Why Laplacian Matrix need normalization and how come the sqrt of Degree Matrix? Why Laplacian matrix needs normalization and how come the sqrt-power of degree matrix? The symmetric normalized Laplacian matrix is defined as
$$\ L = D^{1/2}AD^{-1/2}$$
where L is Laplacian matrix, A is adjacent matrix. Element $A_{ij}$ represents a measure of the similarity between data points with indices $i$ and $j$. D is diagonal matrix, defined as $\ D = \sum\limits_i A_{ij}$. In other words, it is the sum of similarity from node $j$ to all its neighbor node $i$.
I can't figure out some question about the formula:

Why we need to normalize Laplacian Matrix like that? $\ {1/2}$-power ?
Is there any special reason?
Is there any references?
Unnormalized Laplacian serve in the approximation of the minimization of RatioCut, while normalized Laplacian serve in the approximation of the minimization of NCut.
Basically, in the unnormalized case, you optimize an objective relative to the number of nodes in each cluster.
And, in the normalized case, you optimize the objective relative to the volume of each cluster.
The square root comes from this: $f^\top(I-D^{-1}A)f = g^\top(I-D^{-1/2}AD^{-1/2})g$ where $g=D^{1/2}f$.
So, if your optimization method relies on a symmetric PSD matrix, you use the symmetric normalized Laplacian, and then remove the bias by computing $D^{-1/2}g$ to recover $f$.

More insight about RatioCut and NCut.
Let $G=(V,E)$ be a graph with vertex set $V=\{v_1,\dots,v_n\}$ and edge set $E$, $w_{ij}$ denote the positive weight on the edge between $v_i$ and $v_j$. Let $\{C_i:1 \le i \le k\}$ be a disjoint partition of $V$. Define
$$
Cut(C_i:1\le i\le k)\triangleq\frac{1}{2}\sum_{c=1}^k \sum_{i \in C_c,j\in \bar C_c}A_{ij}
$$
$$
RatioCut(A_i:1\le i \le k)\triangleq\sum_{i=1}^k\frac{Cut(A_i,\bar A_i)}{|A_i|}
$$
$$
NCut(A_i:1\le i \le k)\triangleq\sum_{i=1}^k\frac{Cut(A_i,\bar A_i)}{vol(A_i)}
$$
RatioCut:
Let
$$\tag{1}\label{1}
f_i=\begin{cases}\sqrt{|\bar A|/|A|} &amp; \text{if } v_i\in A\\\sqrt{|A|/|\bar A|} &amp; \text{if } v_i \in \bar A\end{cases}
$$
\begin{align*}
f^\top L f &amp; = \frac{1}{2}\sum_{i,j=1}^n w_{ij}(f_i-f_j)^2 \\
&amp; = \frac{1}{2}\sum_{i\in A,j\in \bar A}w_{ij}\left(\sqrt{\frac{|\bar A|}{|A|}}+\sqrt{\frac{|A|}{|\bar A|}}\right) + \frac{1}{2}\sum_{i\in \bar A,j\in A}w_{ij}\left(-\sqrt{\frac{|\bar A|}{|A|}}-\sqrt{\frac{|A|}{|\bar A|}}\right) \\
&amp;=Cut(A,\bar A)\left(\frac{|\bar A|}{|A|}+\frac{|A|}{|\bar A|}+2\right)\\
&amp;=Cut(A,\bar A)\left(\frac{|A|+|\bar A|}{|A|}+\frac{|A|+|\bar A|}{|\bar A|}\right)\\
&amp;=|V| RatioCut(A,\bar A)
\end{align*}
You can also see that $\sum_{i=1}^n f_i=0$, so $f\perp \mathbb 1$.
So minimizing RatioCut is equivalent to the following problem:
$$
\min_{A\subset V}f^\top L f\text{ subject to $f\perp\mathbb 1$, $f_i$ as defined in Eq.\eqref{1}},\|f\|^2=n
$$
NCut:
For this case, define
$$\tag{2}\label{2}
f_i=\begin{cases}
\sqrt{\frac{vol(\bar A)}{vol(A)}} &amp;\text{if $v_i\in A$}\\
\sqrt{\frac{-vol(A)}{vol(\bar A)}} &amp;\text{if $v_i\in \bar A$}\\
\end{cases}
$$
Similar to above, we have $Df\perp \mathbb 1$, $f^\top D f=vol(V)$ and $f^\top L f=vol(V)NCut(A,\bar A)$.
Minimizing NCut is equivalent to
$$
\min_{A\subset V} f^\top L f \text{ subject to $f$ as in Eq.\eqref{2}, $Df\perp \mathbb 1$ and $f^\top Df=vol(V)$}
$$
Substituting $g=D^{1/2}f$ we get
$$
\min_{A\subset V} g^\top D^{-1/2}LD^{-1/2} g \text{ subject to $g=D^{1/2}f$, $f$ as in Eq.\eqref{2}, $g\perp D^{1/2}\mathbb 1$ and $\|g\|^2=vol(V)$}
$$
Then observe that $D^{-1/2}LD^{-1/2}$ is the symmetric normalized Laplacian.
Note that relaxed version of NCut problem is that of minimization of generalized eigenproblem

$Lx = \lambda Dx$

consult the paper by Shi and Malik for a justification.

To solve this problem, i.e. to find eigenvector $x$ and eigenvalue $\lambda$, multiply both sides of this equation by $D^{-1/2}$:

$D^{-1/2}Lx=\lambda D^{1/2}x$

This equation can be rewritten as

$D^{-1/2}LD^{-1/2}D^{1/2}x=\lambda D^{1/2}x$

Setting $y=D^{1/2}x$ you obtain

$D^{-1/2}LD^{-1/2}y=\lambda y$

that is you reduced the generalized eigenproblem $Lx=\lambda Dx$ to the standard eigenproblem $\mathcal{L} y = \lambda y$, where $\mathcal{L} = D^{-1/2}LD^{-1/2}$ is the normalized Laplacian matrix. 

Knowing eigenvector $y$ you can find the eigenvector $x=D^{-1/2}y$.