Let $A$ be a real square matrix of size $n$ by $n$ that have the following eigendecomposition $A = PDP^{-1}$. What can be said about the eigendecomposition of another real square matrix $B$ of size $n+1$ by $n+1$ that "includes" $A$, with the form: $$ B = \begin{bmatrix} c & v^T \\ v & A \end{bmatrix} $$ where $c$ is a real constant and $v$ is a real vector of size $v$?
What if $A$ is symmetric? What if $c = 1$ or $c = 0$?
I wonder is there exists a nested relation between eigenelements of $B$ and those of $A$.
Consider the case of the $1 \times 1$ matrix $A = [1]$. Its only eigenvalue is $1$. In this case, both $v$ and $c$ are numbers, so we're looking at eigenvalues of $$ \pmatrix{ c & v \\ v & 1 } $$ i.e., roots of $$ (c-x)(1-x) - v^2 = 0, $$ which is just $$ x^2 - (c+1)x - v^2 = 0, $$ whose roots are $$ \frac{(c+1) \pm \sqrt{(c+1)^2 + 4v^2}}{2} $$ Letting $u = (c+1)/2$, this becomes $$ x_1, x_2 = u \pm \sqrt{u^2 + v^2} $$
Pick any two numbers $a$ and $b$. If you set $u = \frac{a+b}{2}$, then for any $v$ you get $$ x_1 + x_2 = a + b \\ x_1 - x_2 = \sqrt{\left(\frac{a+b}{2}\right)^2 + v^2} $$ Setting this second thing to be $a-b$ gives you \begin{align} (a - b) &= \sqrt{\left(\frac{a+b}{2}\right)^2 + v^2} \\ (a - b)^2 &= \left(\frac{a+b}{2}\right)^2 + v^2 \\ (a - b)^2 - \left(\frac{a+b}{2}\right)^2 &= v^2 \\ a^2 -2ab + b^2 + \frac{a^2}{2} + ab + \frac{b^2}{2} &= v^2\\ \frac{3a^2}{2} -ab + \frac{3b^2}{2} &= v^2\\ \end{align} so we can let $$ v = \sqrt{\frac{3}{2} (a^2 + b^2) - ab} $$ and we'll have arranged for the two eigenvalues of our matrix to be $a$ and $b$.
In short: Knowing that the only eigenvalue of $A$ is $1$ tells you nothing about the potential eigenvalues of $$ \pmatrix{ c & v \\ v & 1 } $$