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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)

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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.

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  • $\begingroup$ what happens if they have distinct eigenvalues? $\endgroup$ Apr 27, 2020 at 20:32

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