Question

Let \(\rho: M \rightarrow M^{\prime}\) be an \(R\) -linear map. Let \(N\) be a submodule of \(M,\) and let \(\tau: N \rightarrow M^{\prime}\) be the restriction of \(\rho\) to \(N\). Show that \(\tau\) is an \(R\) -linear map and that \(\operatorname{Ker} \tau=\operatorname{Ker} \rho \cap N\).

Short answer

Question: Prove that for a given \(R\)-linear map \(\rho: M \rightarrow M^{\prime}\) and a submodule \(N\) of \(M\), the restriction of \(\rho\) to \(N\) is also an \(R\)-linear map, and the kernel of this restricted map is the intersection of the kernel of \(\rho\) and the submodule \(N\). Solution: We have shown that by defining \(\tau: N \rightarrow M^{\prime}\) as the restriction of \(\rho\) to \(N\) and for any \(n \in N\), \(\tau(n) = \rho(n)\). We then proved that \(\tau\) is an \(R\)-linear map by demonstrating that for all \(x, y \in N\) and \(r \in R\), \(\tau(x+y) = \tau(x) + \tau(y)\) and \(\tau(rx) = r \tau(x)\). Finally, we showed that the kernel of the restricted map \(\tau\) is equal to the intersection of the kernel of \(\rho\) and the submodule \(N\), i.e., \(\operatorname{Ker} \tau=\operatorname{Ker} \rho \cap N\).

Step-by-step solution

Define the restriction of \(\rho\) to \(N\)

Let \(\tau: N \rightarrow M^{\prime}\) be the restriction of \(\rho\) to \(N\), that is, for any \(n \in N\), we have \(\tau(n) = \rho(n)\).

Show that \(\tau\) is \(R\)-linear

To show that \(\tau\) is an \(R\)-linear map, we need to prove that for all \(x, y \in N\) and \(r \in R\), \(\tau(x+y) = \tau(x) + \tau(y)\) and \(\tau(rx) = r \tau(x)\). Let \(x, y \in N\) and \(r \in R\). Then: 1. \(\tau(x+y) = \rho(x+y)\) since \(\tau\) is the restriction of \(\rho\) to \(N\). 2. \(\rho(x+y) = \rho(x) + \rho(y)\) because \(\rho\) is \(R\)-linear. 3. \(\tau(x) + \tau(y) = \rho(x) + \rho(y)\) since \(\tau\) is the restriction of \(\rho\) to \(N\). So, \(\tau\) is \(R\)-linear.

Show that \(\operatorname{Ker} \tau=\operatorname{Ker} \rho \cap N\)

Recall that the kernel of a linear map is the set of all elements that are mapped to the zero element in the codomain. 1. First, let's show that \(\operatorname{Ker} \tau \subseteq \operatorname{Ker} \rho \cap N\). Let \(x \in \operatorname{Ker} \tau\), which means that \(\tau(x) = 0\). We know that \(\tau(x) = \rho(x)\) since \(\tau\) is the restriction of \(\rho\) to \(N\). Therefore, \(\rho(x) = 0\) and \(x \in \operatorname{Ker} \rho\). Moreover, \(x \in N\) by definition of the restriction of \(\rho\) to \(N\). Thus, \(x \in \operatorname{Ker} \rho \cap N\). 2. Now, let's show that \(\operatorname{Ker} \tau \supseteq \operatorname{Ker} \rho \cap N\). Let \(x \in \operatorname{Ker} \rho \cap N\). This means that \(x \in \operatorname{Ker} \rho\) and \(x \in N\). So, \(\rho(x) = 0\). Since \(\tau\) is the restriction of \(\rho\) to \(N\), \(\tau(x) = \rho(x) = 0\), and \(x \in \operatorname{Ker} \tau\). Therefore, \(\operatorname{Ker} \tau=\operatorname{Ker} \rho \cap N\).

Key concepts

Submodules in Linear Algebra

Understanding submodules is crucial when diving into the world of linear algebra. Think of a submodule as a smaller building block within a larger structure.

In the context of an R-module M, a submodule N is a subset that itself forms an R-module when furnished with the same addition and scalar multiplication operations as M. This means that if you pick any elements x and y from N, their sum x+y will also be in N. The same goes for scalar multiplication: for any element x in N and any scalar r from R, the product rx will be in N as well.

Submodules can be thought of as the generalization of the concept of subspaces in vector spaces. Just as subspaces must satisfy certain properties to retain the structure of a vector space, submodules must adhere to the criteria that preserve the structure of an R-module.

Kernel of a Linear Map

Moving on to the kernel of a linear map, this is a concept that serves as a diagnostic tool telling us which elements from the domain get transformed into the zero element of the codomain.

Formally, for a linear map \(\rho: M \rightarrow M'\), the kernel (denoted as \(\operatorname{Ker} \rho\)) is defined as the set of all elements in M that \(\rho\) maps to the zero element in M'. When an element x from M is such that \(\rho(x) = 0\), it's a member of the kernel of \(\rho\).

The kernel gives us deep insights into the structure of the linear map. It is always a submodule of the domain and, in the context of a linear system of equations, the kernel represents the solution space of the homogeneous system. Understanding the kernel helps in analyzing features like injectivity—no non-zero element of the domain will map to the zero element in the codomain if and only if the kernel contains only the zero element.

Linear Map Restriction

Lastly, let's decipher the linear map restriction. In various instances within linear algebra, we're interested in how a particular linear map behaves when we apply it to a subset of its domain.

The restriction of a linear map, say \(\rho\), from a module M to a submodule N, denoted as \(\tau\), is simply \(\rho\)'s behavior observed solely on N. This is akin to narrowing our lens to only capture N within the wider scope of M.

The function \(\tau: N \rightarrow M'\) is now the restricted version of \(\rho\) where for any element n in N, \(\tau(n)\) is equal to \(\rho(n)\). Despite this transformation's apparent simplicity, it holds significance because the restriction \(\tau\) adopts the properties of \(\rho\); namely, it is R-linear when \(\rho\) is R-linear. This corelation preserves the structural integrity of the operation and it's essential in proving related properties such as those involving the kernel, as seen in our exercise.