Find the determinant of $A + I$, where $A$ is a real matrix such that $AA^{\top}=I$ and $\det A<0$.

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 $\det(A)=-1$.

Solution 1:

Because the determinant is multiplicative and $AA^T=I$, we have $$\det(A+I)=\det(A+AA^T)=\det(A(I+A^T))=\det(A)\det(I+A^T).$$ Of course $I+A^T=A^T+I$, and $(A^T+I)^T=(A^T)^T+I^T=A+I$. It follows that $$\det(A)\det(I+A^T)=\det(A)\det(A^T+I)=\det(A)\det(A+I),$$ where we use that $\det(M)=\det(M^T)$ for any matrix $M$. The equalities above show that $$\det(A+I)=\det(A)\det(A+I).$$ But you already noted that $\det(A)=-1$, so then we must have $\det(A+I)=0$.

Solution 2:

We have:

$$ \det(A+I)=\det[A(I+A^T)]=\det(A)\det(I+A^T)=-\det(I+A^T). $$

Now since $\det(M)=\det(M^T)$ for any matrix M, we have:

$$ \det(A+I)=-\det(I+A^T)=-\det(I+A), $$

Therefore $\det(A+I)=0$.

Solution 3:

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 $\det(A+I)=0$.