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Let $A$ be a matrix, $a,b$ its eigenvectors and $\lambda_a, \lambda_b$ their corresponding eigenvalues. Define $a^\prime = Aa$ and $b^\prime = Ab$. Is it true that \begin{equation} \lambda_a = \frac{\Vert a^\prime\Vert}{\Vert a\Vert} \iff \Vert a\Vert = \frac{1}{\sqrt{\lambda_a}} \end{equation} and \begin{equation} \lambda_b = \frac{\Vert b^\prime\Vert}{\Vert b\Vert} \iff \Vert b\Vert = \frac{1}{\sqrt{\lambda_b}} ? \end{equation}

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No. First of all, note that the statements don't even make sense if $\lambda_a=0$ or $\lambda_b=0$. On the other hand, $a'=\lambda_aa$ and therefore$$\frac{\|a'\|}{\|a\|}=\lvert\lambda_a\rvert.$$So, $\lambda_a=\frac{\|a'\|}{\|a\|}\iff\lambda_a\geqslant0$. And, of course, $\lambda_a\geqslant0$ is not equivalent to $\|a\|=\frac1{\sqrt{\lambda_a}}$.

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Hint: $$\frac{||a'||}{||a||}=\frac{||Aa||}{||a||}=\frac{||\lambda_aa||}{||a||}=\frac{|\lambda_a|\cdot||a||}{||a||}=|\lambda_a|$$

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Hint : $$a' = A a = \lambda_a a$$

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