Kernel Principal Component Analysis (PCA) I learn kernel PCA from wikipedia. In this article, the eigen equation is 
\begin{equation}
  N \lambda \vec{\alpha} = \boldsymbol{K} \vec{\alpha}
\end{equation}
where $\lambda$ is the eigen value, $\vec{\alpha}$ is the eigen vector of $\lambda$ and $\boldsymbol{K}$ is the kernel matrix.
Why is the left side multiplied by N? What is the meaning and effect of $N$?
 A: Recall for linear PCA $N$ is the number of data points in the set and comes in the eigen decomposition of the covariance matrix $C$, where $C = \frac{1}{N} \sum_{i=1}^N x_i x_i^\top$ ( we look for eigenvectors $v$ such that $\lambda v = C v$). So we simply multiply through by the constant $N$ to bring it through to the other side. It is similar multiplication by the constant when using a kernel method.
A: The parameter $N$ denotes the number of samples you have, e.g. consider that we have a set $\mathcal{X} = \vec{x}_1, \dots, \vec{x}_N$ in a space $\mathbb{R}^M$, for some $M$. Considering that you use a transformation $\Phi : \mathbb{R}^M \to \textbf{F}$, where $\textbf{F}$ is the space of the maped data, i.e. of $\Phi(\vec{x}_1), \dots, \Phi(\vec{x}_N$), and verifies that $\text{dim}(\textbf{F}) > M$. Since we are only interested in the dot product in this new space, we can use the kernel trick. For this, we consider the Kernel matrix $\boldsymbol{K}$, where $\boldsymbol{K}_{ij}= \Phi(\vec{x}_i)^T \Phi(\vec{x}_j)$. for further details take a look at Support Vector Learning (Schölkopf, 1997).
Now back to the problem, as you suggest, at the end we need to find the solutions to the equation
\begin{equation}
  N \lambda \vec{\alpha} = \boldsymbol{K} \vec{\alpha},
\end{equation}
which can in fact be rewritten as 
\begin{equation}
  \lambda' \vec{\alpha} = \boldsymbol{K} \vec{\alpha}, \tag{1} 
\end{equation}
where $\lambda' = N \lambda$.
(1) has $N$ solutions, given by the eigenvectors $\vec{\alpha}_1, \dots, \vec{\alpha}_N$ and their corresponding eigenvalues $\lambda_1', \dots, \lambda_N'$ (where remember that $\lambda_k' = N \lambda_k$ for $k = 1, \dots, N$). Next, we choose a value $n$ and only consider the first $n$ pairs of eigenvalues and eigenvectors, given that $\lambda_1 \geq \lambda_2 \geq \dots \geq \lambda_n$. Using some algebra, we are then able to project the mapped data and find a manifold in $\textbf{F}$ that accurately represents it.
I have to say that in the literature this aspect is unclear due to some notation abuse (sometimes they use $\lambda_1$ to refer to $\lambda_1'$)
