Eigenvalues of product/sum of two matrices 
Find an example of matrices, $A$ and $B$, with $AB=BA$ and for which $\lambda$ is an eigenvalue of $A$, $\mu$ an eigenvalue of $B$, but $\lambda+\mu$ is not an eigenvalue of $A+B$, and $\lambda \mu$ not an eigenvalue of $AB$.

Can anyone please provide an example of two such matrices?
 A: Ok trying again. Take $$A = \begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 0 & 1\\ 0 & 2 & 0\\ 1 & 0 & 1\end{bmatrix}\,.$$
These matrices commute, neither is diagonal, and neither is triangular.
Eigenvalues of $A$: $-1, 1, 0$.
Eigenvalues of $B$: $2, 2, 0$.
Eigenvalues of $A+B$: $3,2,-1$.
Eigenvalues of $AB$: $2,0,0$.
So take $\lambda = -1$ and $\mu = 2$.
A: $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ has eigenvalue 1,
$B=\begin{pmatrix}0&2\\2&0\end{pmatrix}$ has eigenvalue -2
$A+B=\begin{pmatrix}0&3\\3&0\end{pmatrix}$ does not have eigenvalue $1-2=-1$
$AB=\begin{pmatrix}2&0\\0&2\end{pmatrix}$ does not have eigenvalue $1\cdot-2=-2$
A: I don't understand how this question makes sense the way it is posed here. If each square matrix has dimension $n$, then you have $n^2$ possible products/sums of the individual eigenvalues whereas the matrix product/sum can only have $n$ eigenvalues. So some of these eigenvalue products/sums have to be left out by construction (unless you get exactly $n$ unique numbers out of the possible $n^2$ combinations. Wouldn't it make more sense to ask whether the eigenvalues of the matrix product/sum are always a subset of the possible eigenvalue products/sums?
A: @ hafsah, your sentence "matrices should not be triangular" shows that you did not understand one word about this problem. Indeed, if $A,B$ are complex matrices s.t. $AB=BA$, then $A,B$ are simultaneously triangularizable. Thus there are orderings $(\lambda_i)_i,(\mu_i)_i$ of the eigenvalues of $A,B$ s.t. the eigenvalues of $A+B$ are $(\lambda_i+\mu_i)_i$ and the eigenvalues of $AB$ are $(\lambda_i\mu_i)_i$.
