TITLE: Prove $\rm AB = BA = 0$ if the set of nonzero eigenvalues of $\rm A + B$ is union of set of nonzero eigenvalues of $\rm A$ and $\rm B$.
QUESTION [11 upvotes]: $A$ and $B$ are $n×n$ symmetric matrices, their non-zero eigenvalues are $(\lambda_{1},\ldots,\lambda_{r} )$，$( \mu_{1},\ldots,\mu_{s} )$.  If the nonzero eigenvalues of $A + B$ are  $(\lambda_{1},\ldots,\lambda_{r}，\mu_{1},\ldots,\mu_{s} )$,  show that $AB = BA = 0$
$\bf{Note}$ that the (non-zero) eigenvalues of a matrix form a multiset, so the above is a union of multisets.

REPLY [2 votes]: The paper A simple proof of the generalized Craig-Sakamoto theorem  by Jin Zhang and Jikun Yi cited by Jose Brox contains a minor error. This answer will correct the error.  By the way, the error is actually in the original published paper, and was not introduced by  Jose Brox.
The  cited paper uses the following inequality:  If $X$ and $Y$ are positive semidefinite of the same size $s$, then $\det(X + Y) \ge \det(X) + \det(Y)$.   This is applied as follows:  we are given $X$ and $Y$ positive semidefinite such that  $\det(Y) = \det(X+Y) \ge \det(X) + \det(Y)$.   The desired conclusion is that $X = 0$, but we cannot conclude this, only that $\det(X) = 0$.   We can obtain the desired conclusion using  a sharper inequality:  If $X$ and $Y$ are positive semidefinite,   $Y$ is invertible and $X \ne 0$,  then
  $\det(X +Y) > \det(X) + \det(Y)$.   Proof:
  $$
  X + Y = Y^{1/2}( Y^{-1/2} X Y^{-1/2} + I) Y^{1/2}
   $$
Then $Z   =Y^{-1/2} X Y^{-1/2}$ is non-zero positive semidefinite, with  non-negative eigenvalues, not all zero, $(z_1, \dots, z_s)$.  Then $\det(I + Z) = \prod_j (1 + z_j)  = 1 +  \det(Z) + \sum_J \prod_{j \in J}  z_j$,  where the sum is over non-empty proper subsets $J$ of $\{1, \dots, s\}$. 
Since not all $z_j$ are zero, the sum $\sum_J \prod_{j \in J}  z_j$ is strictly positive. Thus, $\det(I + Z) > 1 + \det(Z)$.  It follows that $\det(X+Y) > \det(X) + \det(Y)$.  
In the application, we have 
$1 = \det(Y) = \det(X+Y)$.  If $X \ne 0$,  we would have
$1 = \det(Y) = \det(X+Y) > \det(X) + \det(Y) \ge \det(Y)$, a contradiction.  Hence $X = 0$.  Apply this with
$X = T_2' T_2$ and $Y = T_4' T_4$, using the notation in the answer by Jose Brox.