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Can this equation $(AX)\cdot Y=Y\cdot (A^{T}X)$ be proved? where $A$ is a symmetric matrix, and $X$, $Y$ are the eigenvectors of $A$.

Eugene Zhang
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MrDi
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    The dot product is defined as $X\cdot Y=X^TY$. So $(AX)\cdot Y=(AX)^TY=X^TA^TY=X^T(A^TY)=X\cdot(A^TY)$. It has nothing to do with $X$ and $Y$ being eigenvectors. – symplectomorphic Jun 01 '15 at 03:57

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$$(AX)\cdot Y = \lambda X\cdot Y$$ $$Y\cdot (A^TX) = Y\cdot X \lambda = \lambda X\cdot Y$$ $$\Rightarrow (AX)\cdot Y = X\cdot(A^T Y)$$

Remember that in a symmetric matrix, $A = A^T$, and that the dot product of vectors is commutative.

  • could this be proved without using eigenvalues – MrDi Jun 01 '15 at 03:50
  • Yes, you could just flip $Y\cdot (A^TX)$ and use the equality $A=A^T$. I figured that since you gave $X$ and $Y$ as eigenvectors that it was a practice problem that was intended to illustrate their usefulness. –  Jun 01 '15 at 03:56