Given that $A$ is orthogonal and $\det(A) > 0$. Find $\det(A+I)$ 
Given that $A$ is orthogonal and $\det(A) > 0$. Find $\det(A+I)$

Tried to find answers online, wolfram alpha, textbook and couldn't find anything of the sorts. Any help or pointers would be appreciated.
 A: There are many possibilities. For example, the identity matrix is orthogonal, and $\det(I+I) = 2^n$, where $n$ is the dimension. On the other hand, $$A = \begin{bmatrix}
 \frac{3}{5} & \frac{4}{5} \\
 -\frac{4}{5} & \frac{3}{5} \\
\end{bmatrix}$$
is an orthogonal matrix with determinant $1$, and $\det(A+I) = \frac{16}{5}$.
In general, the determinant of a matrix is the product of its eigenvalues. The eigenvalues of a $A+I$ are all one more than the eigenvalues of $A$, so $\det(A+I)$ is the product of all the eigenvalues of $A$ increased by $1$. 
The $n$ eigenvalues of $A$ are all either real or come in complex conjugate pairs, and all have absolute value $1$.


*

*In the real case, the eigenvalue is $\pm1$; adding $1$, we contribute a factor of $0$ or $2$ to the determinant of $A+I$.

*In the complex case, the two eigenvalues can be written as $\{e^{i t}, e^{-i t}\}$, and contribute a factor of $(1 + e^{it})(1+e^{-it}) = 2+2\cos t$ to the determinant of $A+I$. This factor is in the range $[0,4]$, but it comes from two eigenvalues at once, so it's the same as each of them contributing a factor in the range $[0,2]$.


So we can show that $\det(A+I)$ is between $0^{n/2}$ and $4^{n/2}$: between $0$ and $2^n$. This is the best we can do.
