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A square matrix with real numbers or elements is said to be an orthogonal matrix, if its transpose is equal to its inverse matrix.

In other words, can say, when the product of a square matrix and its transpose gives an identity matrix, then the square matrix is known as an orthogonal matrix.

For example,

Suppose A is a square matrix with real elements and of n x n order and A^Tis the transpose of A. Then according to the definition, if, A^T = A^-1 is satisfied, then,

A A^T = I

No, there is no such that $\mathit{T}\mathbf{:}{\mathit{R}}^{\mathbf{3}}\mathbf{→}{\mathit{R}}^{\mathbf{2}}$,

$T\left[\begin{array}{c}2\\ 3\\ 0\end{array}\right]=\left[\begin{array}{c}3\\ 0\\ 2\end{array}\right]andT\left[\begin{array}{c}-3\\ 2\\ 0\end{array}\right]=\left[\begin{array}{c}2\\ -3\\ 0\end{array}\right]$

An orthogonal matrix preserves the dot product and hence the orthogonal transformation will also preserve.

$Let{\stackrel{→}{v}}_1=\left[\begin{array}{c}2\\ 3\\ 0\end{array}\right]and{\stackrel{→}{v}}_2\left[\begin{array}{c}-3\\ 2\\ 0\end{array}\right]$.

Now, consider the dot product of them.

$$\begin{gathered} {\stackrel{→}{v}}_2.{\stackrel{→}{v}}_1=0 \\ T\left({\stackrel{→}{v}}_1\right).T\left({\stackrel{→}{v}}_2\right)=6 \end{gathered}$$

Since, both are not same and it is not preserving the dot product and therefore, T cannot be orthogonal.

Hence, the transformation T is not orthogonal.