- #1

- 55

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- Homework Statement
- Find ##m \in \mathbb{R}## so that the following linear operator in ##\mathbb{R}^3## be an isometry: $$F\left(x,y,z\right)=\left(\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+mz,\frac{-1}{\sqrt{6}}x+\frac{2}{\sqrt{6}}y-\frac{1}{\sqrt{6}}z,-\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}z\right).$$

- Relevant Equations
- If ##F## is an isometry, then ##\left\Vert F\left(u\right)\right\Vert =\left\Vert u\right\Vert ##.

Well, my guess is that there is something wrong with the factors chosen, because ##\left\Vert \left(0,1,0\right)\right\Vert =1## and

\begin{align}

\left\Vert F\left(0,1,0\right)\right\Vert &=\left\Vert \left(\frac{1}{\sqrt{3}}\left(0\right)+\frac{1}{\sqrt{3}}\left(1\right)+m\left(0\right),\frac{-1}{\sqrt{6}}\left(0\right)+\frac{2}{\sqrt{6}}\left(1\right)-\frac{1}{\sqrt{6}}\left(0\right),-\frac{1}{\sqrt{2}}\left(0\right)+\frac{1}{\sqrt{2}}\left(0\right)\right)\right\Vert \\&=\left\Vert \left(\frac{1}{\sqrt{3}},\frac{2}{\sqrt{6}},0\right)\right\Vert \\&=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}}\\&=\sqrt{\frac{1}{3}+\frac{4}{3}}\\&=\sqrt{\frac{5}{3}}.

\end{align}

So, am I wright? I mean, it looks like it was some kind of a typing error.

\begin{align}

\left\Vert F\left(0,1,0\right)\right\Vert &=\left\Vert \left(\frac{1}{\sqrt{3}}\left(0\right)+\frac{1}{\sqrt{3}}\left(1\right)+m\left(0\right),\frac{-1}{\sqrt{6}}\left(0\right)+\frac{2}{\sqrt{6}}\left(1\right)-\frac{1}{\sqrt{6}}\left(0\right),-\frac{1}{\sqrt{2}}\left(0\right)+\frac{1}{\sqrt{2}}\left(0\right)\right)\right\Vert \\&=\left\Vert \left(\frac{1}{\sqrt{3}},\frac{2}{\sqrt{6}},0\right)\right\Vert \\&=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}}\\&=\sqrt{\frac{1}{3}+\frac{4}{3}}\\&=\sqrt{\frac{5}{3}}.

\end{align}

So, am I wright? I mean, it looks like it was some kind of a typing error.