TITLE: prove: $A+E$ is not invertible.
QUESTION [5 upvotes]: Suppose square matrix $A$ with order-n, and $A^TA=E,|A|=-1$, prove: $A+E$ is not invertible.
Here, $A^T$ is transpose, $|A|$ is determinant, $E$ is identity matrix.
Prove:

we will show $|A+E|=0$.
\begin{align}|A+E|=\left|A+A^T A\right|=\left|A\left(E+A^T\right)\right|=\left|A\left\|E+A^T\right.\right|=-\left|E+A^T\right|\end{align}
\begin{align}|A+E|=\left|(A+E)^T\right|=\left|A^T+E\right|=\left|E+A^T\right|\end{align}
so we've done.
Any problems? is that all?

REPLY [2 votes]: This way is perhaps more complicated, but it is always good to solve the problem in more than one way:
Orthogonal (unitary) matrices always have eigenvalues lying in $S^1 = \{z \in \mathbb{C} \mid |z| =1\}$. If $A$ is real, then its eigenvalues are either real, or come in complex conjugate pairs. $|A| = \prod_{i}\lambda_i$. So if any of the eigenvalues are complex, $\lambda_i \cdot \bar{\lambda_i} = 1$, so those factors don't contribute. Since all the factors are in $S^1$, the real factors are $\pm 1$. Since the product is $-1$, there must be at least one eigenvalue of $-1$ (in fact, the eigenvalue $-1$ must have odd multiplicity...algebraic and geometric multiplicity are the same, since $A$ is orthogonal). $A+I$ has eigenvalues of $A$ shifted by adding $1$ to each, so $A+I$ has an eigenvalue equal to $-1+1=0$, so it is singular.