Actually the solve the SVM is to solve the following Lagrangian Equation:
If we don't use kernel function, $\langle x^{(i)},x^{(j)}\rangle$ is just the vector vector inner product. The $a_ia_j\langle x^{(j)},x^{(j)}\rangle $ is the same to $\langle a_ix^{(i)}, a_jx^{(j)}\rangle $ in the formula.
But if we use the kernel function. That's totally different, right?
Q: So Can I get the same train result between $\langle a_ix^{(i)}, a_jx^{(j)}\rangle$ and $a_ia_j\langle x^{(j)},x^{(j)}\rangle $?
When using a nonlinear kernel $\alpha_i \alpha_j \langle x^{(i)},x^{(j)}\rangle\neq \langle \alpha_i x^{(i)},\alpha_j x^{(j)}\rangle $. Keep in mind that the support values ($\alpha$'s) are part of the solution of the dual problem.
For nonlinear kernels, the training outcome of $\alpha_i \alpha_j \langle x^{(i)},x^{(j)}\rangle$ is completely different from $\langle \alpha_i x^{(i)},\alpha_j x^{(j)}\rangle$. In fact, for nonlinear kernels, the optimization problem you would get when using $\langle \alpha_i x^{(i)},\alpha_j x^{(j)}\rangle$ is no longer a QP and is probably intractable. There is really no reason for doing this.
Take an RBF kernel for example, with $\kappa(x^{(i)},x^{(j)})=e^{-\gamma \|x^{(i)}-x^{(j)}\|^2}$. In your idea, this becomes: $$\kappa(\alpha_i x^{(i)},\alpha_jx^{(j)})=e^{-\gamma \|\alpha_i x^{(i)}-\alpha_j x^{(j)}\|^2}.$$ You can easily see that solving this to $\alpha$ (and possible $b$) is much more complex!
