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My friend has asked me to find out the relationship between the eigenvalues of $AB$ with eigenvalues of $A$,$B$.Where $A,B$ are any two matrices.

My try:Let $A,B$ are two matrices with $v$ be a common eigenvector.Then $Av=x_1v$ and $Bv=x_2v$,$x_1,x_2$ are the eigenvalues of $A$ and $B$ respectively.Then $ABv=A(x_2v)=x_2(Av)=x_2x_1v$......So $x_1x_2$ is also the eigenvalue of $AB$.

If $A,B$ are $2*2$ matrices let $x_3,x_4,x$ are the other eigenvalues of $A,B,AB$ respectively.As we know $det(AB)=det(A)det(B)$,and the determinant of matrix is the product of it's eigenvalues; I get $x_1x_2x=x_1x_3x_2x_4$$=>x=x_3x_4$.So each eigenvalue of $AB$ is the product of suitable pair of eigenvalues from $A$ and $B$.If I generalize this result I can conclude that if the $n*n$ matrices $A,B$ has same $(n-1)$ linearly independent common eigenvalues then each of the eigenvalues of $AB$ is the product of suitable pair of eigenvalues of $A$ and $B$.

Can I extend this result in more general case?Is the condition "$A$ and $B$ has $n-1$ common linearly independent eigenvectors" is necessary?I will be happy if anyone give me any result almost familer to that result with another condition or any beautiful result concerning eigenvalues of $AB$.

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marked as duplicate by Math Lover, user99914, Claude Leibovici, erfink, Namaste linear-algebra Dec 15 '17 at 23:33

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  • $\begingroup$ @Math Lover thanks to show me that result,I didn't find that post before writing my question. $\endgroup$ – Subhajit Saha Dec 14 '17 at 4:21
  • $\begingroup$ Thanks to show that result but that's not my exact question.I need more general result $\endgroup$ – Subhajit Saha Dec 14 '17 at 4:24
  • $\begingroup$ Can you use line breaks in your text ? $\endgroup$ – zwim Dec 14 '17 at 4:56
  • $\begingroup$ @zwin I have edited it. $\endgroup$ – Subhajit Saha Dec 14 '17 at 5:00