Just had a quick question which I think ought to be simple, but I can't fully convince myself of the result.
Question: If matrices $\mathbf{A} + \mathbf{B} = \mathbf{I}$, ($\mathbf{I}$ is the identity matrix) then do $\mathbf{A}$ and $\mathbf{B}$ have the same eigenvectors? Let us assume that $A$ and $B$ are invertible and that they have real-valued entries.
Context: The context of this question is in graph theory where we have the graph Laplacian $L$, which can be expressed as the difference between the degree matrix $D$ and the adjacency matrix $A$ in: $L = D – A \rightarrow D = L + A$. Then this can be 'normalized' to form symmetric matrices $A_{\text{sym}} = D^{-\frac{1}{2}}AD^{-\frac{1}{2}}$ and $L_{\text{sym}} = D^{-\frac{1}{2}}LD^{-\frac{1}{2}}$ and then the equation becomes:
$$ I = L_{\text{sym}} + A_{\text{sym}} $$
and I have read that $L_{\text{sym}}$ and $A_{\text{sym}}$ can both be diagonalized with the same eigenbasis, so I was wondering why. I suppose the problem simplifies to any two matrices $A$ and $B$, so I just took the context out of the question.
Attempt: From thinking about it, all real vectors $\mathbf{v}$ satisfy $I\mathbf{v} = \mathbf{v}$, so
$$ I\mathbf{v} = \mathbf{v} = (A + B)\mathbf{v} \rightarrow \mathbf{v} = A\mathbf{v} + B\mathbf{v} $$
Then I don't know how to proceed from there – maybe some sort of proof by contradiction (come from an engineering background, some haven't seen these types of proofs since before college/university).
Any help is appreciated. If the question/understanding is wrong, please let me know.
Best Answer
You can write $B=I−A$. So eigenvectors of $A$ are also eigenvectors of $B$ and vice versa.