde{A}\in \mathbb{R}^{(n+1)\times (n+1)}$ in the following way: Pick $i\in \{1,\dots,n\}$ and add a row and a column to $A$ such that for every $j\neq i$: $ a_{n+1,j} = a_{i,j}, ~~ a_{j,n+1} = a_{j,i}$, and $~a_{n+1,n+1} = a_{i,i}, ~a_{i,n+1} = a_{n+1,i} = \frac{1}{2}$. For example, if we pick $i=n$ and denote $A = \begin{pmatrix} A' ~~ b \\ b^\top \alpha \end{pmatrix}$ where $A'\in \mathbb{R}^{(n-1)\times (n-1)},~ b\in \mathbb{R}^{n-1},~ \alpha\in \mathbb{R}$ then:
$\widetilde{A} = \begin{pmatrix} A' ~~ b ~~ b \\ b^\top~~ \alpha ~~ \frac{1}{2} \\ b^\top ~~\frac{1}{2} ~~\alpha\end{pmatrix}$
My questions are the following:
1) Assume that $\lambda$ large enough, will $\widetilde{A}$ be positive definite for every $i$ we chose?
2) will the smallest eigenvalue of $\widetilde{A}$ necessarily be smaller than the least eigenvalue of $A$?
3) Can we get a lower bound on the smallest eigenvalue of $\widetilde{A}$?
4) Is there a strategy to pick the best $i\in\{1,\dots,n\}$ so that the lower bound on the smallest eigenvalue of $\widetilde{A}$ is the largest we can find.
I think that because of the way the row and column are added, the smallest eigenvalue of $\widetilde{A}$ is smaller than the smallest eigenvalue of $A$ by at most $1$, but it might depend on the row $i$ that we pick. Also couldn't find an example of a PD matrix that you add a row and column in this manner and the smallest eigenvalue increases, so I suspect that (2) is also true.