91Ó°ÊÓ

An orthogonal matrix preserves the dot product and hence an orthogonal transformation does the same. For any$\stackrel{âŠ\yen }{v}∈R^n$ we have $\left|\right|L(\stackrel{→}{v})\left|\right|=\left|\right|\stackrel{→}{v}\left|\right|$.

By using the above two observation we have, the angle between two vectors$\stackrel{âŠ\yen }{v}$ and$\stackrel{âŠ\yen }{w}$ is given below.

$$\begin{gathered} ∠\left(\stackrel{r}{v},\stackrel{r}{w}\right)=arc\cos \left(\frac{\stackrel{r}{v},\stackrel{r}{w}}{||\stackrel{r}{v}||.||\stackrel{r}{w}||}\right) \\ =arc\cos \left(\frac{\left(L\stackrel{r}{v}\right),\left(\stackrel{r}{Lw}\right)}{||\stackrel{r}{Lv}||.||\stackrel{r}{Lw}||}\right) \\ =∠\left(\left(\stackrel{r}{Lv}\right),\left(\stackrel{r}{Lw}\right)\right) \end{gathered}$$

Hence, an orthogonal transformation$L:R^n→R^n$ preserves angles.

The converse need not be true. If it just stretches the vectors geometrically the angle will not change but the length will.

For example,

Consider a transformation $T:R^n→R^n,\stackrel{\mathrm{I}}{v}→2\stackrel{\mathrm{I}}{v}$.

For a non-zero vector$\stackrel{âŠ\yen }{v}$.

$$\begin{gathered} ||T\left(\stackrel{r}{v}\right)||=||2\stackrel{r}{n}|| \\ =2||\stackrel{r}{v}|| \\ ≠||\stackrel{r}{v}|| \end{gathered}$$

Therefore, T is not orthogonal.

localid="1659498014544" $$\begin{gathered} ∠\left(\left(T\stackrel{r}{v}\right),\left(T\stackrel{r}{w}\right)\right)=arc\cos \frac{T\left(\stackrel{r}{v}\right).T\left(\stackrel{r}{w}\right)}{||T\left(\stackrel{r}{v}\right)||.||T\left(\stackrel{r}{w}\right)||} \\ =arc\cos \frac{4\stackrel{r}{v}.\stackrel{r}{w}}{4||\stackrel{r}{v}||}||\stackrel{r}{w}|| \\ =arc\cos \frac{\stackrel{r}{v}.\stackrel{r}{w}}{||\stackrel{r}{v}||.||\stackrel{r}{w}||} \\ =∠\left(\stackrel{r}{v}.\stackrel{r}{w}\right) \end{gathered}$$

Consider the diagram below.

Hence, an orthogonal transformation$L:R^n→R^n$preserves angles. And the converse need not to be true.