Question

Use the various characterizations of orthogonal transformations and orthogonal matrices. Find the matrix of an orthogonal projection. Use the properties of the transpose. Which of the matrices in Exercise 1 through 4 are orthogonal? $\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}2& -2& 1\\ 1& 2& 2\\ 2& 1& -2\end{array}\right]$.

Short answer

The Matrix$\frac{1}{3}\left[\begin{array}{ccc}2& -2& 1\\ 1& 2& 2\\ 2& 1& -2\end{array}\right]$ is orthogonal.

Step-by-step solution

Definition of Orthogonal.

A square matrix is orthogonal matrix if $\mathbf{A}\mathbf{\times }{\mathbf{A}}^{\mathbf{T}}\mathbf{=}\mathbf{I}\mathbf{.}$

Verification whether the given matrix is orthogonal.

Let the given matrix is $A=\frac{1}{3}\left[\begin{array}{ccc}2& -2& 1\\ 1& 2& 2\\ 2& 1& -2\end{array}\right]$which is written as$A=\left[\begin{array}{ccc}\frac{2}{3}& \frac{-2}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& \frac{-2}{3}\end{array}\right]$ .

Then, the transpose of the given matrix is$A^T=\left[\begin{array}{ccc}\frac{2}{3}& \frac{1}{3}& \frac{2}{3}\\ \frac{-2}{3}& \frac{2}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}& \frac{-2}{3}\end{array}\right]$ .

Thus, for orthogonal matrix,

$$\begin{gathered} A\times A^T=\left[\begin{array}{ccc}\frac{2}{3}& \frac{-2}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& \frac{-2}{3}\end{array}\right]\times \left[\begin{array}{ccc}\frac{2}{3}& \frac{1}{3}& \frac{2}{3}\\ \frac{-2}{3}& \frac{2}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}& \frac{-2}{3}\end{array}\right] \\ =\left[\begin{array}{ccc}\frac{4}{9}+& \frac{4}{9}+& \frac{1}{9}\\ \frac{2}{9}-& \frac{4}{9}+& \frac{2}{9}\\ \frac{4}{9}-& \frac{2}{9}-& \frac{2}{9}\end{array}\begin{array}{ccc}\frac{2}{9}-& \frac{4}{9}+& \frac{2}{9}\\ \frac{1}{9}+& \frac{4}{9}+& \frac{4}{9}\\ \frac{2}{9}+& \frac{2}{9}-& \frac{4}{9}\end{array}\begin{array}{ccc}\frac{4}{9}-& \frac{2}{9}-& \frac{2}{9}\\ \frac{2}{9}+& \frac{2}{9}-& \frac{4}{9}\\ \frac{4}{9}+& \frac{1}{9}+& \frac{4}{9}\end{array}\right] \\ =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

Since, the matrix satisfies orthogonal condition$A\times A^T=I$ .

Hence,$A=\frac{1}{3}\left[\begin{array}{ccc}2& -2& 1\\ 1& 2& 2\\ 2& 1& -2\end{array}\right]$ is an orthogonal matrix.