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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.

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

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:

\begin{displaymath}\dim \Ker \phi + \dim \Im \phi = \dim V
\end{displaymath}

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

\begin{displaymath}V=\mathbb{R} ^3 \qquad \text{and} \qquad W=\mathbb{R} ^2.
\end{displaymath}

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

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

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


next up previous contents
Next: Linear transformations and matrices. Up: Linear mappings. Previous: Linear mappings.
Noah Dana-Picard
2001-02-26