Tips och lösning till U 13.14d
SamverkanLinalgLIU
Tips 1
Se uppgift 13.13
Tips 2
Se uppgift 13.13
Tips 3
Se uppgift 13.13
Lösning
Enligt Sats 12.22 ges varje vektor entydigt av
\boldsymbol{u}=\underbrace{(\boldsymbol{u}|\boldsymbol{e}_1)\boldsymbol{e}_1+(\boldsymbol{u}|\boldsymbol{e}_2)\boldsymbol{e}_2+\boldsymbol{u}|\boldsymbol{e}_3)\boldsymbol{e}_3} _{P_W(\boldsymbol{u})=\boldsymbol{u}_{\parallel W}} \quad + \underbrace{ (\boldsymbol{u}|\boldsymbol{e}_4)\boldsymbol{e}_4}_{P_{W^\perp}(\boldsymbol{u})=\boldsymbol{u}_{\parallel W^\perp}}.
Om \displaystyle \boldsymbol{u}=(1,0,0,0)^t , så är
P_{W}(\boldsymbol{u}) = \boldsymbol{u}_{\parallel W} = (\boldsymbol{u}|\boldsymbol{e}_1)\boldsymbol{e}_1+(\boldsymbol{u}|\boldsymbol{e}_2)\boldsymbol{e}_2 +(\boldsymbol{u}|\boldsymbol{e}_3)\boldsymbol{e}_3=\frac{1}{22} (21,-2,1,-4)^t
och
P_{W^\perp}(\boldsymbol{u}) = \boldsymbol{u}_{\perp W} = (\boldsymbol{u}|\boldsymbol{e}_4)\boldsymbol{e}_4= \frac{1}{22}(1,2,-1,4)^t.
