Problem 5 - A Computational Introduction to Number Theory and Algebra

Question

EXERCISE 13.5. Let \(\rho_{i}: M_{i} \rightarrow M_{i}^{\prime},\) for \(i=1, \ldots, k,\) be \(R\) -linear maps. Show that the map $$\begin{aligned} \rho: & \boldsymbol{M}_{1} \times \cdots \times \boldsymbol{M}_{k} \rightarrow \boldsymbol{M}_{1}^{\prime} \times \cdots \times \boldsymbol{M}_{k}^{\prime} \\\ \left(\alpha_{1}, \ldots, \alpha_{k}\right) & \mapsto\left(\rho_{1}\left(\alpha_{1}\right), \ldots, \rho_{k}\left(\alpha_{k}\right)\right) \end{aligned}$$ is an \(R\) -linear map.

Step-by-Step Solution

Answer

By following the step-by-step solution process, we have demonstrated that the combined map \(\rho: M_{1} \times \cdots \times M_{k} \rightarrow M_{1}^{\prime} \times \cdots \times M_{k}^{\prime}\) preserves vector addition and scalar multiplication. Consequently, we can conclude that \(\rho\) is an \(R\)-linear map.

Step 1: Define the combined map, \(\rho\)

The given combined map is defined as: $$\rho: M_{1} \times \cdots \times M_{k} \rightarrow M_{1}^{\prime} \times \cdots \times M_{k}^{\prime}$$ $$\rho(\alpha_1, \ldots, \alpha_k) = (\rho_1(\alpha_1), \ldots, \rho_k(\alpha_k))$$ This map takes a tuple of input elements from the product space \(M_{1} \times \cdots \times M_{k}\) and maps each element to the corresponding element in the product space \(M_{1}^{\prime} \times \cdots \times M_{k}^{\prime}\).

Step 2: Check the vector addition property

To show that the map \(\rho\) preserves vector addition, we need to show that, for any \(\textbf{x}=(x_1,\ldots,x_k)\) and \(\textbf{y}=(y_1,\ldots,y_k)\): $$\rho(\textbf{x}+\textbf{y}) = \rho(\textbf{x}) + \rho(\textbf{y})$$ Adding the two tuples \(\textbf{x}\) and \(\textbf{y}\), we have: $$\textbf{x} + \textbf{y} = (x_1+y_1, \ldots, x_k+y_k)$$ Now, let's compute \(\rho(\textbf{x}+\textbf{y})\): $$\rho(\textbf{x} + \textbf{y}) = \rho(x_1+y_1, \ldots, x_k+y_k) = (\rho_1(x_1+y_1), \ldots, \rho_k(x_k+y_k))$$ Using the linearity of individual \(\rho_i'\)s: $$\rho(\textbf{x}+\textbf{y}) = (\rho_1(x_1)+\rho_1(y_1), \ldots, \rho_k(x_k)+\rho_k(y_k))$$ Now, compute \(\rho(\textbf{x})+\rho(\textbf{y})\): $$\rho(\textbf{x})+\rho(\textbf{y}) = (\rho_1(x_1),\ldots,\rho_k(x_k)) + (\rho_1(y_1),\ldots,\rho_k(y_k)) = (\rho_1(x_1)+\rho_1(y_1), \ldots, \rho_k(x_k)+\rho_k(y_k))$$ Since \(\rho(\textbf{x}+\textbf{y}) = \rho(\textbf{x})+\rho(\textbf{y})\), the vector addition property holds.

