Is there a simple way to explain a fundamental representation of $O(N)$? Is there a simple way to explain fundamental representation in Physics? For example, a fundamental representation of $O(N)$?
 A: The fundamental - or defining - representation of $O(N)$ is a representation where the elements of the orthogonal group, which be definition satisfy $R^T\cdot R=\hat 1$, are represented by $N\times N$ matrices acting on vectors in $N$ dimensions.
The fundamental representation of $SU(2)$ is a representation by $2\times 2$ unitary matrices, i.e. matrices of the type
$$
R(a,b,c,d)=\left(\begin{array}{cc}
a&b\\ c &d\end{array}\right)
$$
so that $R^\dagger (a,b,c,d) R(a,b,c,d)=\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)$
The fundamental representation of $SO(3)$ is a representation by $3\times 3$ real matrices, usually factored in the form $R_z(\alpha) R_y(\beta) R_z(\gamma)$, where for instance
$$
R_z(\alpha)=
\left(\begin{array}{ccc}
1&0&0\\
0&\cos\alpha&-\sin\alpha\\
0&\sin\alpha&\cos\alpha 
\end{array}\right)\, .
$$
Note that elements in the orthogonal group also include reflections, meaning the determinant can be $\pm 1$.  The special orthogonal group $SO(N)$ will be the subset of elements of $O(N)$ with determinant $1$.
The dimension of a representation is not enough to identify if it is the defining representation.  For instance, there are two representations of $SU(3)$ of dimension $3$, commonly denoted in HEP by $\textbf{3}$ and $\bar{\textbf{3}}$.  The second is conjugate to the first (implying in particular that the eigenvalues of diagonal generators in $\bar{\textbf{3}}$ are the negatives of those in $\textbf{3}$).  There are also cases where the number of non-equivalent (i.e. not related by a similarity transformation) representations is greater than $2$ so denoting a representation by its dimension can cause labelling headaches.  A clearer labelling scheme is provided by the Dynkin labels of the highest weight of a representation: in this scheme the $\textbf{3}$ and $\bar{\textbf{3}}$ are denoted by $(1,0)$ and $(0,1)$, respectively.
One property of the fundamental representation is that any other representation will occur in some power of the tensor product of this representation.  Thus for instance the representation $\textbf{6}$ or $(2,0)$ of $SU(3)$ will occur in the decomposition of tensor product of $(1,0)\otimes (1,0)\sim \textbf{3}\otimes\textbf{3}$.  Note that other representation can also occur in the tensor product: the representation $(0,1)$ also appears in $(1,0)\otimes (1,0)\sim \textbf{3}\otimes\textbf{3}$.  This holds quite generally so if you wanted to construct a general representation of a group you could do this by tensoring sufficiently many copies of the fundamental and sorting out the pieces until you "discover" the irrep you're interested in (provided it appears at least once.)  This leads to the "tensor method" of constructing representations.
Things are a little more diffuse when not working with classical groups.  The "fundamental" representation of $E(2)$, the Euclidean group in the plane, is actually in terms of $3\times 3$ matrices: there is a $2\times 2$ block for the rotation part, a $1\times 2$ block for the translations, and an extra bit to make the matrix action work out.  Thus, a general element parametrized by a rotation $\theta$ and by a translation vector $(x,y)$ is given by the $3\times 3$ matrix
$$
(\theta,x,y)\mapsto \left(\begin{array}{ccc}
\cos\theta&-\sin\theta&x\\
\sin\theta&\cos\theta& y\\
0&0&1\end{array}\right)
$$
acting on vectors of the form $\left(\begin{array}{c} a\\ b \\ 1\end{array}\right)$
