TITLE: Find the determinant of $A + I$, where $A$ is a real matrix such that $AA^{\top}=I$ and $\det A<0$.
QUESTION [6 upvotes]: Given a real valued matrix $A$ that satisfies $AA^{\top} = I$ and $\det(A)<0$, calculate $\det(A + I)$.

My start :
Since $A$ satisfies  $AA^{\top} = I$, $A$ is a unitary matrix. The determinant of a unitary matrix with real entries is either $+1$ or $-1$. Since we know that $\det(A)<0$, it follows that $\det(A)=-1$.

REPLY [5 votes]: There is another interesting proof:
The product of the eigenvalues of $A$ is the $\det(A)$. Since $A$ is orthogonal all eigenvalues lie on the unit circle. Each non-real eigenvalue comes with its complex conjugate so their product is $1$. So $\det(A)=-1$ implies that the eigenvalue $-1$ has odd multiplicity, hence the corresponding eigenspace has dimension at least $1$. Thus, there is a vector $v\ne 0$ with $Av=-v$. We conclude  $(A+I)v=0$, and finally $\det(A+I)=0$.