Eigenvalues of matrices Let $A$ and $B$ be real matrices .
Suppose that $AB =BA$ and all eigenvalues of $A$ and $B$ real and distinct I.e $$\operatorname{Spec}(A)= \{\lambda_1, \lambda_2,\ldots, \lambda_n\}, \lambda_i\neq \lambda_j
$$
for $i\neq j$ and 
$$\operatorname{Spec}(B)= \{\mu_1, \mu_2,\ldots, \mu_n\}, \mu_i\neq \mu_j
$$
for $i\neq j$
Then show that the eigenvalues of  $A + B$
 is
$$
\lambda_1+\mu_{i_1}, \lambda_2+\mu_{i_2},\ldots,\lambda_n + \mu_{i_n}
$$
where $\{i_1, \ldots,i_n\}$ is a permutation of $\{1, 2,\ldots,n\}$.
 A: Hint: Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda_1$. Then we have 
$$
A(Bv)=B(Av)=B(\lambda_1v)=\lambda_1(Bv)
$$
which means that $Bv$ is also an eigenvector of $A$ with eigenvalue $\lambda_1$. What does that say about the relationship between $v$ and $Bv$?
A: The answer upstairs is quite a good hint.by Schur's decomposition,we know every matrix $M\in M_n(C)$ is similar to an upper triangular matrix.we only need to prove that there is a matrix $P$ such that $PAP^{-1}=UPPERTRIANGULAR_1$,$PBP^{-1}=UPPERTRIANGULAR_2$.In other words,they can be transfromed into upper triangular matrix synchronizely.
By the hint upstairs,we know there exists a matrix $P_1$,such that
   $P_1AP_1^{-1}$is in the form 
$A^{'}=\begin{pmatrix}
   \lambda_{1}&*&*&*\\
   0 &*&*&*\\
   .&.&.&.\\
   0&0&0&*
  \end{pmatrix}$
$P_1BP_1^{-1}$ in the form
  $B^{'}=\begin{pmatrix}
   \mu_{1}&*&*&*\\
   0 &*&*&*\\
   .&.&.&.\\
   0&0&0&*
  \end{pmatrix}$.
The first column of P is the common eigenvector of $A,B$.(Use the hint upstairs to show the common eigenvector always exists.)
Now notice $A^{'}B^{'}=P_1ABP_1^{-1}=P_1BAP_1^{-1}=B^{'}A^{'}$
We write $A^{'}$ in the form:
$A^{'}=\begin{pmatrix}
  \lambda_1&*\\
  0&A_1
 \end{pmatrix}$
Also we write $B_1$ in the form:
$B^{'}=\begin{pmatrix}
   \mu_1&*\\
   0&B_1
   \end{pmatrix}$
Please show $A^{'}B^{'}=B^{'}A^{'}$ implies $A_1B_1=B_1A_1$
Since the order of your matrices A,B are finite.Do the same procedure on $A_1,B_1$.Actually it has been proved.(Please finish it yourself.)
