eigenvalues of sum of tensor product of matrices Lets say $A, B$ are  rank 1 hermitian real matrices. It follows immediately that $A\otimes A$ and $B\otimes B$ individually are rank 1. My question is how does the largest eigenvalue of $A+B$ relate to the largest eigenvalue of $A\otimes A+B\otimes B$. I have phrased the question as if A,B are 2 by 2 matrices, but I would like to know the intuition in general for arbitrary dimensions. Any references would also be appreciated!
 A: As $A$ is rank-1 Hermitian, its only nonzero eigenvalue is $a=\operatorname{tr}(A)$. Define $b$ similarly for $B$.
Weyl's inequality states that if $X$ and $Y$ are Hermitian matrices of the same sizes, and $\lambda_k(\cdot)$ denotes the $k$-th smallest eigenvalue of a matrix, then
$$
\lambda_k(X)+\lambda_\min(Y)\le\lambda_k(X+Y)\le\lambda_k(X)+\lambda_\max(Y).\tag{1}
$$
It follows that
$$
\max\left\{\lambda_\max(X)+\lambda_\min(Y), \ \lambda_\max(Y)+\lambda_\min(X)\right\}\le\lambda_\max(X+Y)\le\lambda_\max(X)+\lambda_\max(Y).\tag{2}
$$
Now, as $a$ is the only nonzero eigenvalue of $A$, we have $\lambda_\max(A\otimes A)=a^2$ (even when $a$ is negative) and $\lambda_\min(A\otimes A)=0$ and similarly for $B\otimes B$. So, $(2)$ gives
$$
\max\{a^2,b^2\}\le\lambda_\max(A\otimes A+B\otimes B)\le a^2+b^2\tag{3}
$$
and both bounds are sharp: the lower bound is attained when $A=\operatorname{diag}(a,0,0,\ldots,0)$ and $B=\operatorname{diag}(0,b,0,\ldots,0)$, while the upper bound is attained if take $B=\operatorname{diag}(b,0,0,\ldots,0)$ instead.
Note that the eigenvalues of $A\otimes A+B\otimes B$ are not solely determined by $a$ and $b$. More specifically, by spectral decomposition, we can always write $A=auu^\ast$ and $B=bww^\ast$ for some unit vectors $u,w$. Let $w=cu+sv$ for some unit vector $v\perp u$. By picking some appropriate phase angles for $w$ and $v$, we may assume that $c$ and $s$ are real nonnegative and hence $s=\sqrt{1-c^2}$. So, by a change of basis, we may assume that
$$
A=\pmatrix{a&0\\ 0&0}\oplus0,\quad B=b\pmatrix{c^2&sc\\ sc&s^2}\oplus0.
$$
It follows that $c = \sqrt{\dfrac{\operatorname{tr}(ABA)}{aba}}$. Therefore, the eigenvalues of $A+B$ and $A\otimes A+B\otimes B$ are completely determined by $a,b$ and $c$.
A: This is not a full answer, but I wish it can help.
Let $\lambda_{max}$ be the largest eigenvalue of $A+B$,
and  $\Lambda_{max}$ be the largest eigenvalue of $C=A \otimes A + B \otimes B$.
I have done extensive simulations (see Matlab program and accompanying figure: stars for $\Lambda_{max}$, circles for $\lambda_{max}$): something general and interesting can be conjectured about a certain threshold situated here around $s=1.75$, yielding broadly two cases:


*

*either $s\leq\lambda_{max}\leq\Lambda_{max}$ or the reverse situation:

*$\Lambda_{max}\leq\lambda_{max}\leq s$.

*But this classification suffers exceptions that should be analysed in a more thorough manner.
Did you have a similar conjecture ?

The part of analysis I have done is as follows (but I have not enough time to devote to it).
The term hermitian can be replaced here by symmetric $n \times n$.
$A$ and $B$ being rank one, their resp.  spectra are $\{\lambda_A,0,0,\cdots,0\}$ and $\{\lambda_B,0,0,\cdots,0\}$ ($n-1$ zeros) with $\lambda_A,\lambda_B$ real because they are symmetric. 
$A+B$, in general is thus a symmetric rank 2 matrix, meaning that it has two non zero real eigenvalues.
The spectra of $A \otimes A$ and $B \otimes B$ are thus resp. $\{\lambda_A^2,0,0,\cdots,0\}$ and $\{\lambda_B^2,0,0,\cdots,0\}$ (with $n^2-1$ zeros),  meaning that they are both rank 1 matrices.
Their sum $C=A \otimes A + B \otimes B$ is thus a rank 2 matrix in general.
Here is the Matlab program and (below) the graphical display
clear all;close all;hold on
i=complex(0,1);
for k=1:50
U=rand(3,1);
A=U*U';
KA=kron(A,A);
V=rand(3,1);
B=V*V';
KB=kron(B,B);
eKAB=eig(KA+KB);
I=find(abs(eKAB)>0.01);
LKAB=eKAB(I)';
eAB=eig(A+B);
I=find(abs(eAB)>0.0001);
LAB=eAB(I)';
plot(k*i+max(LAB),'or');
plot(k*i+max(LKAB),'*b');
plot(k*i+[max(LAB),max(LKAB)])
end;

