Hallo Simon,
ich schreibe x,y,z für die indizierten xi
f(\(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\)) = \(\begin{pmatrix} x-y+z \\ 0 \\ y \\ x+z \end{pmatrix}\)
f( k*\(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\)+ \(\begin{pmatrix} u \\ v \\ w \end{pmatrix}\)) = f(\(\begin{pmatrix} kx+u \\ ky+v \\ kz+w \end{pmatrix}\) ) = \(\begin{pmatrix} kx+u-ky-v+kz+w \\ 0 \\ ky+v \\ kx+u+kz+w \end{pmatrix}\)
= \(\begin{pmatrix} kx-ky+kz \\ 0 \\ ky \\ kx+kz \end{pmatrix}\) + \(\begin{pmatrix} u-v+w \\ 0 \\ v \\ u+w \end{pmatrix}\) = f( \(\begin{pmatrix} kx \\ ky \\ kz \end{pmatrix}\)) + f(\(\begin{pmatrix} u \\ v \\ w \end{pmatrix}\))
= k * f(\(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\)) + f(\(\begin{pmatrix} u \\ v \\ w \end{pmatrix}\))
→ f ist eine lineare Abbildung
Kern(f) = { r * \(\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}\) | r∈ℝ }
Bild(f) = { r * \(\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}\) + s * \(\begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}\) + t * \(\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}\) | r,s,t ∈ ℝ }
Bild(f) = { λ * \(\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}\) + μ * \(\begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}\) | λ,μ ∈ ℝ }
Gruß Wolfgang