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 $\rm B$. $A$ 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.
Edit:
As stated this is not true. Take
 $$A=\begin{bmatrix}2 &amp;0 &amp;0\\0&amp;1 &amp;0 \\ 0 &amp; 0&amp;0\end{bmatrix} \text{ and } B= \begin{bmatrix}0 &amp;0 &amp;0\\0&amp;1 &amp;0 \\ 0 &amp; 0&amp;1\end{bmatrix}$$ 
then $AB \ne 0$, but $A,B, A+B$ fulfill the hypothesis.
So you can either assume that $A,B$ are positive semi-definite as orangeskid did or assume that the multiplicities add up as I will do:

What do we know about a diagonalizable matrix $M$? By looking at eigenspaces with non zero eigenvalue and the eigenspace with eigenvalue zero we get the decomposition
$$k^n= \operatorname{im} M \oplus \ker M$$

They are an isomorphism when restricted to their image. 

By counting eigenvalues we know that
$$\operatorname{rank}{A}+ \operatorname{rank}{B} = \operatorname{rank}{(A+B)}.$$

This gives us $\dim\text{im}{A} + \dim \text{im}{B} =\dim \text{im}{(A+B)}$ by looking at the dimension of subspaces we know that the intersection must be $0$ and thus
$$\operatorname{im}{A} \oplus \operatorname{im}{B} =\operatorname{im}{(A+B)}.$$

In other word we get a decomposition 
$$k^n = \operatorname{im}{A} \oplus \operatorname{im}{B} \oplus \ker (A+B).$$

How does 
$\ker (A+B)$
looks like? We know that $\ker(A) \cap \ker(B) \subset \ker (A+B)$, but since $\operatorname{im}{A} $ and $\operatorname{im}{B} $ only intersect at zero we get equiality.

Thus 
$$k^n = \operatorname{im}{A} \oplus \operatorname{im}{B} \oplus( \ker(A) \cap \ker(B)),$$

hence we can assume that $A+B$ is an isomorphism and diagonalize $A$ and $B$ on their images.

Let $O$ be orthonormal such that $A$ is diagonalized on $\operatorname{im} A$ and let $U$ be orthonormal such that $B$ is diagonalized on $\operatorname{im} B$

Choose appropriate $M,N,C$  such that
$$A+B=\begin{bmatrix}M &amp;C\\ C^T&amp; N\end{bmatrix} $$ 

then 

$$T= \begin{bmatrix}O^T &amp;0\\ 0&amp; U^T\end{bmatrix}   \begin{bmatrix}M &amp;C\\ C^T&amp; N\end{bmatrix} \begin{bmatrix}O &amp;0\\ 0&amp; U\end{bmatrix} = \begin{bmatrix}O^T M O &amp; O^T CU\\  U^T C^TO&amp; U^TNU  \end{bmatrix} . $$

Now let us consider the trace of the square of $T$
\begin{align*} \operatorname{tr}(T^2) &amp;=  \operatorname{tr} (O^T M^2O + O^TC C^TO)+ \operatorname{tr} (U^T N^2U + U^TC^T CU)  \\
&amp;=  \sum \lambda_i + \operatorname{tr}(CC^T) + \sum \mu_i + \operatorname{tr}(C^TC)
\end{align*}

where 
$$ \operatorname{tr}(T^2) = \operatorname{tr}(A+B)^2 =  \sum \lambda_i + \sum \mu_i $$
and $\operatorname{tr}(CC^T) &gt;0,$
thus we can conclude that
$$ C=0.$$

Hence $T$ is diagonal, thus we have  $\operatorname{im}A \subset \ker B$ and $\operatorname{im}B \subset \ker A$.
As said by user60589 we need that 

$$
r + s \leq n.
$$
$A,B$ and $A+B$ can be diagonalized, thus
$$
A = UD_{A}U^{T}
$$

$$
B =VD_{B}V^{T}
$$
where $U, V$ consist of eigenvectors to $A, B$ respectively. Algebraic and geometric manifold are the same because 
we are studying symmetric matrices. Eigenvectors of different eigenvalues are always orthogonal to each other if we consider symmetric matrices because
$$
\lambda_{i} \langle x_{i}, x_{j} \rangle = \langle A x_{i}, x_{j} \rangle = \langle x_{i},A x_{j} \rangle = \lambda_{j} \langle x_{i}, x_{j} \rangle
$$
, from this and  $\lambda_{i} \neq \mu_{j}$ for all $i,j$ it follows 
$$
V^{T}U = 0
$$
simply because $V,U$ consist of eigenvectors which must by assumption all be orthogonal to each other.
Thus
$$
BA\boldsymbol{v}=VD_{B}V^{T}UD_{A}U^{T}\boldsymbol{v} = 0
$$

What about the case $\lambda_{i} = \mu_{j}$ for some $i,j$ ?
So we have 2 linear independent eigenvectors $\boldsymbol{v} \neq \boldsymbol{w}$ to one eigenvalue $\lambda$ (with algebraic manifold 2). So we can choose 2 orthogonal eigenvectors $\boldsymbol{v},\boldsymbol{l}$ replacing our previous ones:
$$ \boldsymbol{l} = \boldsymbol{w}- \frac{ \langle \boldsymbol{v}, \boldsymbol{w} \rangle}{||\boldsymbol{v}||^{2}}\boldsymbol{v} $$
which is orthogonal to $\boldsymbol{v}$.

The eigenvalue equations read now:
$$
(A+B)\boldsymbol{v} = \lambda \boldsymbol{v} 
$$
$$
(A+B)\boldsymbol{l}=(A+B)\boldsymbol{w}-(A+B)\frac{ \langle \boldsymbol{v}, \boldsymbol{w} \rangle}{||\boldsymbol{v}||^{2}}\boldsymbol{v} = \lambda \left(\boldsymbol{w} +\frac{ \langle \boldsymbol{v}, \boldsymbol{w} \rangle}{||\boldsymbol{v}||^{2}} \boldsymbol{v} \right) =\lambda \boldsymbol{l}.
$$
From here we can use the argumentation given above.