Bestimmen sie den kern und das bild

Question

Determine the core and the image:

f2: R3->R2 : f2 (x1, x2, x3)=(2x1-x2, 3x1+x3)

Answer 1

Kern:  2x1-x2 = 0  und  3x1+x3 = 0

    offenbar Rang 2, also mit x1=s

alle (  s ;  2s ;   3s )  mit s aus IR 

dim(Kern) = 1 hat zur Folge

Bild = IR^2.

Comments

(  s ;  2s ;   -3s ) oder?

also mit -3s, weil ja 3S +X3 = 0

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Oh ja, hatte das minus vergessen.

Answer 2

the function describes a linear transformation. The corresponding transformation matrix regarding the canoncial basis of \(\mathbb{R}^3\) is:

$$ A_{f_2} = \begin{pmatrix} 2 & -1 & 0 \\ 3 & 0 & 1 \end{pmatrix} $$

Regarding this matrix you can easily deduce that the image of \(f_2\) is \(\mathbb{R}^2\) and that the kernel is the subset \( \left \{ t \cdot \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} | t \in \mathbb{R} \right \} \subset \mathbb{R^3} \).

Regards