I'm trying to prove the following result:
Let $(\lambda,x)$ be an eigenvalue-eigenvector pair of an $ n \times n$ matrix $\mathbf A$, it follows that there exists a vector $y$ such that $$ \operatorname{adj} (\lambda \mathbf I_n - \mathbf A) = x \cdot y'.$$
I think somehow this should follow from the fact that $$\det(\lambda \mathbf I_n - \mathbf A)=0$$ by definition of $\lambda$. It follows that $$ \operatorname{rank} (\lambda \mathbf I_n - \mathbf A) \le n-1.$$ If $\operatorname{rank} (\lambda \mathbf I_n - \mathbf A) \le n-2$, the result is trivial since $\operatorname{adj} (\lambda \mathbf I_n - \mathbf A) = \mathbf 0$. So we are left with the case in which $ \operatorname{rank} (\lambda \mathbf I_n - \mathbf A) \le n-1,$ and therefore $$ \operatorname{rank}(\operatorname{adj} (\lambda \mathbf I_n - \mathbf A)) = 1.$$ I haven't been able to proceed from here.
Let $\mathbf{B}=\lambda\mathbf{I}_n-\mathbf{A}$, $\mathbf{C}=\mathrm{adj}\ \mathbf{B}$. If $\mathrm{rank}\mathbf{B}=n-1$, then the solution space for the equation $\mathbf{B}x=0$ is one dimensional, i.e. a constant multiple of $x$. Since $\mathbf{B}\cdot\mathbf{C}=0$, all the column vector of $\mathbf{C}$ are solutions for the equation. The result follows.