$(\lambda, x)$ is a simple (with multiplicity 1) eigenpair of $A\in \mathbb C_n$ with $x^Hx=1$, $H$ denotes Hermitian.
Use Schur decomposition to show that there exists a nonsingular matrix $(x\ \ X)$ such that $$ (x\ X)^{-1}A(x\ X)=\left(\matrix{\lambda \ \ \ \ 0\\ \ 0\ \ \ M}\right) $$ where $M\in\mathbb C_{n-1}$
Hint: Suppose for some invertible $P$ we have $$ P^{-1}AP = \pmatrix{\lambda&b^H\\ 0&M} $$ for some $b\in\mathbb{C}^{n-1}$ and $M\in M_{n-1}(\mathbb{C})$. For any vector $u\in\mathbb{C}^{n-1}$, we have $$ \pmatrix{1&-u^H\\ 0&I}\pmatrix{\lambda&b^H\\ 0&M}\pmatrix{1&u^H\\ 0&I} =\pmatrix{\lambda&b^H - u^H(M-\lambda I)\\ 0&M}. $$ Since $\lambda$ is not an eigenvalue of $M$ (why?), we can choose $u$ such that $b^H - u^H(M-\lambda I)=0$ (why?).