12.4 Product sapce

1. Suppose \(T: V \to W\) is a function. The graph of \(T\) is defined by \(\text{Graph}(T) = \{ (v, T(v)) \mid v \in V \} \subseteq V \times W\) Prove that \(T\) is a linear map \(\iff \text{Graph}(T)\) is a subspace of \(V \times W.\)

   \(\implies\) Suppose \(T \in L(V, W)\), let’s check that \(\text{Graph}(T)\) is a subspace of \(V \times W.\)

   Closed by addition: If \((v, T(v)), (w, T(w)) \in \text{Graph}(T)\), \((v, T(v)) + (w, T(w)) = (v + w, T(v) + T(w)) \quad \text{(since T is linear)}.\) Thus, \((v + w, T(v + w)) \in \text{Graph}(T).\)

   Closed by scalar multiplication: If \((v, T(v)) \in \text{Graph}(T), \lambda \in \mathbb{F},\) \(\lambda (v, T(v)) = (\lambda v, \lambda T(v)) \quad \text{(since T is linear)}.\) Thus, \((\lambda v, T(\lambda v)) \in \text{Graph}(T).\)

   So \(\text{Graph}(T)\) is a subspace.

   \(\impliedby\) Suppose \(\text{Graph}(T)\) is a subspace of \(V \times W.\)

   Since \(\text{Graph}(T)\) is closed by addition, if \((v, T(v)), (w, T(w)) \in \text{Graph}(T),\) \((v, T(v)) + (w, T(w)) = (v + w, T(v) + T(w)).\) By definition of \(\text{Graph}(T), T(v) + T(w) = T(v + w).\)

   Since \(\text{Graph}(T)\) is closed by scalar multiplication, if \(\lambda \in \mathbb{F}\) and \((v, T(v)) \in \text{Graph}(T),\) \(\lambda (v, T(v)) = (\lambda v, \lambda T(v)).\) By definition of \(\text{Graph}(T), \lambda T(v) = T(\lambda v).\)

   With \((1)\) and \((2), T \in L(V, W).\)

2. Suppose \(W_1, \dots, W_n\) are vector spaces. Prove that \(L(V, W_1 \times \cdots \times W_n)\) and \(L(V, W_1) \times \cdots \times L(V, W_n)\) are isomorphic vector spaces.

   Consider the projections: \(\Pi_k: W_1 \times \cdots \times W_n \to W_k\), defined by \(\Pi_k(w_1, \dots, w_n) = w_k \quad \text{for } k = 1, \dots, n.\) Clearly, \(\Pi_k\) is a linear map.

   Now we define \(\mathcal{I}: L(V, W_1 \times \cdots \times W_n) \to L(V, W_1) \times \cdots \times L(V, W_n)\) by \(\mathcal{I}(\phi) = (\Pi_1 \circ \phi, \dots, \Pi_n \circ \phi).\) This is well-defined since if \(\phi: V \to W_1 \times \cdots \times W_n\), \(\Pi_k \circ \phi: V \to W_k\) is a composition of linear maps.

   Let us check that \(\mathcal{I}\) is an isomorphism.

   • \(\mathcal{I}\) is a linear map: Let \(\phi_1, \phi_2 \in L(V, W_1 \times \cdots \times W_n)\), \(\lambda \in \mathbb{F}\), \(\mathcal{I}(\phi_1 + \lambda \phi_2) = (\Pi_1 \circ (\phi_1 + \lambda \phi_2), \dots, \Pi_n \circ (\phi_1 + \lambda \phi_2)) = (\Pi_1 \circ \phi_1 + \lambda \Pi_1 \circ \phi_2, \dots, \Pi_n \circ \phi_1 + \lambda \Pi_n \circ \phi_2).\) Thus, \(\mathcal{I}(\phi_1 + \lambda \phi_2) = (\Pi_1 \circ \phi_1, \dots, \Pi_n \circ \phi_1) + \lambda (\Pi_1 \circ \phi_2, \dots, \Pi_n \circ \phi_2) = \mathcal{I}(\phi_1) + \lambda \mathcal{I}(\phi_2).\) Therefore, \(\mathcal{I}\) is a linear map.
   • \(\mathcal{I}\) is injective: It is enough to check \(\text{Null}(\mathcal{I}) = \{ 0 \}.\) Let \(\phi \in \text{Null}(\mathcal{I})\), so \(\mathcal{I}(\phi) = 0\), i.e., \((\Pi_1 \circ \phi, \dots, \Pi_n \circ \phi) = (0, \dots, 0).\) This implies \(\Pi_k \circ \phi = 0\) for all \(k = 1, \dots, n\), and hence \(0=\Pi_{k}\circ\phi\left(v)=\Pi_{k}(w_1,\dots,w_{n}\right)=w_{k}\) since \(\phi(v)=(w_1,...,w_n)\)

   Thus, \(\phi\left(v\right)=0\Rightarrow\phi=0\), so \(\text{Null}(\mathcal{I}) = \{ 0 \}\), which means \(\mathcal{I}\) is injective. * \(\mathcal{I}\) is surjective: Let \((\psi_1, \dots, \psi_n) \in L(V, W_1) \times \cdots \times L(V, W_n).\) We want to find \(\phi \in L(V, W_1 \times \cdots \times W_n)\) such that \(\mathcal{I}(\phi) = (\psi_1, \dots, \psi_n).\)

   Define \(\phi: V \to W_1 \times \cdots \times W_n\) by \(\phi(v) = (\psi_1(v), \dots, \psi_n(v)).\)

   Check: \(\phi\) is a linear map since \(\psi_k\) are linear maps. And \(\mathcal{I}(\phi) = (\Pi_1 \circ \phi, \dots, \Pi_n \circ \phi) = (\psi_1, \dots, \psi_n).\)

   Since \(\mathcal{I}\) is linear, injective, and surjective, it is an isomorphism.

Definition: \(\Pi: V \to V / U, \Pi(v) = v + U\) is the quotient map.

Remark: \(\Pi\) is a surjective linear map and \(\text{Null}(\Pi) = U.\) Therefore, \(\dim(V / U) = \dim(V) - \dim(U).\)

1. Suppose \(U\) is a subspace of \(V\) such that \(V / U\) is finite-dimensional. Give an isomorphism between \(V\) and \(U \times V / U.\)

   Let \(\{v_1 + U, \dots, v_n + U\}\) be a basis for \(V / U.\) So, \(\forall v \in V\), \(\exists \lambda_1, \dots, \lambda_n \in \mathbb{F}\) such that \(v + U = \lambda_1(v_1 + U) + \dots + \lambda_n(v_n + U) = \sum_{i=1}^n \lambda_i(v_i + U).\)

   By the definition of addition and scalar multiplication in \(V / U\), \(v + U = \left( \sum_{i=1}^n \lambda_i v_i \right) + U.\)

   Since \(v + U = \left( \sum_{i=1}^n \lambda_i v_i \right) + U\), \(v - \sum_{i=1}^n \lambda_i v_i \in U.\)

   Thus, \(v - \sum_{i=1}^n \lambda_i v_i \in U.\)

   Now we can define \(\Phi: V \to U \times V / U\) by \(\Phi(v) = \left(v - \sum_{i=1}^n \lambda_i v_i, \sum_{i=1}^n \lambda_i(v_i + U)\right),\) where \(v + U = \sum_{i=1}^n \lambda_i(v_i + U).\)

   Check: \(\Phi\) is a linear map. Exercise: Verify that \(\Phi\) is an isomorphism.