Mathematics-Online course: Linear Algebra-Matrices-Linear Maps-Linear Maps

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	Mathematics-Online course: Linear Algebra - Matrices - Linear Maps
	Linear Maps

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Let be -vector spaces. A map $\alpha: V \longmapsto W$ is called linear map, if for all $v_1,v_2,v \in V$ and $\lambda \in K$ the following holds true:

(i): $\alpha(v_1+v_2)=\alpha(v_1)+\alpha(v_2)$
(ii): $\alpha(\lambda v)=\lambda \alpha(v)$

As illustrated in the following figure, a rotation is a linear mapping.

$\includegraphics[width=.4\linewidth]{b_drehung_bild1}$ $\includegraphics[width=.4\linewidth]{b_drehung_bild2}$

The sum and the vectors form a triangle the shape of which remains unchanged under rotation. That is, it does not matter whether we form the sum before or after rotation. It is evident that a dilatation by factor $\lambda$ commutes with any rotation.

In an anologous way we can illustrate that a reflection is also a linear mapping.

$\includegraphics[width=.4\linewidth]{b_spiegelung_bild1}$ $\includegraphics[width=.4\linewidth]{b_spiegelung_bild2}$

However, a translation of points in the plane is not linear. Neither additivity nor homogeneity are satisfied. For example, for

$\displaystyle \alpha:\ (x_1,x_2)\mapsto (x_1+1,x_2)$

and

$\displaystyle v_1 = (1,0),\, v_2 = (0,1),\,\lambda=2$

we have

$\displaystyle \alpha(v_1+v_2) = (2,1)$	$\displaystyle \ne$	$\displaystyle (3,1) = \alpha(v_1) + \alpha(v_2)$
$\displaystyle \alpha(\lambda v) = (3,0)$	$\displaystyle \ne$	$\displaystyle (4,0) = \lambda \alpha(v)\, .$

(Authors: Burkhardt/Höllig/Hörner)

The following table provides some examples of mappings of real-valued functions. For each mapping it is indicated which of the two conditions of linearity is/are satisfied.

Funktion	additivity	homogeneity
$f\mapsto f^\prime$	X	X
$f\mapsto \vert f\vert$	-	-
$f\mapsto \int_0^1 f$	X	X
$f\mapsto \max f$	-	-
$f\mapsto f(0)$	X	X
$f\mapsto (\max f+\min f)/2$	-	X

Observe mapping $\alpha$ which assigns to each complex-valued function the corresponding real part. This mapping is additive but not homogeneous. Here we have $\alpha(\mathrm{i}f)\neq \mathrm{i}\alpha(f)$ .

(Authors: Burkhardt/Höllig/Hörner)

automatically generated 4/21/2005