Chapter 3: Problem 3
Prove that \(\mathcal{A}^{2}\), the derived algebra of \(\mathcal{A}\), is indeed an algebra.
Short Answer
Expert verified
Yes, given all the described properties, it can indeed be confirmed that \(\mathcal{A}^{2}\) is an algebra.
Step by step solution
01
Establish Properties
First, state the properties that make an object an algebra. For our case, it is sufficient to show that \(\mathcal{A}^{2}\) is closed under addition and scalar multiplication, and that the product in \(\mathcal{A}^{2}\) is bilinear. In particular, we should show that for all \(a, b, c \in \mathcal{A}^{2}\) and any scalar \(r\), the following holds: \(a * (b + c) = a*b + a*c\), \((a + b) * c = a*c + b*c\), and \(r*(a*b) = (r*a)*b = a*(r*b)\).
02
Show Closure Under Addition and Scalar Multiplication
If \(a, b \in \mathcal{A}^{2}\), then the sum \(a + b\) is also in \(\mathcal{A}^{2}\). Likewise, for any scalar \(r\), the product \(r*a\) is in \(\mathcal{A}^{2}\). These results come directly from the fact that \(\mathcal{A}^{2}\) is a subset of \(\mathcal{A}\), and \(\mathcal{A}\) is an algebra.
03
Show Bilinearity
Firstly, for \(a, b, c \in \mathcal{A}^{2}\), we have \(a * (b + c) = a*b + a*c\), because both operations are being performed in \(\mathcal{A}\) and so the product is bilinear. Secondly, for any \(a, b, c \in \mathcal{A}^{2}\), we have \((a + b) * c = a*c + b*c\), and so the product is left-distributive in \(\mathcal{A}^{2}\). Lastly, for any scalar \(r\) and any \(a, b \in \(\mathcal{A}^{2}\)), we have \(r*(a*b) = (r*a)*b = a*(r*b)\), and so this is consistent with scalar multiplication in \(\mathcal{A}\) and so \(\mathcal{A}^{2}\) is closed under scalar multiplication.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Closure Under Addition
In mathematics, the concept of closure under addition is fundamental when considering algebraic structures. An object, like a subset of a vector space or an algebra, is said to be closed under addition if the addition of any two elements within it remains inside the set.
This is a crucial property to prove when trying to establish that a set forms an algebra.
In the case of the derived algebra, denoted as \(\mathcal{A}^2\), we must check that if \(a\) and \(b\) are elements of \(\mathcal{A}^2\), then their sum \(a + b\) is also in \(\mathcal{A}^2\).
This is a crucial property to prove when trying to establish that a set forms an algebra.
In the case of the derived algebra, denoted as \(\mathcal{A}^2\), we must check that if \(a\) and \(b\) are elements of \(\mathcal{A}^2\), then their sum \(a + b\) is also in \(\mathcal{A}^2\).
- This involves showing that the addition operation retains the defined structure and properties of the original set.
- By confirming this property, we ensure that \(\mathcal{A}^2\) indeed preserves the closure under addition, pivotal for algebraic structure.
Scalar Multiplication
Scalar multiplication is an operation on an algebraic set or structure where each element is multiplied by a scalar (a constant value), which could be a real number or other defined unit.
The purpose is to determine whether the set is closed under this operation, meaning any element of the set multiplied by the scalar should still remain within the set.
For \(\mathcal{A}^2\), showing closure under scalar multiplication might look like this: if \(a\) is in \(\mathcal{A}^2\) and \(r\) is any scalar, then \(r * a\) should also be in \(\mathcal{A}^2\).
The purpose is to determine whether the set is closed under this operation, meaning any element of the set multiplied by the scalar should still remain within the set.
For \(\mathcal{A}^2\), showing closure under scalar multiplication might look like this: if \(a\) is in \(\mathcal{A}^2\) and \(r\) is any scalar, then \(r * a\) should also be in \(\mathcal{A}^2\).
- This property confirms that the structure adheres to algebra's rules, maintaining consistency even under external scaling.
- Ensuring closure under scalar multiplication upholds scaling transformations which are vital in numerous applications, such as transformations in geometry and physics.
Bilinearity
Bilinearity is a property that involves two operations being linear in each of the variables separately.
In the context of algebra, it's essential for establishing how operations such as multiplication interact within the structure.
Proving bilinearity in \(\mathcal{A}^2\) involves checking conditions like:
In the context of algebra, it's essential for establishing how operations such as multiplication interact within the structure.
Proving bilinearity in \(\mathcal{A}^2\) involves checking conditions like:
- For any elements \(a, b, c\) in \(\mathcal{A}^2\), \(a * (b + c) = a * b + a * c\).
- It also involves demonstrating that \((a + b) * c = a * c + b * c\), a key aspect indicating left-distributive law.
- For a scalar \(r\) and elements \(a, b\), it is required that \(r * (a*b) = (r*a)*b = a*(r*b)\).