Relationship between eigenvalues of $A^TA$ and $B=\left[\begin{array}{cc} 0 & A^T\\ A & 0\end{array}\right]$ 
Let $A \in \mathbb{R}^{n\times n}$ and $\mu_1, \mu_2, \dots, \mu_n$ be
  the eigenvalues of $A^TA$. Let
$$B := \left[\begin{array}{cc} 0 & A^T\\ A & 0\end{array}\right]$$
What is the relationship between the eigenvalues of $A^TA$ and $B$?

Using SVD decomposition, I can solve this question. I want to know whether there is another solution.
 A: Since $\mathrm A^{\top} \mathrm A$ is positive semidefinite, its eigenvalues, $\mu_1, \dots, \mu_n$, are nonnegative.
The characteristic polynomial of $\mathrm B$ is
$$\det (s \mathrm I_{2n} - \mathrm B) = \det \begin{bmatrix} s \mathrm I_{n} & -\mathrm A^{\top}\\ -\mathrm A & s \mathrm I_{n} \end{bmatrix} = \det (s^2 \mathrm I_{n} - \mathrm A^{\top} \mathrm A)$$
Thus, the $2n$ eigenvalues of $\mathrm B$ are $\pm\sqrt\mu_1, \dots, \pm\sqrt\mu_n$.
A: Let $A^TAv = \mu_k v$, $v\ne 0$. My idea was to use a vector of the form $\pmatrix{ v \\ Av}$, then $Bv$ is of the same shape.
With scalars $a,b$
$$
B \pmatrix{ av \\ bAv} = \pmatrix{ bA^TAv\\ aAv} = \pmatrix{ \mu_k b v\\ a Av}.
$$
In order to obtain eigenvectors of $B$ to some eigenvalue $\lambda$, it must hold $\mu_k b=\lambda a$ and $a = \lambda b$.
The choice $b=0$ is impossible: this would imply $a\ne 0$ (to obtain non-zero eigenvector) and $\lambda=0$, a contradiction to $a=\lambda b$.
Thus, we can set $b=1$, $a=\lambda$, $\mu_k = \lambda^2$. This shows that the vector
$$
\pmatrix{ \pm \sqrt \mu_k v\\Av}
$$ 
is an eigenvector of $B$ to the eigenvalue $\pm \sqrt\mu_k$.
This construction yields all eigenvectors to non-zero eigenvalues of $B$. Since the ranks of $A$, $A^T$, $A^TA$ are the same, one can construct eigenvectors of $B$ to the eigenvalue zero by
$$
\pmatrix{v\\0}, \pmatrix{0\\w}
$$
with $Av=0$, $A^Tw=0$. Thus it is possible to construct a basis of eigenvectors of $B$ with the help of the eigenbasis of $A^TA$ plus a basis of $\ker A^T$.
