TITLE: Can the matrices $A$ and $I+A$ have the same determinant?
QUESTION [29 upvotes]: Let $A\in\mathbb R^{n\times n}$ be an arbitrary matrix. Can $A$ and $I+A$ have the same determinant, if not how to prove it? 
Furthermore, can $A$ and $I+A$ have the same eigenvalues?

REPLY [2 votes]: Here is a simple way to construct examples of matrices $A$ such that $\det(A)=\det(I+A)$ in arbitrary large dimension $n$.
Take any monic polynomial $p(X) = a_0 + a_1X + \dots + a_{n-1}X^{n-1} + X^n$ such that $p(0)=p(-1)$ (there are plenty of them, for instance $X^n+X^{n-1}$) and consider its companion matrix
$$
A=\begin{pmatrix}  0 & \cdots & 0& -a_{0} \\  1 & \cdots & 0 & -a_{1}\\ \vdots &\ddots & \vdots &\vdots \\ 0 &\cdots & 1 & -a_{n-1} \end{pmatrix}
$$
By construction, the characteristic polynomial of $A$ is $p(X)$, hence $\det(A) = (-1)^n p(0)$ and $\det(I+A)=(-1)^n p(-1).$
If you don't like zeros and ones, take any matrix that is similar to $A$.