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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.

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  • $\begingroup$ Hint: What would an orthogonal basis for such a product look like? $\endgroup$ Mar 27 at 18:20
  • $\begingroup$ The conventional name for your “non-strictly positive definite inner product” is semi-inner product. $\endgroup$
    – user1551
    Mar 27 at 18:52

1 Answer 1

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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 $A$.

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  • $\begingroup$ Oh, you want a positive semidefinite form that is not definite, which is particular is necessarily degenerate. Usually the term "inner product" requires nondegeneracy (and often, but not always, it requires positive definiteness). I'll update my answer. $\endgroup$ Mar 27 at 19:16

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