Let $\mathcal{X}$ represent your input space i.e the space where your data points resides. Consider a function $\Phi:\mathcal{X} \rightarrow \mathcal{F}$ such that it takes a point from your input space $\mathcal{X}$ and maps it to a point in $\mathcal{F}$. Now, let us say that we have mapped all your data points from $\mathcal{X}$ to this new space $\mathcal{F}$. Now, if you try to solve the normal linear svm in this new space $\mathcal{F}$ instead of $\mathcal{X}$, you will notice that all the earlier working simply look the same, except that all the points $x_i$ are represented as $\Phi(x_i)$ and instead of using $x^Ty$ (dot product) which is the natural inner product for Euclidean space, we replace it with $\langle \Phi(x), \Phi(y) \rangle$ which represents the natural inner product in the new space $\mathcal{F}$. So, at the end, your $w^*$ would look like,
$$
w^*=\sum_{i \in SV} h_i y_i \Phi(x_i)
$$
and hence,
$$
\langle w^*, \Phi(x) \rangle = \sum_{i \in SV} h_i y_i \langle \Phi(x_i), \Phi(x) \rangle
$$
Similarly,
$$
b^*=\frac{1}{|SV|}\sum_{i \in SV}\left(y_i - \sum_{j=1}^N\left(h_j y_j \langle \Phi(x_j), \Phi(x_i)\rangle\right)\right)
$$
and your classification rule looks like: $c_x=\text{sign}(\langle w, \Phi(x) \rangle+b)$.
So far so good, there is nothing new, since we have simply applied the normal linear SVM to just a different space. However, the magic part is this -
Let us say that there exists a function $k:\mathcal{X}\times\mathcal{X}\rightarrow \mathbb{R}$ such that $k(x_i, x_j) = \langle \Phi(x_i), \Phi(x_j) \rangle$. Then, we can replace all the dot products above with $k(x_i, x_j)$. Such a $k$ is called a kernel function.
Therefore, your $w^*$ and $b^*$ look like,
$$
\langle w^*, \Phi(x) \rangle = \sum_{i \in SV} h_i y_i k(x_i, x)
$$
$$
b^*=\frac{1}{|SV|}\sum_{i \in SV}\left(y_i - \sum_{j=1}^N\left(h_j y_j k(x_j, x_i)\right)\right)
$$
For which kernel functions is the above substitution valid? Well, that's a slightly involved question and you might want to take up proper reading material to understand those implications. However, I will just add that the above holds true for RBF Kernel.
To answer your question, "Is the situation so that all the support vectors are needed for the classification?"
Yes. As you may notice above, we compute the inner product of $w$ with $x$ instead of computing $w$ explicitly. This requires us to retain all the support vectors for classification.
Note: The $h_i$'s in the final section here are solution to dual of the SVM in the space $\mathcal{F}$ and not $\mathcal{X}$. Does that mean that we need to know $\Phi$ function explicitly? Luckily, no. If you look at the dual objective, it consists only of inner product and since we have $k$ that allows us to compute the inner product directly, we don't need to know $\Phi$ explicitly. The dual objective simply looks like,
$$
\max \sum_i h_i - \sum_{i,j} y_i y_j h_i h_j k(x_i, x_j) \\
\text{subject to : } \sum_i y_i h_i = 0, h_i \geq 0
$$
This reasoning is essentially that of Sycorax's answer, but no need to resort to that theorem:
Consider two distinct points $x$ and $y$. For $\theta<0$, their Gram matrix is
$$
\begin{bmatrix}
k(x, x) & k(x, y) \\
k(x, y) & k(y, y)
\end{bmatrix}
=
\begin{bmatrix}
1 & \alpha \\
\alpha & 1
\end{bmatrix}
$$
where $\alpha = k(x, y) = \exp\left( - \frac{\theta}{2} \lVert x - y \rVert^2 \right) = \exp\left( \tfrac12 \lvert{\theta}\rvert \lVert x - y \rVert^2 \right) > 1$, since the argument to $\exp$ is strictly positive.
The characteristic polynomial of this Gram matrix gives $(\lambda - 1)^2 - \alpha^2 = 0$, so that $\lvert \lambda - 1 \rvert = \alpha$, and
the eigenvalues of this matrix are $1 + \alpha$ and $1 - \alpha$. Since $\alpha > 1$, that second eigenvalue is negative, and the kernel is not psd.
Best Answer
I figured out what is needed to be done. Actually, it's something simple, but its seems I had a matlaboid bug... Here is the code and the resulting figure for the "XOR" binary classification problem.
where the function
just produces the kernel function (RBF kernel), $k(\mathbf{x},\mathbf{x}')=\operatorname{exp}\left(-\gamma\lVert\mathbf{x}-\mathbf{x}'\rVert^2\right)$.
Here is the result for the XOR problem. Here $\gamma=4$.