$$B := \begin{pmatrix} A+nI && -E\\ -E^T&& nI\end{pmatrix}$$ where $E$ is the all-ones matrix. If the eigenvalues of $n \times n$ matrix $A$ are known, is it possible to find the eigenvalues of $B$?
I find the characteristic polynomial of $B$ to be
$$\begin{pmatrix} xI-(A+nI) && -E\\ -E^T&& (x-n)I\end{pmatrix}$$
I found that
$$\det(xI-B)= \det(x-n)I\times \{(xI-(A+nI)-E(x-n)^{-1}E^T\}$$
but cant find the characteristic polynomial of $B$ in terms of $A$. Can I get some help here?
No. E.g. $A$ has a zero spectrum when it is the zero matrix or the nilpotent Jordan block $J_2(0)$, but the characteristic polynomials of $\pmatrix{0&-E\\ -E&0}$ and $\pmatrix{J_2(0)&-E\\ -E&0}$ are $x^4-4x^2$ and $x^4-4x^2-2x$ respectively.
In general, if $A$ is $m\times m$ and its characteristic polynomial is $f$, then \begin{aligned} \det\left((y+n)I-B\right) &=\det\pmatrix{yI-A&-E\\ -E&yI}\\ &=\det\left((yI-A)(yI)-E^2\right)\\ &=\det\left(y(yI-A)-mee^T\right)\\ &=\det\left(y(yI-A)\right)-me^T\operatorname{adj}\left(y(yI-A)\right)e\\ &=y^mf(y)-my^{m-1}e^T\operatorname{adj}(yI-A)e, \end{aligned} where $e$ denotes the all-one vector. It follows that $$ \det(xI-B)=(x-n)^mf(x-n)-m(x-n)^{m-1}e^T\operatorname{adj}\left((x-n)I-A\right)e. $$ If $A$ happens to be a diagonal matrix, the expression $e^T\operatorname{adj}\left((x-n)I-A\right)e$ above can be rewritten as $(x-n)^mf(x-n)-m(x-n)^{m-1}f'(x-n)$, but in general, it is not solely determined by the eigenvalues of $A$.