Eigenvalues and eigenvectors of operators on $H \oplus H$ $K$ is a compact and self-adjoint operator on a Hilbert space $H$ and $L$ is a compact operator on $H$. Consider the operators $A(v,w)=(Kw,Kv)$ and $B(v,w)=(Lw,L
^*v).$
If the eigenvalues and eigenvectors of $K$ and $L^*L$ are known, what can can we say about the eigenvalues and eigenvectors of $A$ and $B$?
I know all the basic theorems regarding compact and self-adjoint operators, but it got me nowhere. Any ideas?
 A: If $\lambda$ is an eigenvalue of $A$ and $(v,w)$ is an eigenvector,
$$
(\lambda v,\lambda w)=A(v,w)=(Kw,Kv). 
$$
So $$ \lambda^2v=\lambda Kw=K(\lambda w)=K^2v
$$
(if $v=0$, we can do a similar trick with $w$). 
This shows that the eigenvalues of $A$ are contained in the set
$$
\{\pm\lambda:\ \lambda\in\sigma(K)\}
$$
(because the eigenvalues of $K^2$ are the squares of the eigenvalues of $K$). 
Conversely, if $\lambda$ is an eigenvalue of $K$ with eigenvector $v$, then 
$$
A(v,v)=(Kv,Kv)=\lambda(v,v). 
$$
And 
$$
A(v,-v)=(-Kv,Kv)=(-\lambda)(v,-v). 
$$
So the set $\{\pm\lambda: \lambda\in \sigma(K)\}$ is contained in the eigenvalues of $A$. In summary, 
$$
\sigma(A)=\{\pm\lambda:\ \lambda\in\sigma(K)\}. 
$$

In the case of $B$, we can apply something similar. Note that the eigenvalues of $L^*L$ are all real and non-negative. 
If $B(v,w)=\lambda(v,w)$, then
$$
(\lambda v,\lambda w)=B(v,w)=(Lw,L^*v).
$$
So 
$$
L^*Lw=\lambda L^*v=\lambda^2 w.
$$
Thus $\lambda^2$ is an eigenvalue of $L^*L$. If $w=0$, we can play a similar game to obtain that $\lambda^2$ is an eigenvalue of $LL^*$, and so it is also an eigenvalue of $L^*L$. 
Conversely, if $\lambda^2$ is an eigenvalue of $L^*L$, then $|\lambda|$ is a singular value of $L$. Let $L=USV$ be the singular value decomposition, i.e. $U,V$ are unitaries and $S$ is diagonal with the singular values of $L$ in the diagonal. So there exists nonzero $v$ with $Sv=|\lambda|\,v$. Then
$$
B(Uv,V^*v)=(LV^*v,L^*Uv)=(USv,V^*Sv)=|\lambda|\,(Uv,V^*v).
$$
And
$$
B(Uv,-V^*v)=(-LV^*v,L^*Uv)=(-USv,V^*Sv)=-|\lambda|\,(Uv,-V^*v).
$$
So both $|\lambda|$ and $-|\lambda|$ are eigenvalues of $B$. In other words, 
$$
\sigma(B)=\{\pm \lambda:\ \lambda^2\in\sigma(L^*L)\}=\{\pm\lambda^{1/2}:\ \lambda\in\sigma(L^*L)\}.
$$

Although not asked in the question, one can have $0$ in the spectrum even though it may not be an eigenvalue. Because all operators involved are compact, in that case $0$ will be a limit of eigenvalues. So the computations above show that if $0$ is in the spectrum of $A$, then it will be in the spectrum of $K$; and viceversa. And the same for $B$ and $L^*L$. 
