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$Ae=\lambda e$

$e^Te=1$

$A$ is a real matrix. $\lambda, e$ are real.

$(A+\Delta A)(e+\Delta e)=(\lambda+\Delta \lambda)(e+\Delta e)$

Neglecting small terms,

$\Delta Ae +A \Delta e=\Delta \lambda e+\Delta e \lambda \tag{1}$

$e^T \times (1)$

$e^T\Delta Ae +e^TA\Delta e=e^T\Delta \lambda e+e^T\Delta e \lambda $

$e^T\Delta Ae +\lambda e^T\Delta e=e^T\Delta \lambda e+e^T\Delta e \lambda $

$e^T\Delta Ae =e^T\Delta \lambda e $

$\Delta \lambda=e^T\Delta Ae \tag{2}$

Is this correct?

How to derive for $\Delta e?$

I have seen somewhere that the eigenvector is almost unchanged for small perturbation in a matrix. How to prove it?

I am looking to prove that

when $\Delta A$ is small, $\Delta e \rightarrow 0$

Update : This thread has an answer, but does it solve my problem? What happens when eigenvalues are same?

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