Tips och lösning till U 13.14b
SamverkanLinalgLIU
Tips 1
Hej 1
Tips 2
Hej 2
Tips 3
Hej 3
Lösning
Enligt Sats 12.14 kan varje godtycklig vektor \displaystyle \boldsymbol{u} skrivas entydigt som en linjärkombination av ON-basen
\displaystyle \underline{\boldsymbol{e}} via
\boldsymbol{u}=(\boldsymbol{u}| \boldsymbol{e}_1)\boldsymbol{e}_1+(\boldsymbol{u} | \boldsymbol{e}_2)\boldsymbol{e}_2
+(\boldsymbol{u} | \boldsymbol{e}_3)\boldsymbol{e}_3+(\boldsymbol{u}
|\boldsymbol{e}_4)\boldsymbol{e}_4,
där talen \displaystyle (\boldsymbol{u}|\boldsymbol{e}_j) , \displaystyle j=1,2,3,4 är koordinaterna till vektorn \displaystyle \boldsymbol{u} i basen \displaystyle \underline{\boldsymbol{e}} .
Vektorn \displaystyle \boldsymbol{u}=(1,0,0,0)^t har koordinaterna
(\boldsymbol{u}|\boldsymbol{e}_1)=\frac{1}{\sqrt2}((1,0,0,0)^t|(1,0,1,0)^t)=\frac{1}{\sqrt2},
(\boldsymbol{u}|\boldsymbol{e}_2)=\frac{1}{\sqrt3}((1,0,0,0)^t|(-1,1,1,0)^t)=-\frac{1}{\sqrt3},
(\boldsymbol{u}|\boldsymbol{e}_3)=\frac{1}{\sqrt{33}} ((1,0,0,0)^t|(-2,-4,2,3)^t)=-\frac{2}{\sqrt{33}},
och
(\boldsymbol{u}|\boldsymbol{e}_4) = \frac{1}{\sqrt{22}} ((1,0,0,0)^t|(1,2,-1,4)^t)=\frac{1}{\sqrt{22}},
Därmed ges \displaystyle \boldsymbol{u} i basen \displaystyle \underline{\boldsymbol{e}} av
\boldsymbol{u}=\frac{1}{\sqrt2}\boldsymbol{e}_1-\frac{1}{\sqrt3}\boldsymbol{e}_2-\frac{2}{\sqrt{33}}\boldsymbol{e}_3+\frac{1}{\sqrt{22}}\boldsymbol{e}_4 =\underline{\boldsymbol{e}} \left(\begin{array}{c}1/\sqrt{2} \\-1/\sqrt{3}\\-2/\sqrt{33}\\ 1/\sqrt{22}\end{array}\right).
