Question

Let b be a fixed element of an algebra B. Consider the linear transformation \(\mathbf{T}_{b}: \mathcal{B} \rightarrow \mathcal{B}\) given by \(\mathbf{T}_{b}(\mathbf{x})=\mathbf{x} \mathbf{b} .\) Using the dimension theorem. show that if \(\mathcal{B} \mathbf{b}=\mathcal{B}\), then \(\operatorname{ker} \mathbf{T}_{b}=\mathbf{0}\).

Short answer

If the multiplication by \(b\) maps the algebra \(B\) onto itself, this shows that the linear transformation \(T_{b}\) is surjective. By using the Dimension theorem, it can be concluded that \(ker(T_b) = 0\), which means the kernel of the transformation consists only of the zero vector.

Step-by-step solution

Show that the image of the transformation is the entire space B

Given that \(\mathcal{B} \mathbf{b}=\mathcal{B}\), it implies that for any element \(x\) in \(\mathcal{B}\), \(x \cdot b = y\), where \(y\) is also in \(\mathcal{B}\). Therefore, any element in \(\mathcal{B}\) can be expressed as the product of some element in \(\mathcal{B}\) and \(b\). This shows that \(\mathbf{T}_{b}\) is surjective, which means the image of the transformation under \(\mathbf{T}_{b}\) is the entire space \(B\).\n

Apply the Dimension theorem

According to the Dimension theorem, the dimension of the vector space \(B\) (which is the domain of the transformation \(\mathbf{T}_{b}\)) is equal to the dimension of the image under the transformation \(\mathbf{T}_{b}\) and the dimension of the kernel (null space) of \(\mathbf{T}_{b}\). That is, if the dimension of \(B\) is \(n\), it follows that \(n = dim(\mathbf{T}_{b}(B)) + dim(ker(\mathbf{T}_{b}))\).

Prove that the kernel of the transformation is zero

Since, as we discovered in Step 1, the image of the transformation is the entire space \(B\), that means that the dimension of the image of \(T_b\) is \(n\). Substituting this into the equation from the Dimension theorem (from Step 2) gives us \(n = n + dim(ker(T_b))\). The only way this equation is possible is if \(dim(ker(T_b)) = 0\), which implies that the kernel of the transformation consists of the zero vector alone, i.e., \(ker(T_b) = 0\).

Key concepts

Dimension Theorem

The Dimension Theorem is a crucial concept in linear algebra that relates different dimensions involved in a linear transformation. It's a formula that connects the dimensions of the domain, the image, and the kernel (also known as null space) of a transformation. The theorem states that the dimension of the domain such as a vector space, is equal to the sum of the dimension of the image of the transformation, and the dimension of its kernel.

Mathematically, it can be represented as: \[ \text{dim}(V) = \text{dim}(\text{Im}(T)) + \text{dim}(\text{Ker}(T)) \] where:

• \( \text{dim}(V) \) is the dimension of the domain space \( V \).
• \( \text{dim}(\text{Im}(T)) \) is the dimension of the image of \( T \).
• \( \text{dim}(\text{Ker}(T)) \) is the dimension of the kernel of \( T \).

This theorem ensures that every element from the domain either maps to an element in the image or contributes to the kernel. If the kernel has dimension zero, this confirms that \( T \) is injective (one-to-one).

Kernel of Transformation

The Kernel of a Transformation, also known as the null space, is one of the central ideas when analyzing linear transformations. The kernel consists of all elements in the domain that the transformation maps to the zero vector in the codomain.

Whenever you encounter a transformation \( \mathbf{T} \), its kernel can be defined as:\[ \text{Ker}(\mathbf{T}) = \{ \mathbf{x} \in V \mid \mathbf{T}(\mathbf{x}) = \mathbf{0} \} \]

• If \( \text{Ker}(\mathbf{T}) \) contains only the zero vector, it implies that the transformation is injective, meaning there are no non-trivial elements that map to zero.
• If the kernel contains more than just the zero vector, the transformation has multiple elements mapping to zero, indicating non-injectivity.

Understanding the kernel is vital because it shows how much freedom is lost when applying the transformation. For example, in this exercise, the kernel being zero means the transformation retains maximum information.

Surjective Function

A Surjective Function, or onto function, is a concept in which every element in the codomain is mapped to by at least one element in the domain. For linear transformations, being surjective means that the image of the transformation covers the entire codomain.

In the context of this exercise, \( \mathbf{T}_{b} \) being surjective means the image of \( \mathbf{T}_{b} \) is the entire space B. Therefore, for every element \( y \) in B, there is an \( x \) in A such that \( \mathbf{T}_{b}(x) = y \).

• This property is essential for establishing dimension equivalence between the image of \( T \) and the codomain \( B \).
• If the transformation is both injective (from the kernel) and surjective, it is called a bijection.

Knowing whether a function is surjective helps determine the coverage of potential outputs and confirms the transformation's exhaustiveness.

Algebra

Algebra involves a wide range of concepts and structures, essential for solving linear transformation problems. Specifically, within the context of this problem, it is crucial to understand algebraic systems and spaces, like vector spaces.

An algebra in this context is considered a vector space that is closed under multiplication, meaning any two elements in the algebra can be multiplied to give another element within the same algebra. The specific transformation \( \mathbf{T}_{b}(\mathbf{x}) = \mathbf{x} \mathbf{b} \) in this exercise showcases how algebraic structures are used in defining and analyzing transformations.

• Algebra equips us with the tools to handle equations, manipulate expressions, and explore transformations satisfactorily.
• It provides the framework for ensuring the properties like closure, associativity, and distributivity hold, which are crucial for defining and working within vector spaces.

Combining algebraic principles with transformation properties allows us to deeply understand and validate the functionality and outputs of a given transformation.