Eigenvalues of the sum of two matrices: one diagonal and the other not.

Two diagonalizable matrices $A$ and $B$ will commute if and only if they are simultaneously diagonalizable; i.e., we can write $$A = P^{-1} D_A P$$ and $$B=P^{-1} D_B P$$ where $D_A$ and $D_B$ are diagonal, and the elements of $D_A$ and $D_B$ represent the eigenvalues of $A$ and $B$, respectively.

Pointed out by Ian in the comments, a nondiagonalizable matrix commutes with some matrices, in particular itself.

Now, observe the following for simultaneously diagonalizable $A$ and $B$: $$A+B = P^{-1}D_AP + P^{-1}D_BP=P^{-1}(D_A+D_B)P.$$ Consequently, the eigenvalues of $A+B$ are given by the elements on the diagonal of $D_A+D_B$.

Note: considering you example, $A$ and $B=\alpha I$, $A$ and $B$ are simultaneously diagonalizable.

Extra note: It is important to recognize that for the general case that I have mentioned, there is some ambiguity as to precisely what the eigenvalues are. More precisely, the equality $$\lambda_{A+B,i} = \lambda_{A,i}+\lambda_{B,i}$$ does not hold in general. However, given a $\lambda_{A+B,i}$, it is always possible to find a $\lambda_{A,j}$ and $\lambda_{B,k}$ such that $$\lambda_{A+B,i} = \lambda_{A,j}+\lambda_{B,k}.$$ I'm not sure if anyone else can elaborate on this.