Question

Let \(\omega \in \mathcal{L}(\mathcal{V})\) be defined by \(\omega(\mathbf{a})=-\mathbf{a}\) for all \(\mathbf{a} \in \mathcal{V}\). Is \(\omega\) an involution of \(V ?\) Now suppose that \(V\) is an algebra. Is \(\omega\) so defined an involution of the algebra \(V\) ? Recall that an involution of an algebra must be a homomorphism of that algebra.

Short answer

Given \( \omega: \mathbf{a} \mapsto -\mathbf{a} \), \( \omega \) is an involution of the vector space \( V \), but not an involution of the algebra \( V \).

Step-by-step solution

Checking Involution over Vector Space

An involution is a function that is its own inverse. Hence, to check involution over the vector space \( V \), let's see if applying \( \omega \) twice to any vector \( \mathbf{a} \) in \( V \) returns \( \mathbf{a} \). As per the given exercise, \( \omega(\mathbf{a}) = -\mathbf{a} \). Applying \( \omega \) again, \( \omega(\omega(\mathbf{a})) = \omega(-\mathbf{a}) = -(-\mathbf{a}) = \mathbf{a} \). Hence, \( \omega \) is indeed an involution of the vector space \( V \).

Checking Involution over Algebra

As the exercise states, an involution of an algebra must also be a homomorphism of that algebra. Homomorphism preserves the operation, meaning if \( V \) is an algebra with multiplication denoted as '•', then \( \omega(\mathbf{a} • \mathbf{b}) = \omega(\mathbf{a}) • \omega(\mathbf{b}) \) for all \( \mathbf{a}, \mathbf{b} \) in \( V \). Using the given definition of \( \omega \), \( \omega(\mathbf{a} • \mathbf{b}) = -(\mathbf{a} • \mathbf{b}) \), whereas \( \omega(\mathbf{a}) • \omega(\mathbf{b}) = -\mathbf{a} • -\mathbf{b} = \mathbf{a} • \mathbf{b} \). Clearly, these two expressions are not equal for all \( \mathbf{a}, \mathbf{b} \) in \( V \). Hence, \( \omega \) is not an involution of the algebra \( V \).

Key concepts

Involution of Vector Space

In the realm of linear algebra, the concept of an involution refers to a function that when applied twice returns the initial input. It's akin to looking in a mirror where your reflection also holds a mirror, creating a reflection of a reflection, which brings you back to the original image.

For example, if we have a function \( \omega \), it is an involution if \( \omega(\omega(\mathbf{a})) = \mathbf{a} \) for all vectors \( \mathbf{a} \) in the vector space \( \mathcal{V} \). In simpler terms, applying \( \omega \) twice should reverse any changes made during the first application.

In the given exercise, the function \( \omega(\mathbf{a}) = -\mathbf{a} \) has been defined. Upon applying \( \omega \) twice to any vector \( \mathbf{a} \) (i.e., \( \omega(\omega(\mathbf{a})) \)), we end up with the original vector \( \mathbf{a} \) because a double negation cancels out. This characteristic perfectly aligns with our definition of involution in a vector space. Hence, we confirm that \( \omega \) is indeed an involution of the vector space \( \mathcal{V} \).

Homomorphism

In algebra, the term 'homomorphism' signifies a structure-preserving map between two algebraic structures, such as groups, rings, or in our case, algebras. This concept is analogous to a translator who accurately converts a message from one language to another without altering its meaning.

For a function to be considered a homomorphism, it must conserve the algebraic operations of the structures it connects. This means that an operation performed before the function is applied should yield the same result as performing the function on each operand before the operation.

In mathematical notation, a function \( \phi \) is a homomorphism if \( \phi(x \circ y) = \phi(x) \circ \phi(y) \) for a given operation \( \circ \). The problem in the exercise checks whether our involution \( \omega \) also behaves as a homomorphism for the algebra \( \mathcal{V} \) with its algebraic operations. We find that while \( \omega \) is an involution for the vector space, it doesn't satisfy the homomorphism property for the algebraic operation of multiplication, as the negative signs do not cancel out. As a result, \( \omega \) cannot be considered a homomorphism for the algebra, highlighting the nuanced difference between these terms.

Algebraic Operations

When discussing algebraic structures, the term 'algebraic operations' refers to the fundamental operations such as addition, subtraction, multiplication, and sometimes division that are defined within that structure. It's similar to the basic rules of a board game that dictate how players can move their pieces.

In our context, an algebra is a vector space that comes with an additional multiplication operation. This operation adheres to certain properties, like associativity and distributivity, to ensure consistency within the algebraic framework.

The involution \( \omega \) from the exercise negates vectors in the vector space \( \mathcal{V} \). While this operation works on individual vectors, it does not translate well when we consider the multiplication operation that comes with \( \mathcal{V} \) being an algebra. The exercise demonstrates that the multiplication of two negated vectors doesn't correspond to the negation of their product, \( -(\mathbf{a} \cdot \mathbf{b}) eq (-\mathbf{a}) \cdot (-\mathbf{b}) \), thus \( \omega \) cannot preserve the algebraic operation of multiplication in this context.

Understanding algebraic operations is crucial, as they are the building blocks of more complex concepts in algebra and provide a foundation for how elements within algebraic systems interact with one another.