SamverkanLinalgLIU

Låt \displaystyle \boldsymbol{u}=\underline{\boldsymbol{e}}X_1=\underline{\boldsymbol{e}}\begin{pmatrix}{a_1}\\{b_1}\\{c_1}\end{pmatrix} och \displaystyle \boldsymbol{v}=\underline{\boldsymbol{e}}X_2=\underline{\boldsymbol{e}}\begin{pmatrix}{a_2}\\{b_2}\\{c_2}\end{pmatrix}. Vi behöver summan

\displaystyle \boldsymbol{u}+\boldsymbol{v}=\underline{\boldsymbol{e}}\begin{pmatrix}{a_1}\\{b_1}\\{c_1}\end{pmatrix}+\underline{\boldsymbol{e}}\begin{pmatrix}{a_2}\\{b_2}\\{c_2}\end{pmatrix}=\underline{\boldsymbol{e}}\begin{pmatrix}{a_1+a_2}\\{b_1+b_2}\\{c_1+c_2}\end{pmatrix}

och

<center>\displaystyle \lambda\boldsymbol{u}=\lambda\underline{\boldsymbol{e}}{e}\rvekt{a_1}{b_1}{c_1}=\underline{\boldsymbol{e}}\rvekt{\lambda a_1}{\lambda b_1}{\lambda c_1}.

    Avbildningen \displaystyle G är inte linjär, ty 

<center>\displaystyle 1.\quad G(\boldsymbol{u}+\boldsymbol{v})\neq G(\boldsymbol{u})+G(\boldsymbol{v})\qquad\qquad 2.\quad G(\lambda\boldsymbol{u})\neq\lambda G(\boldsymbol{u}).

    T.ex., följer av~(\ref{C445}) att
    <center>\displaystyle G(\lambda\boldsymbol{u})=G\left(\underline{\boldsymbol{e}}\rvekt{\lambda a_1}{\lambda b_1}{\lambda c_1}\right)
                          =\underline{\boldsymbol{e}}\rvektc{\lambda a_1\cdot\lambda c_1}{\lambda^2b_1^2}{\lambda b_1+\lambda c_1}

=\lambda\underline{\boldsymbol{e}}\rvektc{\lambda a_1c_1}{\lambda b_1^2}{b_1+c_1}\neq \lambda G(\boldsymbol{u}).