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?

• Pick for A and B two matrices that are really easy to calculate with that satisfy the conditions. Which ones did you pick? Do they work? If not, why not? Commented Mar 22, 2014 at 13:53

$A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ 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 better answer than mine, obviously! Commented Mar 22, 2014 at 15:37
• Nice one.${{}}$ Commented Mar 22, 2014 at 15:53

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 $\mu = 2$.

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?

@ 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 $(\lambda_i\mu_i)_i$.