Question

Question: Consider an $\mathbf{n}\mathbf{\times }\mathbf{n}$ matrix A. Show that A is an orthogonal matrix if (and only if) A preserve the dot product, meaning that$\left(A\overrightarrow{x}\right)\mathbf{=}\mathbf{.}\left(A\overrightarrow{y}\right)\mathbf{=}\overrightarrow{\mathbf{x}}\mathbf{.}\overrightarrow{\mathbf{y}}$ for allrole="math" localid="1659499729556" $\overrightarrow{\mathbf{x}}$ and$\overrightarrow{\mathbf{y}}$ in ${\mathit{R}}^{\mathbf{n}}$.

Short answer

A is orthogonal if and only if A preserves the dot product.

Step-by-step solution

Consider the theorem.

If $\overrightarrow{\mathbf{v}}\mathbf{}\mathbf{and}\mathbf{}\overrightarrow{\mathbf{w}}$are two (column) vectors in, then

$\begin{array}{cc}\stackrel{\overrightarrow{\mathbf{v}}\mathbf{.}\overrightarrow{\mathbf{w}}}{\mathbf{\uparrow }}& \stackrel{\mathbf{=}{\overrightarrow{\mathbf{v}}}^{\mathbf{T}}\overrightarrow{\mathbf{v}}}{\mathbf{\uparrow }}\\ \mathbf{}& \frac{\mathbf{\left(}\mathbf{dot}\mathbf{\right)}\mathbf{\left(}\mathbf{matrix}\mathbf{\right)}}{\mathbf{product}}\mathbf{}\end{array}$

Observe that if A is orthogonal, then for any $\overrightarrow{x},\overrightarrow{y}\in R^n$

$$\begin{gathered} \left(A\overrightarrow{x}\right).\left(A\overrightarrow{y}\right)={\left(A\overrightarrow{x}\right)}^T\left(A(\overrightarrow{x}\right)\left(theorem5.3.6\right) \\ =\left({\overrightarrow{x}}^T{A}^T\right)\left(A\overrightarrow{y}\right)\left(theorem5.3.9\left(c\right)\right) \\ ={\overrightarrow{x}}^T\left(A^{T}A\right)\overrightarrow{y}\left(summary5.3.8\left(l\right)-\left(lv\right)\right) \\ ={\overrightarrow{x}}^T\left(l_n\right)\overrightarrow{y} \\ ={\overrightarrow{x}}^T\overrightarrow{y}\left(theorem5.3.6\right) \\ =\overrightarrow{x}.\overrightarrow{y} \end{gathered}$$

thus, for any $\overrightarrow{x},\overrightarrow{y}\in R^n$,$\left(A\overrightarrow{x}\right).\left(A\overrightarrow{y}\right)=\overrightarrow{x}.\overrightarrow{y}$

Conversing them

By converse them,

$$\begin{gathered} {||A\overrightarrow{x}||}^2=\left(A\overrightarrow{x}\right).\left(A\overrightarrow{x}\right) \\ =\overrightarrow{x}\overrightarrow{x} \\ ={||\overrightarrow{x}||}^2 \end{gathered}$$

Thus, it is written as $||A\overrightarrow{x}||=||\overrightarrow{x}||$.

Hence, A is orthogonal.