Kernel and Image of a linear mapping.

Definition 6.2.1   Let V and W be two vector spaces and let $\phi : V \longrightarrow W$ be a linear mapping. The kernel of $\phi$, denoted $\Ker \phi$ is the subset of V formed of the vectors whose image is the zero vector in W.

$\Ker \phi = \{ \overrightarrow{u} \in V; \; \phi ( \overrightarrow{u} ) = \overrightarrow{0} \}$.

Proposition 6.2.2   $\Ker \phi$ is a vector subspace of V.

Definition 6.2.3   Let V and W be two vector spaces and let $\phi : V \longrightarrow W$ be a linear mapping. The image of $\phi$, denoted $\Im \phi$ is the subset of W formed of the vectors having a preimage in V.

$$\Im \phi = \{ \overrightarrow{w}\in W \; \vert \; \exists \ov... ...\in V, \; \phi ( \overrightarrow{v} ) = \overrightarrow{w}\}.$$

Proposition 6.2.4   $\Im \phi$ is a vector subspace of W.

Theorem 6.2.5   Let V and W be two vector spaces of finite dimension and let $\phi : V \longrightarrow W$ be a linear mapping. Then:

$$\dim \Ker \phi + \dim \Im \phi = \dim V$$

Example 6.2.6   Consider the following vector spaces over $\mathbb{R}$:

$$V=\mathbb{R} ^3 \qquad \text{and} \qquad W=\mathbb{R} ^2.$$

For $\overrightarrow{u} =(x,y,z) \in V$, we define $\phi (x,y,z)=(x+y,y-z)$. Then:

$$\Ker \phi = \text{Span} ( (1,-1,1) ) \qquad \text{and} \qquad \Im \phi = \mathbb{R} ^2,$$

thus $\dim \Ker \phi = 1$ and $\dim Im \phi =2$ and this verifies Th.  2.5.