Find the characteristic polynomial of the matrix $M$ in terms of characteristic polynomial of $N$.
$M=\begin{pmatrix}N+V & U\\U^T& I\end{pmatrix}$ where $V=\left[ \begin{array}{ccccc} n & 0 & \ldots & 0\\ 0 & 0 & \ldots & 0\\ \ldots& \ldots &\ldots &\ldots\\ \ldots& \ldots &\ldots &\ldots\\ \ldots& \ldots &\ldots &\ldots\\ 0 & 0 & \ldots & 0\\ \end{array} \right]_{n\times n} $ and $U=\left[ \begin{array}{ccccc} 1 & 1 & \ldots & 1\\ 0 & 0 & \ldots & 0\\ \ldots& \ldots &\ldots &\ldots\\ \ldots& \ldots &\ldots &\ldots\\ \ldots& \ldots &\ldots &\ldots\\ 0 & 0 & \ldots & 0\\ \end{array} \right]_{n\times n}$.
MY TRY:
I used Schur Complement formula.
We have $\det(xI-M)=\begin{pmatrix}xI-N-V & -U\\-U^T& (x-1)I\end{pmatrix}$
Thus $\det(xI- M)=\det((x-1)I_n)\times \det ((xI-N-V)-U\frac{I}{x-1}U^{T})$
Since $UU^T=V$ we get
$\det \biggl((xI-N-V)-U\frac{I}{x-1}U^{T}\biggr)=\det \biggl((xI-N-V)-\frac{V}{x-1}\biggr)$
But I cant proceed further. How do I express the characteristic polynomial of the matrix $M$ in terms of characteristic polynomial of $N$ from here?
Can someone please help me out?
You have already rewritten the formula as $$ \det((x-1)I_n)\cdot\det((x I - N - V) - V/(x-1))= (x-1)^n \det\left((xI - N) - \frac x{x-1}V \right). $$ Now, note that we can express $\frac x{x-1}V$ in the form $V = uv^T$, where $u = (1,0\dots,0)^T$ and $v = (n\frac x{x-1},0,\dots,0)^T$. Applying the matrix determinant lemma yields $$ \det((x I - N) - uv^T) = (1 + v^T(x I - N)^{-1}u)\det(x I - N) \\ = \left(1 + \frac{nx}{x-1}[(x I - N)^{-1}]_{1,1}\right)\det(x I - N), $$ where $[A]_{i,j}$ denotes the $i,j$ entry of the matrix $A$. Putting everything together, we have $$ \det(x I- M) = (x-1)^n \left(1 + \frac{nx}{x-1}[(x I - N)^{-1}]_{1,1}\right)\det(x I - N). $$