Tips och lösning till U 13.12e
SamverkanLinalgLIU
Tips 1
Hej 1
Tips 2
Hej 2
Tips 3
Hej 3
Lösning
Ortogonala projektionen \displaystyle P_{W}(\boldsymbol{u}) ges av
\begin{array}{rcl}
P_{W}(\boldsymbol{u})&=&(\boldsymbol{u}|\boldsymbol{e}_1)\boldsymbol{e}_1+(\boldsymbol{u}|\boldsymbol{e}_2)\boldsymbol{e}_2\\
&=&\frac{1}{2}\left(\left(\begin{array}{r}x_1\\x_2\\x_3\\x_4\end{array}\right)\bigg|\left(\begin{array}{r}1\\1\\1\\1\end{array}\right)\right)
\frac{1}{2}\left(\begin{array}{r}1\\1\\1\\1\end{array}\right)
+\frac{1}{2}\left(\left(\begin{array}{r}x_1\\x_2\\x_3\\x_4\end{array}\right)\bigg|\left(\begin{array}{r}1\\-1\\1\\-1\end{array}\right)\right)
\frac{1}{2}\left(\begin{array}{r}1\\-1\\1\\-1\end{array}\right)\\
&=&\frac{1}{2}\left(\begin{array}{rr}x_1+x_3\\x_2+x_4\\x_1+x_3\\x_2+x_4\end{array}\right)\\
\end{array}
Sista uttrycket kan skrivas som en matrisprodukt så att
P_{W}(\boldsymbol{u})=\frac{1}{2}\left(\begin{array}{rrrr}1&0&1&0\\0&1&0&1\\1&0&1&0\\0&1&0&1\end{array}\right)
\left(\begin{array}{r}x_1\\x_2\\x_3\\x_4\end{array}\right).
T.ex., så är
P_{W}\left(\begin{array}{r}0\\4\\4\\0\end{array}\right)
=\frac{1}{2}\left(\begin{array}{rrrr}1&0&1&0\\0&1&0&1\\1&0&1&0\\0&1&0&1\end{array}\right)
\left(\begin{array}{r}0\\4\\4\\0\end{array}\right)
=\left(\begin{array}{r}2\\2\\2\\2\end{array}\right).
