# An example of non-strictly positive-definite inner product

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.

• Hint: What would an orthogonal basis for such a product look like? Mar 27 at 18:20
• The conventional name for your “non-strictly positive definite inner product” is semi-inner product. Mar 27 at 18:52

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 $$A$$.