Question

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be a linear transformation that preserves lengths; that is, \(\left\| {T\left( {\bf{x}} \right)} \right\| = \left\| {\bf{x}} \right\|\) for all x in \({\mathbb{R}^n}\).

1. Show that T also preserves orthogonality; that is, \(T\left( {\bf{x}} \right) \cdot T\left( {\bf{y}} \right) = 0\) whenever \({\bf{x}} \cdot {\bf{y}} = 0\).
2. Show that the standard matrix of T is an orthogonal matrix.

Short answer

1. It is proved that \(T\) preserves orthogonality.
2. It is proved that the standard matrix of T is an orthogonal matrix.

Step-by-step solution

The Pythagoras Theorem

The vectors \({\bf{u}}\) and \({\bf{v}}\) are said to beorthogonal such that if \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\).

Show that T preserves orthogonality

Consider that \({\bf{x}} \cdot {\bf{y}} = 0\). According to the Pythagoras Theorem, \({\left\| {\bf{x}} \right\|^2} + {\left\| {\bf{y}} \right\|^2} = {\left\| {{\bf{x}} + {\bf{y}}} \right\|^2}\). The linear transformation T retains lengths and is linear.

\({\left\| {T\left( {\bf{x}} \right)} \right\|^2} + {\left\| {T\left( {\bf{y}} \right)} \right\|^2} = {\left\| {T\left( {{\bf{x}} + {\bf{y}}} \right)} \right\|^2} = {\left\| {T\left( {\bf{x}} \right) + T\left( {\bf{y}} \right)} \right\|^2}\)

The above equation demonstrates that \(T\left( {\bf{x}} \right)\) and \(T\left( {\bf{y}} \right)\) is orthogonal, according to the Pythagoras Theorem.

Therefore, \(T\) preserves orthogonality.

Hence, it is proved that \(T\) preserves orthogonality.

Show that the standard matrix of T is orthogonal

It is known that thesquare invertible matrix\(U\)is an orthogonal matrix, such that, \({U^{ - 1}} = {U^T}\). This square matrix contains orthonormal columns. The square matrix with orthonormal columns is clearly orthogonal.

The standard matrix of \(T\) is represented by \(\left[ {\begin{array}{*{20}{c}}{T\left( {{e_1}} \right)}& \ldots &{T\left( {{e_n}} \right)}\end{array}} \right]\), where \({e_1}, \ldots ,{e_n}\) are the columns of the identity matrix. Then, the orthonormal set is \(\left[ {\begin{array}{*{20}{c}}{T\left( {{e_1}} \right)}& \ldots &{T\left( {{e_n}} \right)}\end{array}} \right]\) since \(T\) retains both orthogonality and lengths (since the columns of the identity matrix produce an orthonormal set). Furthermore, the square matrix contains orthonormal columns that is known as the orthogonal matrix.

Hence, it is proved that the standard matrix of T is orthogonal.