I'm struggling with my understanding of eigenvalue equations.
I am trying to prove that if λ is an eigenvalue of a diagonalizable matrix A with right and left eigenvectors x and y, then by perturbing the system such that λ + $\delta \lambda $ is an eigenvalue of A+$\delta \lambda $ one can obtain:
$$\delta \lambda = \frac{y^{H}\delta Ax}{y^{H}x}$$
where H is the Hermitian conjugate.
I'm struggling to see how a perturbed eigenvalue equation can be expanded to obtain the $\delta \lambda $.
So far I have that:
$$ (A + \delta A )(x + \delta x) = (\lambda + \delta \lambda )(x + \delta x) $$
and
$$ (A + \delta A )(y + \delta y) = (\lambda + \delta \lambda )(y + \delta y) $$
What throws me off is where do the $ \delta y $ and $\delta x$ terms disappear to in the result, and where the $ y^{H} $ term comes from and also the significance of the 'right' and 'left' eigenvectors.
The trick here is not to use $y$ since it complicates the derivation. Expanding the perturbed equation, we get $$ Ax + A\delta x + \delta A x +\delta A \delta x = \lambda x + \lambda \delta x + \delta \lambda x + \delta \lambda \delta x $$ Since $Ax = \lambda x$, we get $$ A\delta x + \delta A x \approx \lambda \delta x + \delta \lambda x $$ where we have neglected second order terms. Pre-multiplying the above by $x^H$, we obtain $$ x^H A\delta x + x^H\delta A x \approx \lambda x^H \delta x + \delta \lambda x^H x $$
Since, $x^H x= 1$ (we assume that $x$ is normalised), we reduce the above equation to $$ x^H A \delta x + x^H\delta A x \approx \lambda x^H \delta x + \delta \lambda $$ Thus, $$ \delta \lambda \approx x^H (A -\lambda I)\delta x + x^H \delta A x . $$ If $A$ is Hermitian, then the first term in the rhs vanishes.
By pre-multiplying, the second displayed equation by $y^H$, we also get $$ y^H(A-\lambda I) \delta x + y^H \delta A x \approx \delta \lambda y^H x $$ which gives the requested result $$ \delta \lambda = \frac{y^H\delta Ax}{y^Hx} $$ since $ y^H(A-\lambda I) = 0$.