Task with combination of spectrums of matrices Here (Represent the common term of the sequence as minus one in polynomial degree), I start to post unsolved problems from my old notebook.
So, here is the next one:
Let $A,B$ be heremitian matrices. Combination of spectrums of matrix $A$ and $B$, besides eigenvalues that equal zero , is equal to spectrum of matrix $A+B$, besides eigenvalues that equal zero.
Here what is this mean: Let $\sigma_A=({0,0,1,2})$ - spectrum of matrix $A$ and $\sigma_B=({0,2,3,0})$ - spectrum of matrix $B$. Then spectrum of matrix $A+B$ due to the problem must be $\sigma_{A+B}=(1,2,2,3)$.
So, the question is to prove that if this property is performed, then $AB=0$.
My atetemts was made in the direction, that such matrices $A$ and $B$ will always be zero matrices, so it always will be that $AB=0$. Trying to show this analytically, I've used the fact, that hermitian ,atrices has real eigenvalues and spectrum of matrices is invariant, so we can work only with diagonal matrices $A,B$ and $A+B$. But I can't get a strictly reasoned solution.
This problem I've found at NCUMC competition in 2015(http://mathdep.ifmo.ru/ncumc/wp-content/uploads/2017/01/NCUMC_15_problems.pdf)
Any help will be appreciated! Thanks in advance!
 A: Attempt:
Suppose that $A,B$ are Hermitian matrices of size $n$ such that the spectrum of $A + B$ is the concatenation of the non-zero elements of the spectra of $A$ and $B$. Because all eigenvalues of $A+B$ are non-zero, it must hold that $A + B$ is invertible. However, because $A$ and $B$ have $n$ non-zero eigenvalues all together, we have
$$
\operatorname{rank}(A) + \operatorname{rank}(B) = n.
$$
Let $U = \operatorname{im}(A)$ and $V = \operatorname{im}(B)$. We note that $\Bbb C^n = \operatorname{im}(A+B) \subseteq U + V$, which means that $U + V = \Bbb C^n$. With that, we note that the dimension formula yields
$$
\dim(U) + \dim(V) = \dim(U + V) + \dim(U \cap V) \implies\\
\operatorname{rank}(A) + \operatorname{rank}(B) = \dim(\Bbb C^n) + \dim(U \cap V) \implies\\
n = n + \dim (U \cap V) \implies \dim(U \cap V) = 0.
$$
That is, we have
$$
U+V = \Bbb C^n, \quad U \cap V = \{0\}. 
$$
Now, because $A$ and $B$ are Hermitian, we have $U^\perp = \ker(A)$ and $V^\perp = \ker(B)$. From the above, conclude that
$$
U^\perp + V^\perp = (U \cap V)^\perp = \Bbb C^n, \quad 
U^\perp \cap V^\perp = (U + V)^\perp = \{0\}.
$$
