Lösning till övning 3

SamverkanLinalgLIU

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Rad 1: Rad 1:
Låt <math>\boldsymbol{u}=\underline{\boldsymbol{e}}X_1=\underline{\boldsymbol{e}}\begin{pmatrix}{a_1}\\{b_1}\\{c_1}\end{pmatrix}</math> och <math>\boldsymbol{v}=\underline{\boldsymbol{e}}X_2=\underline{\boldsymbol{e}}\begin{pmatrix}{a_2}\\{b_2}\\{c_2}\end{pmatrix}</math>.
Låt <math>\boldsymbol{u}=\underline{\boldsymbol{e}}X_1=\underline{\boldsymbol{e}}\begin{pmatrix}{a_1}\\{b_1}\\{c_1}\end{pmatrix}</math> och <math>\boldsymbol{v}=\underline{\boldsymbol{e}}X_2=\underline{\boldsymbol{e}}\begin{pmatrix}{a_2}\\{b_2}\\{c_2}\end{pmatrix}</math>.
Vi behöver summan
Vi behöver summan
-
<center><math>\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}</center></math>
+
<center><math>\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}</math></center>
och
och
<center><math>
<center><math>
Rad 7: Rad 7:
a_1}\\{\lambda b_1}\\{\lambda
a_1}\\{\lambda b_1}\\{\lambda
c_1}\end{pmatrix}.
c_1}\end{pmatrix}.
-
</center></math>
+
</math></center>
Avbildningen <math>G</math> är inte linjär, ty
Avbildningen <math>G</math> är inte linjär, ty
<center><math>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}).</center></math>
<center><math>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}).</center></math>
Rad 13: Rad 13:
<center><math>G(\lambda\boldsymbol{u})=G\left(\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1}\\{\lambda b_1}\\{\lambda c_1}\end{pmatrix}\right)=G\left(\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1}\\{\lambda b_1}\\{\lambda c_1}\end{pmatrix}\right)=\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1\cdot\lambda c_1}\\{\lambda^2b_1^2}\\{\lambda b_1+\lambda c_1}\end{pmatrix}
<center><math>G(\lambda\boldsymbol{u})=G\left(\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1}\\{\lambda b_1}\\{\lambda c_1}\end{pmatrix}\right)=G\left(\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1}\\{\lambda b_1}\\{\lambda c_1}\end{pmatrix}\right)=\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1\cdot\lambda c_1}\\{\lambda^2b_1^2}\\{\lambda b_1+\lambda c_1}\end{pmatrix}
-
=\lambda\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1c_1}\\{\lambda b_1^2}\\{b_1+c_1}\end{pmatrix}\neq\lambda G(\boldsymbol{u}).</center></math>
+
=\lambda\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1c_1}\\{\lambda b_1^2}\\{b_1+c_1}\end{pmatrix}\neq\lambda G(\boldsymbol{u}).</math></center>

Versionen från 15 augusti 2008 kl. 08.41

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

\displaystyle

\lambda\boldsymbol{u}=\lambda\underline{\boldsymbol{e}}\begin{pmatrix}{a_1}\\{b_1}\\{c_1}\end{pmatrix}=\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1}\\{\lambda b_1}\\{\lambda c_1}\end{pmatrix}.

Avbildningen \displaystyle G är inte linjär, ty

\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 att

<center>\displaystyle G(\lambda\boldsymbol{u})=G\left(\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1}\\{\lambda b_1}\\{\lambda c_1}\end{pmatrix}\right)=G\left(\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1}\\{\lambda b_1}\\{\lambda c_1}\end{pmatrix}\right)=\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1\cdot\lambda c_1}\\{\lambda^2b_1^2}\\{\lambda b_1+\lambda c_1}\end{pmatrix}

=\lambda\underline{\boldsymbol{e}}\begin{pmatrix}{\lambda a_1c_1}\\{\lambda b_1^2}\\{b_1+c_1}\end{pmatrix}\neq\lambda G(\boldsymbol{u}).
Den här artikeln är hämtad från http://wiki.sommarmatte.se/wikis/samverkan/linalg-LIU/index.php/L%C3%B6sning_till_%C3%B6vning_3