Give an example of non-strictly positive-definite inner product on an arbitrary vector space.
By non-strictly positive-definite inner product, I mean that $||X||=0$ does not necessarily imply $X=0$.
As you may know, $||X||=\sqrt{\langle X,X\rangle}$.
My attempt: I considered $\langle.,.\rangle:\mathbb{R}*\mathbb{R}\to\mathbb{R}$ as an inner product for vector space $\mathbb{R}$ over field $\mathbb{R}$ and defined $\langle u,v\rangle = 0$ for all vectors. Clearly $||u||=0$ for all vectors.
But I want a non-trivial example.
Thank for your time.
Example 1 The symmetric bilinear form $\Phi : \Bbb R^2 \times \Bbb R^2 \to \Bbb R$ defined by $$\Phi(X, Y) := X_1 Y_1$$ is positive semidefinite, as $$\Phi(X, X) = X_1^2 \geq 0,$$ but it is not definite, as $\Phi\left(\pmatrix{0\\1}, \pmatrix{0\\1}\right) = 0$.
In fact, Sylvester's Law of Inertia implies that any $2$-dimensional real vector space equipped with symmetric bilinear form that is positive semidefinite but not definite is isometric to this example.
Example 2 More generally, for any vector space $\Bbb V$ and any nonzero linear functional $\alpha \in \Bbb V^*$, i.e., linear map $\alpha : \Bbb V \to \Bbb R$, the symmetric bilinear form $$\Phi(X, Y) := \alpha(X) \alpha(Y)$$ is positive semidefinite but not zero, and it is not definite if $\dim \Bbb V > 1$. Example 1 is the special case $\Bbb V = \Bbb R^2$ and $\alpha(X) = X_1$.
Example 3 For any symmetric matrix $A \in M_n(\Bbb R)$, the symmetric bilinear map $\Bbb R^n \times \Bbb R^n \to \Bbb R$ defined by $$\langle X, Y \rangle := Y^\top A X$$ in a symmetric bilinear form. This bilinear form is positive semidefinite if all of the eigenvalues of $A$ are nonnegative, and it is not positive definite if it moreover has at least $1$ zero eigenvalue. Example 1 is the special case $n = 2$ and $A = \pmatrix{1&\cdot\\\cdot&0}$.
In general if a symmetric bilinear form $\Phi$ on a vector space $V$ is positive semidefinite but not definite, it is degenerate, that is, there is a vector $X \in V$ such that $\Phi(X, Y) = 0$ for all $Y \in V$, that is, such that the linear functional $\Phi(X, \,\cdot\,)$ is the zero functional. We sometimes call the set of such $X$ the kernel of $\Phi$, as it is the kernel of the linear map $\operatorname{Sym}^2 V^* \to V^*$, $\Phi \mapsto \Phi(X, \,\cdot\,)$. In Example $1$, $\ker \Phi = \operatorname{span}\left\{\pmatrix{0\\1}\right\}$, in Example $2$, $\ker \Phi = \ker \alpha$, and in Example $3$, $\ker \Phi = \ker A$, that is, the $0$-eigenspace of $A$.