Problem 23 - Differential Equations and Linear Algebra

Question

On \(\mathbb{R}^{2}\), define the operation of addition by $$\left(x_{1}, y_{1}\right) \oplus\left(x_{2}, y_{2}\right)=\left(x_{1} x_{2}, y_{1} y_{2}\right).$$ Do axioms (A5) and (A6) in the definition of a vector space hold? Justify your answer.

Step-by-Step Solution

Answer

The given operation satisfies axiom (A5), as the additive identity is \( (1, 1) \). However, axiom (A6) does not hold, since we cannot find an additive inverse for the vector \( (0, 0) \). Therefore, the given operation does not form a proper vector space according to the definition.

Step 1: A5: Existence of an additive identity

We need to find a vector \( (e_{1}, e_{2}) \) such that for any vector \( (x, y) \), we have \[(x, y) \oplus (e_{1}, e_{2}) = (x, y)\] Using the given operation, this means \[(x e_{1}, y e_{2}) = (x, y)\] This implies that \( e_{1} = 1 \) and \( e_{2} = 1 \). Therefore, the additive identity is \( (1, 1) \), and axiom (A5) holds for the given operation.

Step 2: A6: Existence of an additive inverse

We need to find a vector \( (x_{2}, y_{2}) \) such that for any vector \( (x_{1}, y_{1}) \), we have \[(x_{1}, y_{1}) \oplus (x_{2}, y_{2}) = (1, 1)\] Using the given operation, this means \[(x_{1} x_{2}, y_{1} y_{2}) = (1, 1)\] This implies that \( x_{2} = \frac{1}{x_{1}} \) and \( y_{2} = \frac{1}{y_{1}} \). Therefore, the additive inverse exists for all vectors except for \( (0, 0) \). Since zero is an element of the vector space, we cannot find an additive inverse for it, and axiom (A6) does not hold for the given operation. In conclusion, the given operation satisfies axiom (A5) but does not satisfy axiom (A6), so it does not form a proper vector space according to the definition.

Key Concepts

Additive Identity

In the context of vectors, the concept of additive identity is crucial for forming a vector space. Basically, an additive identity is a special vector that, when added to any vector, returns the original vector unchanged. For instance, in the traditional sense with real numbers, zero acts as this identity element because adding zero to any number results in the same number.

In our exercise, we have a unique addition operation given by \((x_1, y_1) \oplus (x_2, y_2) = (x_1 x_2, y_1 y_2)\). Here, finding the additive identity means determining a vector \((e_1, e_2)\) so that for any vector \((x, y)\), \((x, y) \oplus (e_1, e_2) = (x, y)\).

This leads us to the equations \(x e_1 = x\) and \(y e_2 = y\), which imply \(e_1 = 1\) and \(e_2 = 1\). Therefore, the vector \((1, 1)\) serves as the additive identity in this scenario. It satisfies the condition for all vectors, so this operation fulfills the axiom for the existence of an additive identity.

Additive Inverse

The concept of an additive inverse is another vital trait for defining a vector space. This term refers to a vector that, when added to a given vector, results in the additive identity. Consider real numbers, where for any number \(a\), the additive inverse is \(-a\) because \(a + (-a) = 0\).

In our problem, using the operation \((x_1, y_1) \oplus (x_2, y_2) = (x_1 x_2, y_1 y_2)\), we seek the vector \((x_2, y_2)\) that makes \((x_1, y_1) \oplus (x_2, y_2) = (1, 1)\).

After calculations, we find that \(x_2 = \frac{1}{x_1}\) and \(y_2 = \frac{1}{y_1}\). This means, for any vector, except \((0, 0)\), there exists an additive inverse, allowing us to return to the additive identity \((1, 1)\).

However, a big gap arises here. When \((x_1, y_1) = (0, 0)\), we can't find any \((x_2, y_2)\) to satisfy the equation, due to division by zero. Hence, since every vector must have an inverse, axiom (A6) is violated for this vector operation.

Vector Operations

Vector operations often include addition and scalar multiplication among vectors. Here, we have defined a new addition operation: \((x_1, y_1) \oplus (x_2, y_2) = (x_1 x_2, y_1 y_2)\). This is different from the typical vector addition \((x_1 + x_2, y_1 + y_2)\).

Let's analyze some key points about this operation:

• This operation multiplies corresponding components of two vectors instead of adding them.
• It maintains some properties, like the existence of an additive identity (as seen with \((1, 1)\)).
• However, the lack of a universal additive inverse, particularly failing with \((0, 0)\), leads to problems.

These properties help determine whether a set equipped with an operation can be a vector space. In our case, this unique addition operation limits compliance with all vector space axioms, particularly failing to satisfy axiom (A6). Understanding these details provides insight into how altering vector operations can influence broader mathematical structures.