2.2 Image and Kernel
Definition : Let E, F be two K-vector spaces and \( f \in L(E, F) \): The image of \( f \), denoted by \( \text{Im}(f) \), is the vector subspace of \( F \) defined by: \[ \text{Im}(f) = f(E) = \{ y \in F \mid \exists x \in E, \ y = f(x) \} \] Definition: Let E, F be two K-vector spaces and \( f \in L(E, F) \): The kernel of \( f \), denoted by \( \ker(f) \), is the vector subspace of \( E \) defined by: \[ \ker(f) = f^{-1}(\{0_F\}) = \{ x \in E \mid f(x) = 0_F \} \]
Example 2.4 Consider the linear map \[ f : \mathbb{R}^3 \to \mathbb{R}^2 \] defined by \[ (x, y, z) \mapsto (x + y, y - 2z). \] Let us determine the image and the kernel of \( f \): **Kernel of \( f \):** Let \( u = (x, y, z) \in \mathbb{R}^3 \) such that \( f(u) = 0_{\mathbb{R}^2} \). We have: \[ f(u) = 0_{\mathbb{R}^2} \Leftrightarrow (x + y, y - 2z) = (0,0). \] This gives the system: \[ \begin{cases} x + y = 0 \\ y - 2z = 0 \end{cases} \] Solving for \( x \) and \( y \): \[ \begin{cases} x = -y \\ y = 2z \end{cases} \] Thus, \[ u = (x, y, z) = (-2z, 2z, z) = z(-2, 2, 1). \] Therefore, \( \ker(f) \) is the vector subspace of \( \mathbb{R}^3 \) spanned by the vector \( X = (-2, 2, 1) \): \[ \ker(f) = \text{span}(X = (-2, 2, 1)). \] \
Image of \( f \): Let \( u = (x, y, z) \in \mathbb{R}^3 \). We have: \[ f(u) = (x + y, y - 2z) \] Rewriting it in terms of basis vectors: \[ f(u) = (x, 0) + (y, y) + (0, -2z) \] \[ = x(1, 0) + y(1, 1) + z(0, -2). \] Thus, \( \text{Im}(f) \) is the vector subspace of \( \mathbb{R}^2 \) spanned by the vectors \( Y = (1,0) \), \( Z = (1,1) \), and \( T = (0,-2) \): \[ \text{Im}(f) = \text{span}(Y = (1,0), Z = (1,1), T = (0,-2)). \] We observe that \( \text{Im}(f) = \mathbb{R}^2 \).
Proposition 2.3:
Let \( E, F \) be two \( K \)-vector spaces and \( f \in L(E, F) \):
- \( f \) is injective if and only if \( \ker(f) = \{0_E\} \).
- \( f \) is surjective if and only if \( \text{Im}(f) = F \).
\( **Proof** 1) (\(\Rightarrow\)) Suppose that \( f \) is injective. Let \( u \in \ker(f) \). Then, \[ f(u) = 0_F = f(0_E). \] Since \( f \) is injective, it follows that \( u = 0_E \). Thus, \( \ker(f) = \{0_E\} \).
(\(\Leftarrow\)) Suppose that \( \ker(f) = \{0_E\} \). Let \( u, v \in E \) such that \( f(u) = f(v) \). Then, \[ f(u) = f(v) \Leftrightarrow f(u) - f(v) = 0_F \] \[ \Leftrightarrow f(u - v) = 0_F. \] Thus, \[ u - v \in \ker(f) = \{0_E\}. \] It follows that \( u = v \), proving injectivity.
2) \( f \) is surjective if and only if \[ \forall y \in F, \ \exists x \in E \text{ such that } y = f(x). \] \[ \Leftrightarrow f(E) = F \] \[ \Leftrightarrow \text{Im}(f) = F. \] \)
Rank of a Linear Mapping:
Definition:Let \( E \) and \( F \) be two finite-dimensional vector spaces over \( K \), and let \( f \in L(E, F) \). The rank of \( f \), denoted by \( \operatorname{rank}(f) \), is the natural integer defined by: \[ \operatorname{rg}(f) = \dim (\operatorname{Im} f) \]
theorem :(Rank Theorem) Let \( E \) and \( F \) be two finite-dimensional vector spaces over \( K \), and let \( f \in L(E, F) \). We have: \[ \operatorname{rank}(f) = \dim E - \dim (\ker f) \]