Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$? 
Question: Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?

I can prove that it is not possible for $n=1,2$, but I am not sure for the general case.
Case $n=1$: $a^2v=-v$ $\implies$ $(a^2+1)v=0$ $\implies$ $v=0$.
Case $n=2$: Write $A=\begin{pmatrix}a&b \\ c&d\end{pmatrix}$. Then, $A^T A=\begin{pmatrix}a^2+c^2&ab+cd\\ ab+cd&b^2+d^2\end{pmatrix}$, so
$$
\begin{align}
\det(A^TA+1) 
&= \begin{vmatrix}a^2+c^2+1&ab+cd\\ ab+cd&b^2+d^2+1\end{vmatrix}\\
&= 1+a^2+b^2+c^2+d^2+(ad-bc)^2\\
&\neq 0.
\end{align}
$$
Now, can we generalize to all $n$?

suppose $\lambda,$ necessarily real, is an eigenvalue of the symmetric matrix $A^\top A$ and $x \neq 0$ an eigenvector. then $$A^\top A x = \lambda x \tag 1$$  multiplying by $x^\top$ on the left, we get $$\lambda x^\top x =x^\top A^\top Ax = (Ax)^\top Ax \implies\lambda = \frac{|Ax|^2}{|x|^2} \ge 0.  $$
