Proof that eigenvector corresponding to simple eigenvalue is continuous Let $\lambda$ be a simple eigenvalue of $A \in L(C^n)$ and let $x$ be the corresponding eigenvector. Then for $E \in L(C^n)$,  $A+E$ has an eigenvalue $\lambda(E)$ and an eigenvector $x(E)$ such that as $E\longrightarrow 0$,
$\lambda(E) \longrightarrow \lambda$ and $x(E) \longrightarrow x$.
This is 3.1.3 (Chapter 3, page 46) from Numerical Analysis: a second course by J. Ortega. I am using the book  for self study.
The continuity of the eigenvalue is clear, but I do not understand the proof for the eigenvector bit.
Could someone help me understand what is going on ? I looked at Continuity of a simple eigenvalue and its corresponding eigenvector
 but that does not answer my question. I would like to have a proof that does not invoke implicit/inverse function theorem.
 A: Let us first note that this statement is not completely correct as it stands. This is mainly because eigenvectors are only unique up to a scalar multiple. So for a sequence of matrices $(E_n)_n$ with $E_n \rightarrow 0$, you could choose $x(E_n)$ in such a way that $\Vert x(E_n) \Vert = n$. In this case, your convergence certainly does not hold.
For simplicity note that by subtracting $\lambda \cdot \rm{id}$ from $A$, we can assume $\lambda = 0$.
The above means that you have to normalize the $x(E)$ in a suitable way.
For this, let us assume $\varphi(x) \neq 0$ for some linear functional $\varphi$ on $\Bbb{C}^n$. You could take e.g. $\varphi(z) = z_j$ for a suitable $j$.
Let us assume (by rescaling) that $x(E)$ is chosen so that $\varphi(x(E)) = \varphi(x)$ if $\varphi(x(E)) \neq 0$.
Now assume that your claim is false. Then there is a sequence $(E_n)_n$ with $E_n \rightarrow 0$ but $x(E_n) \not\rightarrow x$. Choosing a suitable subsequence (and denoting that by $(E_n)_n$ again), we can assume $\Vert x(E_n) - x(E)\Vert > \varepsilon$ for all $n$.
Let us now consider the "normalized" vectors $y(E_n) := \frac{x(E_n)}{\Vert x(E_n) \Vert}$. By restricting to a subsequence, we ca assume (by Bolzano Weierstraß) that $y(E_n) \rightarrow y$ for some $y \in \Bbb{C}^n$ with $\Vert y \Vert = 1$.
Note that $(A + E_n) y_n = \lambda_n y_n \rightarrow \lambda y = 0$ (because you already know that $\lambda_n \rightarrow \lambda$, as you say), i.e. $A y_n = (A + E_n) y_n - E_n y_n \rightarrow 0$, because $E_n \rightarrow 0$ and $(y_n)_n$ is bounded.
This shows $Ay = 0$, i.e. $y = c \cdot x$ for some $c \in \Bbb{C}$ (because $x$ is the only eigenvector for the eigenvalue $0$). We have $c \neq 0$ because $\Vert y \Vert = 1$.
This shows $\varphi(y_n) \rightarrow \varphi(y) = c \cdot \varphi(x) \neq 0$ and thus $\varphi(y_n) \neq 0$ and hence $\varphi(x(E_n)) \neq 0$ for $n$ large enough and thus (by our normalization of the $x(E_n)$) that $\varphi(x(E_n)) = \varphi(x)$ for $n$ large.
Finally, 
$$\Vert x(E_n)\Vert = \frac{\varphi(x(E_n))}{\varphi(y_n)} = \frac{\varphi(x)}{\varphi(y_n)} \rightarrow \frac{\varphi(x)}{\varphi(y)}.$$
We conclude $x(E_n) = \Vert x(E_n) \Vert \cdot y_n \rightarrow \frac{\varphi(x)}{\varphi(y)} \cdot y = \frac{\varphi(x)}{c \varphi(x)} \cdot c x = x$. This contradicts our assumption $\Vert x(E_n) - x\Vert > \varepsilon$ for all $n$.
