Eigenvalues of commutative matrices

Question

Let be $ A,B\in \mathbb{K}^{n\times n} $ diagonalizable with $ A\cdot B=B\cdot A $. What can you say about the eigenvalues of the matrix $ A+B $?

My conjecture is: If $ \lambda $ is an eigenvalue of $ A $ and $ \mu $ is an eigenvalue of $ B $ than $ \lambda+\mu $ is an eigenvalue of $ A+B $.

My idea:

At first I have an eigenvalue $ \lambda\in \mathbb{K} $ of $ A $ with eigenvector $ v\in \mathbb{K}^n $. Than $ A\cdot B\cdot v=B\cdot A\cdot v=B\cdot \lambda\cdot v=\lambda\cdot B\cdot v $.

I want to try to show that $ A $ and $ B $ have the same eigenvectors. But from here I get stuck. If $ A $ and $ B $ have the same eigenvectors than both have the same eigenspaces or eigenspaces have the same dimension. I have no information about the eigenspaces.

You might as well take $A=D$ already diagonal; once $A,B$ commute, so do $P^{-1}AP$ and $P^{-1} BP $ — Will Jagy, May 08 '21 at 00:05
meanwhile, a matrix that commutes with a diagonal matrix is block diagonal — Will Jagy, May 08 '21 at 00:11
Hmm but there are commutative matrices which $ A $ and $ B $ are not diagonal. This for example: $ \begin{pmatrix}1&-1\5&-4 \end{pmatrix} $ and $ \begin{pmatrix}1&1\-5&6 \end{pmatrix} $ and they are diagonalizable. — hallo007, May 08 '21 at 00:22
You need to be careful in how you state your conjecture. As an example, suppose $A = \text{diag}(0,1,2,\ldots,n-1)$ and $B = \text{diag}(0,n,2n,\ldots,n^2-n)$. If $\lambda$ is an eigenvalue of $A$ and $\mu$ is an eigenvalue of $B$, then $\lambda+\mu$ could be any of the $n^2$ numbers $0,1,2,\ldots,n^2-1$. But not all of these $n^2$ numbers are eigenvalues of $A+B = \text{diag}(0,n+1,2n+2,\ldots,n^2-1)$. — JimmyK4542, May 08 '21 at 02:49

Eigenvalues of commutative matrices

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