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For two matrices $A$ and $B$ if the eigen vector of $AB$ is $X$ and that of $BA$ is $Y$, What is the relation between $X$ and $Y$? Can I prove $BX =Y$?

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2 Answers 2

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If $v$ is an eigenvector of $AB$ to the eigenvalue $\lambda\neq0$, then $Bv\ne0$ and $$\lambda Bv=B(ABv)=(BA)Bv,$$ which means $Bv$ is an eigenvector for $BA$ with the same eigenvalue.

But if $\lambda=0$ then $0=\det(AB)=\det(BA)$, so $0$ is an eigenvalue of $BA$ as well.

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  • $\begingroup$ As @egreg said, consider the case when $\lambda = 0$ and $B v = 0$, as in $B = \begin{bmatrix} 1 & 0\\0 & 0\end{bmatrix}$ and $v = (0,1)^{t}$. $\endgroup$ Jan 9, 2014 at 11:36
  • $\begingroup$ Good fix! +1, now. $\endgroup$
    – egreg
    Jan 9, 2014 at 11:37
  • $\begingroup$ @egreg Thanks for the remark $\endgroup$ Jan 9, 2014 at 11:38
  • $\begingroup$ @user127001, but the question was about eigenvectors, see my answer. $\endgroup$ Jan 9, 2014 at 11:45
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I take $X$ to be the set of eigenvectors for $AB$, and similarly for $Y$.

If $B$ is invertible then $AB v = \lambda v$ implies $B A (B v) = \lambda (B v)$. Conversely, $B A u = \lambda u$ implies $A B (B^{-1} u) = \lambda (B^{-1} u)$, so $B X = Y$.

If $A$ is invertible, a similar argument shows that $A Y = X$.

Now note the example $$ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}. $$ Here $A$ is invertible, while $B$ is not. We have $$ A B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \qquad B A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}. $$ So $X$ consists of the multiples of $e_2$, and $Y$ of the multiples of $e_{1}$, but $B X = \{ 0 \} \ne Y$. (However, $A Y = X$.)

Consider also a case when neither matrix is invertible, $$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}. $$ Then $AB = 0$, so $X = V$, the whole space, while $BA = A$, so $Y$ consists of the multiples of $e_1$ only.

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