0
$\begingroup$

i am interested if there is some relationship between eigenvalues of commutative matrices?or for any two matrix $A$ and $B$,we have following equation

$A*B=B*A$

for example

A=[1 2;3 4]

A =

     1     2
     3     4

and second matrix

B=[7 10; 15 22]

B =

     7    10
    15    22

we have

A*B

ans =

    37    54
    81   118

and

B*A

ans =

    37    54
    81   118

now i have several questions,just in simple please help me to clarify it,what is relationship between eigenvalues of $A$ and $B$?or between $A*B$ and $A$?or between product of these two matrix and matrix $B$?as i know for some given two matrices

$e^{A+B}=e^{A}*e^{B}$

when $A*B=B*A$

because we can represent matrix exponential for diagonalisable matrix using eigenvalues,can we see some relationship between commute matrices?thanks in advance

$\endgroup$
0
$\begingroup$

Commuting matrices have the same set of non zero eigenvalues (counted with multiplicity).

This is because commuting set of matrices are simultaneously trianguralizable. The eigenvalues of an upper triangular matrix are the diagonal elements. Eigenvalues of $AB$ are just the product of eigenvalues of $A$ with those of $B$.

$\endgroup$
  • $\begingroup$ but different value right? how can i understand same set of non zero eigenvalues? $\endgroup$ – dato datuashvili Feb 1 '14 at 18:20
  • 1
    $\begingroup$ Non zero eigenvalues are the same for $AB$ and $BA$. The eigenvalues are the products of eigenvalues of $A$ and $B$. For example if $A$ has an eigenvalue 2, and $B$ has an eigenvalue $4$, then $AB$ has an eigenvalue $8$, and so on. Hope that helps. $\endgroup$ – voldemort Feb 1 '14 at 18:25
  • $\begingroup$ and does linear combination of commute matrices would be also commute with original matrices?? $\endgroup$ – dato datuashvili Feb 1 '14 at 18:26
  • $\begingroup$ Yes :). Even more is true. If $P,Q$ are polynomials, and $AB=BA$ then $P(A)Q(B)=Q(B)P(A)$. $\endgroup$ – voldemort Feb 1 '14 at 18:30
  • 1
    $\begingroup$ The first sentence is very wrong. Did you mean to say eigenvectors (which are nonzero by definition)? Even then, the sets are not always the same. $\endgroup$ – Marc van Leeuwen Mar 13 '15 at 13:13

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.