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 $n$?
 A: Hint: $A^TA$ is positive semidefinite.
A: No, it is not possible. 
Suppose that $v$ is an eigenvector of $A^TA$ associated with eigenvalue $\lambda$.  We then have 
$$
\langle A^TAv, v \rangle  = \langle \lambda v, v \rangle = \lambda \|v\|^2
$$
However,
$$
\langle A^TAv, v \rangle  = 
\langle Av,A v \rangle = \|Av\|^2
$$
So, we have
$$
\lambda \|v\|^2 = \|Av\|^2 \implies\\
\lambda = \frac{\|Av\|^2}{\|v\|^2} \geq 0
$$
Thus, the eigenvalues of $A^TA$ are non-negative.

If we consider the entrywise transpose (as opposed to the conjugate transpose), then we can find such a complex matrix $A$. In particular, take
$$
A = \pmatrix{i&0\\0&i} = iI
$$
then we have $A^TA = A^2 = -I$ with eigenvalue $-1$.
A: According to wiki, the matrix $A^TA$ is positive definite for any non-singular $A$. The proof there is easily modified to show that singular $A$ give $A^TA$ positive semi-definite.
A: You may try to show that $A^TA+I$ is invertible. Hint: if $(A^TA+I)x=0$, consider $x^T(A^TA+I)x$. Try to rewrite it as a sum of squares and infer that $x=0$.
A: 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 $$\lambda x^\top x =x^\top A^\top Ax = (Ax)^\top Ax \implies\lambda = \frac{|Ax|^2}{|x|^2} \ge 0.  $$
