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If $A$ and $B$ are both $n\times n$ matrices, and $\lambda_A$ and $\lambda_B$ are the eigenvalues for those matrices with respect to the same eigenvector $v$, then does $ABv = BAv$?

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Since scalar multiplication commutes with matrix multiplication, this is not too hard to show. Always go back to the definitions. In this case, $\lambda, v$ form an eigenvector-eigenvalue pair of a matrix $X$ exactly if $Xv=\lambda v$.

Take the left hand side:

$$A(Bv)=A\lambda_Bv=\lambda_B Av = \lambda_B\lambda_A v$$

Taking the right hand side:

$$B(Av)=B\lambda_Av=\lambda_A Bv = \lambda_A\lambda_B v$$

And now of course, $\lambda_A\lambda_B v = \lambda_B\lambda_A v$.

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