Suppose I have a vector space $V$ over $\mathbb{R}$. A scalar product on $V$ is a symmetric bilinear form, so such that $\langle v,w\rangle = \langle w,v\rangle~ \forall v,w \in V$.
To my understanding, the kernel $V^{\perp}$ of a scalar product $\langle \cdot ,\cdot\rangle $ is defined as the set of vectors $v \in V$ such that $\langle v,w\rangle = 0$ for every $w \in V$.
Furthermore, the scalar product is non-degenerate if $V^{\perp} = {O}$; it is degenerate otherwise.
Starting from these definitions, I'm having a hard time visualizing the kernel and degenerate scalar products. I can see that, for example, if the space we are considering is $\mathbb{R}^{2}$ and we are considering the canonical scalar product as $\langle v,w\rangle =\cos(\theta )\left \| v \right \|\left \| w \right \|$, the kernel would be only the null vector, making it therefore a non-degenerate scalar product.
But besides this example, I can't see others. The kernel contains those vectors that scalar-multiplied by any vector of the subspace yields $0$. What is an example of "non-empty" kernel? In $\mathbb{R}^2$? $\mathbb{R}^3$?
Hope the question is clear and excuse me if the answer may be trivial, I just want to get my bearings straight.
On 2-space, define $\langle\langle x, y \rangle \rangle = x_1 y_1$. That's symmetric, bilinear, etc., but it's degenerate, as the vector $(0, 1)$ is in the kernel.