Question

Let \(\mathbf{b}\) be a symmetric bilinear form. Show that the kernel of \(\mathbf{b}_{*}: \mathcal{V} \rightarrow\) \(V^{*}\) consists of all vectors \(\mathbf{u} \in \mathcal{V}\) such that \(\mathbf{b}(\mathbf{u}, \mathbf{v})=0\) for all \(\mathbf{v} \in \mathcal{V}\). Show also that in the \(\mathbf{b}\) -orthonormal basis \(\left\\{\mathbf{e}_{j}\right\\}\), the set \(\left\\{\mathbf{e}_{i} \mid \mathbf{b}\left(\mathbf{e}_{i}, \mathbf{e}_{i}\right)=0\right\\}\) is a basis of ker \(\mathbf{b}\), and therefore the set of linearly independent isotropic vectors is the nullity of \(\mathbf{b}\).

Short answer

The kernel of a symmetric bilinear form \(\mathbf{b}_{*}\) contains all vectors \(\mathbf{u}\) such that \(\mathbf{b}(\mathbf{u}, \mathbf{v}) = 0\) for all \(\mathbf{v}\). Furthermore, the set $\left\{\mathbf{e}_{i} \mid \mathbf{b}\left(\mathbf{e}_{i},\mathbf{e}_{i}\right)=0\right\}\$ is a basis of ker \(\mathbf{b}\) and the set of linearly independent isotropic vectors is the nullity of \(\mathbf{b}\).

Step-by-step solution

Proof of Kernel

If \(\mathbf{u}\) is part of the kernel of \(\mathbf{b}_{*}\), then the linear map: \(\mathbf{b}_{*}(\mathbf{u})\) applied to any vector \(\mathbf{v}\) equals to zero. \(\mathbf{b}_{*}(\mathbf{u})(\mathbf{v}) = \mathbf{b}(\mathbf{u}, \mathbf{v}) = 0\). As \(\mathbf{v}\) was arbitrary, this means \(\mathbf{b}(\mathbf{u}, \mathbf{v}) = 0\) for all \(\mathbf{v} \in \mathbb{V}\). Hence every vector \(\mathbf{u}\) in the kernel has this property for all \(\mathbf{v}\) in \(\mathbb{V}\). Conversely if \(\mathbf{u}\) in \(\mathbb{V}\) creates \(\mathbf{b}(\mathbf{u}, \mathbf{v}) = 0\) for all \(\mathbf{v}\), this means the map \(\mathbf{b}_{*}(\mathbf{u})\) equals to zero for any vector, hence \(\mathbf{u}\) is in the kernel. Hence the kernel includes all vectors with this property.

Proof of Orthonormal basis

Let's denote \(\mathcal{I}\) to be the set of \(\mathbf{e}_{i}\) such that \(\mathbf{b}(\mathbf{e}_{i}, \mathbf{e}_{i}) = 0\). If \(\mathbf{u}\) is a linear combination of the basis vectors \(\mathbf{e}_{i}\) in \(\mathcal{I}\), then it will be in the kernel of \(\mathbf{b}_{*}\) since \(\mathbf{b}(\mathbf{u}, \mathbf{e}_{j}) = 0\) for all \(\mathbf{e}_{j}\). Hence \(\mathcal{I}\) is a subset of the kernel. Precisely, \(\mathcal{I}\) is a basis of it, because it consists of linearly independent isotropic vectors. As the kernel is always a subspace, \(\mathcal{I}\) will also be a subspace and therefore the set of linearly independent isotropic vectors is the nullity of \(\mathbf{b}\), concluding the proof.

Key concepts

Kernel of a Bilinear Form

Understanding the kernel of a bilinear form is essential in linear algebra as it helps identify the set of vectors that map to zero under the bilinear form. For a symmetric bilinear form \( \mathbf{b} \), defined on a vector space \( \mathcal{V} \) over a field, the kernel consists of all vectors \( \mathbf{u} \) such that \( \mathbf{b}(\mathbf{u}, \mathbf{v}) = 0 \) for every vector \( \mathbf{v} \) in \( \mathcal{V} \).

This concept is analogous to finding the null space in matrix algebra, where the kernel consists of all vectors that result in the zero vector after being transformed by a given matrix. The kernel captures the idea of 'invisible' effects in transformations, being the collection of elements that completely 'disappear' under the bilinear map. The importance of this subset of vectors cannot be overstated, as it provides insights into the structure and dimensions of vector spaces and their mappings.

Isotropic Vectors

In the context of bilinear forms, isotropic vectors play a pivotal role. An isotropic vector \( \mathbf{u} \) is one that, when evaluated against itself under a bilinear form \( \mathbf{b} \), yields zero: \( \mathbf{b}(\mathbf{u}, \mathbf{u}) = 0 \).

When dealing with a symmetric bilinear form, isotropic vectors are particularly interesting as they reveal 'self-orthogonal' properties. This means that each isotropic vector is orthogonal to itself with respect to the bilinear form, which is a unique characteristic compared to standard vector dot products where only the zero vector is orthogonal to itself. Understanding the behavior of these vectors can lead to valuable insights in studying the geometry and invariants of the provided form, specifically in the realms of quadratic forms and differential geometry.

Orthonormal Basis

Establishing an Orthonormal Framework

An orthonormal basis of a vector space \( \mathcal{V} \) is a set of vectors that are both orthogonal (mutually perpendicular) and normalized (each having unit length) with respect to a given bilinear form \( \mathbf{b} \).

Forming an orthonormal basis is akin to setting up a perfectly aligned coordinate system where measurements can be made with ease and precision. The concept is fundamental in simplifying problems and understanding vector spaces more intuitively since vectors in such a basis are non-interfering, serving as the simplest 'building blocks' of the space. This perspective not only provides clarity but also facilitates computations such as projections, rotations, and determining vector components.

Nullity of a Bilinear Form

The nullity of a bilinear form is a measure similar to the concept of dimension but focuses on the volume of the kernel rather than the whole space. Specifically, the nullity of a symmetric bilinear form \( \mathbf{b} \) is the dimension of the kernel of \( \mathbf{b} \), which is the subspace containing all vectors that \( \mathbf{b} \) maps to zero. The nullity thus quantifies the 'size' of the space wherein \( \mathbf{b} \) exhibits trivial action.

Understanding nullity is critical; it dovetails with rank-nullity theorems, impacting the understanding of linear transformations and matrices. In practical terms, the nullity can reveal symmetries and conservation laws within physical systems, providing a quantitative tool to probe into the nature of interactions and invariant properties.