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

• What, specifically, are you looking to understand about the problem? – DBPriGuy Jan 6 '17 at 23:42
• Show that $A,B$ can be simultaneously diagonalised. – copper.hat Jan 6 '17 at 23:48
• – DBPriGuy Jan 6 '17 at 23:51
• Btw, that is some pretty grotesque (is that a contradiction?) formatting :-). – copper.hat Jan 6 '17 at 23:58
• @copper.hat an "oxymoron", I suppose – Omnomnomnom Jan 7 '17 at 0:25

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 $Bv$?

• One needs to do a special case for the case when $Bv=0$, though. – egreg Jan 7 '17 at 11:36
• @egreg That just means that $\mu_{i_1} = 0$, methinks. – Arthur Jan 7 '17 at 11:44
• Yes, of course, but one should be aware of it. – egreg Jan 7 '17 at 13:02
• İ don't know really what can we say about that relationship – user401187 Jan 10 '17 at 11:15
• What does that mean or how can I use it – user401187 Jan 10 '17 at 11:15

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.)

• Ok thanks a lot i will try – user401187 Jan 7 '17 at 10:41