Step 3: Check scalar multiplication property

To show that the map \(\rho\) preserves scalar multiplication, we need to show that, for any \(\textbf{x}=(x_1,\ldots,x_k)\) and scalar \(r \in R\): $$\rho(r\textbf{x}) = r\rho(\textbf{x})$$ Consider the tuple \(r\textbf{x}\): $$r\textbf{x} = (rx_1, \ldots, rx_k)$$ Now, compute \(\rho(r\textbf{x})\): $$\rho(r\textbf{x}) = \rho(rx_1, \ldots, rx_k) = (\rho_1(rx_1), \ldots, \rho_k(rx_k))$$ Using the linearity of individual \(\rho_i'\)s: $$\rho(r\textbf{x}) = (r\rho_1(x_1), \ldots, r\rho_k(x_k))$$ Now, compute \(r\rho(\textbf{x})\): $$r\rho(\textbf{x}) = r(\rho_1(x_1),\ldots, \rho_k(x_k)) = (r\rho_1(x_1), \ldots, r\rho_k(x_k))$$ Since \(\rho(r\textbf{x}) = r\rho(\textbf{x})\), the scalar multiplication property holds. As both vector addition and scalar multiplication properties hold for the given map \(\rho\), it is indeed an \(R\)-linear map.

Key Concepts

Vector Addition

When working with vector spaces, vector addition is a fundamental operation. It allows us to combine two vectors to create a new one. For any two vectors, say \( \mathbf{x} = (x_1, x_2, \ldots, x_k) \) and \( \mathbf{y} = (y_1, y_2, \ldots, y_k) \), vector addition means adding the corresponding elements from each vector. The result is another vector:

• The sum of these two vectors is \( \mathbf{x} + \mathbf{y} = (x_1+y_1, x_2+y_2, \ldots, x_k+y_k) \).

This operation ensures that the new vector belongs to the same vector space as the original vectors.
For mappings such as \( \rho \) defined in the exercise, vector addition plays a critical role. To establish that the map is linear, it must respect vector addition. You can see that \( \rho(\mathbf{x} + \mathbf{y}) = \rho(\mathbf{x}) + \rho(\mathbf{y}) \) holds, confirming that the operation preserves the structure of addition in the vector space. This property is vital for the validity of \( R \)-linear maps.

Scalar Multiplication

Scalar multiplication in vector spaces is the process of changing the magnitude of a vector by a scalar, which we often denote by \( r \). When you multiply a scalar \( r \) with a vector \( \mathbf{x} = (x_1, x_2, \ldots, x_k) \), you multiply each component of the vector by that scalar:

• The product is \( r\mathbf{x} = (rx_1, rx_2, \ldots, rx_k) \).

This operation, like vector addition, must be preserved under any linear map. In the context of the map \( \rho \) provided in the exercise, it's crucial that \( \rho(r\mathbf{x}) = r\rho(\mathbf{x}) \).
This equality ensures that the mapping \( \rho \) maintains consistency in scaling vectors. It's one of the two defining properties that confirm \( \rho \) as a linear map. The maintenance of scalar multiplication under the mapping process ensures that linearity is not lost when vectors are scaled, enhancing the predictability and utility of such linear mappings.

R-Linear Maps

An \( R \)-linear map, often simply referred to as a linear map, is a functional construct between two modules over a ring \( R \) that preserves vector addition and scalar multiplication. This means for any linear map \( \rho \), and for all vectors \( \mathbf{x} \), \( \mathbf{y} \) in the domain, and scalar \( r \) in \( R \):

• \( \rho(\mathbf{x} + \mathbf{y}) = \rho(\mathbf{x}) + \rho(\mathbf{y}) \) for vector addition.
• \( \rho(r\mathbf{x}) = r\rho(\mathbf{x}) \) for scalar multiplication.

These properties ensure that the map is consistent with the operations in vector spaces.
In the exercise, you are dealing with a composite \( R \)-linear map \( \rho \) that acts on Cartesian products \( M_1 \times \cdots \times M_k \). It's crucial to verify that this map respects linearity across both vector addition and scalar multiplication. The mapping \( \rho \) is assembled from component maps \( \rho_i \) that are individually \( R \)-linear. This means each part of the product respects the linear operations, confirming the overall linearity of the composite map \( \rho \). Linear maps like these are essential in many areas of mathematics, as they simplify complex relationships between vector spaces.