# Eigenvectors of a sum of PSD matrices

Lets say I have two real-valued positive-semidefinite matrices $$A$$ and $$B$$ with eigenvectors given by $$v_i$$ and $$w_i$$, respectively. Does this tell me anything about the eigenvectors of $$C = A + B$$

I am particularly interested in the case where A and B are matrices related to graphs (e.g. adjacency matrices, laplacians etc)

No, it doesn't. $$A$$ and $$B$$ could be the identity matrix $$I$$ (for which any sequences of nonzero vectors are eigenvectors). And then any sequence of nonzero vectors are eigenvectors of $$C=2I$$ as well.
Things might be a little more interesting if you assumed that $$A$$ and $$B$$ have distinct eigenvalues.
EDIT: We may assume $$v_i$$ and $$w_i$$ are orthonormal bases. Suppose you want the members of orthonormal basis $$u_i$$ to be eigenvectors of $$A+B$$. These vectors form the columns of orthogonal matrices $$S, T, U$$ respectively, where $$D_A = S^\top A S$$, $$D_B = T^\top B T$$ and $$D_C = U^\top C U$$ are all diagonal with diagonal entries the eigenvalues of $$A, B, C$$ respectively. But you need $$A+B = C$$, i.e. $$S D_A S^\top + T D_B T^\top = U D_C U^\top$$, or equivalently $$U^\top (S D_A S^\top + T D_B T^\top) U$$ is diagonal. This matrix is certainly symmetric, but if the matrices are $$n \times n$$ there are $$n(n-1)/2$$ entries above the diagonal that must be $$0$$. There are $$n(n-1)/2$$ conditions to satisfy with only $$2n$$ free variables (the eigenvalues of $$A$$ and $$B$$). There are always two linearly independent solutions where $$A$$ and $$B$$ are multiples of the identity. But if $$n$$ is large enough we might not have a solution other than those.
For example, I tried the $$6 \times 6$$ matrices $$S = \frac{1}{15} \left[ \begin {array}{cccccc} -4&3&-8&6&-8&6\\ -3&- 4&-6&-8&-6&-8\\ -8&6&-4&3&8&-6\\ - 6&-8&-3&-4&6&8\\ -8&6&8&-6&-4&3\\ -6&-8&6&8&-3&-4\end {array} \right], \ T = \frac{1}{15} \left[ \begin {array}{cccccc} 4&8&8&-3&-6&-6\\ 8&4& -8&-6&-3&6\\8&-8&4&-6&6&-3\\ 3&6& 6&4&8&8\\ 6&3&-6&8&4&-8\\ 6&-6&3&8 &-8&4\end {array} \right],\ U = I$$ and found that indeed the only solutions are where $$D_A$$ and $$D_B$$ are multiples of the identity. So in this case the basis $$\{u_i\}$$ is not going to be arbitrary. On the other hand, to state explicitly what bases are possible is not going to be simple.