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QUESTION

Check that the vectors

$(1,-1,2,-1), (-2,2,3,2), (1,2,0,-1), (1,0,0,1)$

form an orthogonal basis for $\textbf{R}^4$. Express each of the
following vectors as a linear combination of these basis elements.

\begin{description}

\item[(a)]
$(1,1,1,1)$

\item[(b)]

$(\sqrt{2},-3\sqrt{2},5\sqrt{2},-\sqrt{2})$

\item[(c)]

$(-\frac{1}{3},\frac{2}{3},-\frac{1}{3},\frac{4}{3})$

\end{description}



ANSWER It is straightforward to check that \textbf{u.v}=0 for each
of the six distinct pairs. Four mutually orthogonal vectors in
$\textbf{R}^4$ must form a basis.

Now if
$\textbf{v}=\lambda_1\textbf{v}_1+\lambda_2\textbf{v}_2+\ldots+
\lambda_n\textbf{v}_n$ where
$\textbf{v}_1,\textbf{v}_2,\ldots,\textbf{v}_n$ is an orthogonal
basis, then

$$\lambda_r=\frac{\textbf{v.v}_r}{\textbf{v}_r.\textbf{v}_r}
=\frac{(\textbf{v.v}_r)}{|\textbf{v}_r|^2},$$

The coefficients only are given below.

\begin{description}

\item[(a)]

$\ds\frac{(1,1,1,1).(1,-1,2,-1)}{7}=\frac{1}{7}$,

$\ds\frac{(1,1,1,1).(-2,2,3,2)}{21}=\frac{5}{21}$,

$\ds\frac{(1,1,1,1).(1,2,0,-1)}{6}=\frac{1}{3}$,

$\ds\frac{(1,1,1,1).(1,0,0,1)}{2}=1$

\item[(b)]

$\ds\frac{(\sqrt{2},-3\sqrt{2},5\sqrt{2},-\sqrt{2}).(1,-1,2,-1)}{7}=\frac{15\sqrt{2}}{7}$,

$\ds\frac{(\sqrt{2},-3\sqrt{2},5\sqrt{2},-\sqrt{2}).(-2.2.3.2)}{21}=\frac{5\sqrt{2}}{21}$,

$\ds\frac{(\sqrt{2},-3\sqrt{2},5\sqrt{2},-\sqrt{2}).(1,2,0,-1)}{6}=\frac{-4\sqrt{2}}{6}=\frac{-2\sqrt{2}}{3}$,

$\ds\frac{(\sqrt{2},-3\sqrt{2},5\sqrt{2},-\sqrt{2}).(1,0,0,1)}{2}=\frac{1}{2}$,

\item[(c)]

$\ds\frac{(-\frac{1}{3},\frac{2}{3},-\frac{1}{3},\frac{4}{3}).(1,-1,2,-1)}{7}=-\frac{3}{7}$,

$\ds\frac{(-\frac{1}{3},\frac{2}{3},-\frac{1}{3},\frac{4}{3}).(-2,2,3,2)}{21}=\frac{11}{63}$,

$\ds\frac{(-\frac{1}{3},\frac{2}{3},-\frac{1}{3},\frac{4}{3}).(1,2,0,-1)}{6}=-\frac{1}{18}$,

$\ds\frac{(-\frac{1}{3},\frac{2}{3},-\frac{1}{3},\frac{4}{3}).(1,0,0,1)}{2}=\frac{1}{2}$,

\end{description}



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