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

  • $\begingroup$ What, specifically, are you looking to understand about the problem? $\endgroup$ – DBPriGuy Jan 6 '17 at 23:42
  • 2
    $\begingroup$ Show that $A,B$ can be simultaneously diagonalised. $\endgroup$ – copper.hat Jan 6 '17 at 23:48
  • $\begingroup$ Related? mathoverflow.net/questions/4224/eigenvalues-of-matrix-sums $\endgroup$ – DBPriGuy Jan 6 '17 at 23:51
  • 1
    $\begingroup$ Btw, that is some pretty grotesque (is that a contradiction?) formatting :-). $\endgroup$ – copper.hat Jan 6 '17 at 23:58
  • 2
    $\begingroup$ @copper.hat an "oxymoron", I suppose $\endgroup$ – 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 $Bv$?

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.