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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 $\{1, 2,\ldots,n\}$.

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    What, specifically, are you looking to understand about the problem? – 2017-01-06
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    Show that $A,B$ can be simultaneously diagonalised. – 2017-01-06
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    Related? http://mathoverflow.net/questions/4224/eigenvalues-of-matrix-sums – 2017-01-06
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    I really don't have any ideas just I know that the eigenvalues of AB are equal to the eigenvalues of BA – 2017-01-06
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    What can you say about the eigenvectors? Big hint :-). – 2017-01-06
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    Btw, that is some pretty grotesque (is that a contradiction?) formatting :-). – 2017-01-06
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    All those curly braces, `\left`, `\right` and `\mathrm` to clean up. – 2017-01-06
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    @copper.hat an "oxymoron", I suppose – 2017-01-07
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    Note that any commuting matrices can be simultaneously upper triangularized – 2017-01-07

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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 $Bv$?

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    One needs to do a special case for the case when $Bv=0$, though. – 2017-01-07
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    @egreg That just means that $\mu_{i_1} = 0$, methinks. – 2017-01-07
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    Yes, of course, but one should be aware of it. – 2017-01-07
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    İ don't know really what can we say about that relationship – 2017-01-10
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    What does that mean or how can I use it – 2017-01-10
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    @user401187 Since all eigenvalues of $A$ are distinct, and $v$ and $Bv$ are eigenvectors corresponding to the same eigenvalue of $A$, they must be _parallel_. That means that $v$ is also an eigenvector for $B$ with eigenvalue $\mu_{i_1}$ for some $i_1$. – 2017-01-10
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    İs that enough ? – 2017-01-10
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    @user401187 We have, $(A+B)v = Av+Bv = \lambda_1 v + \mu_{i_1}v = (\lambda_1+\mu_{i_1})v$, which is what we wanted. Now do the same for every other eigenvalue of $A$, and you're done. – 2017-01-10
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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.)

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    Ok thanks a lot i will try – 2017-01-07