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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.

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    $\begingroup$ Any symmetric matrix gives rise to an inner product via $<u,v>\equiv uMv$. Simply choose a matrix that is not full rank. $\endgroup$ – Derek Elkins Jan 29 '16 at 12:18
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    $\begingroup$ The simplest such example would be defined as $\langle v,w \rangle =0$ for all $v$ and $w$. $\endgroup$ – Pierre-Guy Plamondon Jan 29 '16 at 12:20
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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.

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  • $\begingroup$ I see. Just to see if I understood: the kernel in this case would be the set of all vectors of the form $(0,i),(i,0)$ for $i \in \mathbb{R}$ ? $\endgroup$ – Easymode44 Jan 29 '16 at 14:06
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    $\begingroup$ Nope. The vector $v = (1,0)$ is not in the kernel, because $\langle\langle v,v\rangle\rangle = 1$. But all vectors of the form $(0, x)$ are in the kernel, yes. $\endgroup$ – John Hughes Jan 29 '16 at 22:35

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