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Orthogonal Transformations#

Definition: Orthogonal Transformation

Let \((V, \langle \cdot, \cdot \rangle)\) be a real inner product space.

An endomorphism \(f: V \to V\) is orthogonal if it preserves the real inner product:

\[\langle f(v), f(w) \rangle = \langle v, w \rangle \qquad \forall v,w \in V\]

Theorem: Length Preservation \(\iff\) Orthogonal Transformation

Let \((V, \langle \cdot, \cdot \rangle)\) be a real inner product space.

An endomorphism \(f: V \to V\) is an orthogonal transformation if and only if it preserves the induced norm:

\[||f(v)|| = ||v|| \qquad \forall v \in V\]
Proof

We need to prove two things:

Proof of (I):

Since \(f\) is an orthogonal transformation, we have

\[\langle f(v), f(v) \rangle = \langle v, v\rangle\]
\[||f(v)||^2 = ||v||^2\]

Since \(||f(v)||\) and \(||v||\) are non-negative by definition, we have:

\[||f(v)|| = ||v||\]

Proof of (II):

We apply the polarization identity to \(\langle f(v), f(w) \rangle\):

\[\langle f(v), f(w) \rangle = \frac{1}{2}(||f(v)+f(w)||^2-||f(v)||^2-||f(w)||^2)\]

Since \(f\) is linear, we have:

\[\langle f(v), f(w) \rangle = \frac{1}{2}(||f(v+w)||^2-||f(v)||^2-||f(w)||^2)\]

Since \(f\) preserves the induced norm, we have:

\[\langle f(v), f(w) \rangle = \frac{1}{2}(||v+w||^2-||v||^2-||w||^2)\]

The right-hand side is just the polarization identity for \(\langle v, w \rangle\):

\[\langle f(v), f(w) \rangle = \frac{1}{2}(||v+w||^2-||v||^2-||w||^2) = \langle v, w \rangle\]

Theorem: Orthonormal Basis to Orthonormal Basis

Let \((V, \langle \cdot, \cdot \rangle)\) be a finite-dimensional real inner product space and let \(f: V \to V\) be an endomorphism.

If \(b_1, \dotsc, b_n\) is an orthonormal basis of \(V\) and \(f\) is an orthogonal transformation, then \(f(b_1), \dotsc, f(b_n)\) is also an orthonormal basis of \(V\).

Conversely, if \(b_1, \dotsc, b_n\) and \(f(b_1), \dotsc, f(b_n)\) are both orthonormal bases of \(V\), then \(f\) is an orthogonal transformation.

Proof

TODO

Theorem: Bijectivity of Orthogonal Transformations

Every orthogonal transformation on a finite-dimensional inner product space is bijective.

Proof

TODO

Theorem: Adjoint as Inverse of Orthogonal Transformation

Let \((V, \langle \cdot, \cdot \rangle)\) be a real inner product space and let \(f: V \to V\) be an orthogonal transformation.

If \(f\) is bijective and its adjoint \(f^{\ast}\) exists, then \(f\)'s inverse is \(f^{\ast}\):

\[f^{-1} = f^{\ast}\]
Proof

TODO

Theorem: Determinants of Orthogonal Transformations

Let \((V, \langle \cdot, \cdot \rangle)\) be a finite-dimensional real inner product space.

If \(f: V \to V\) is an orthogonal transformation, then its determinant is either \(+1\) or \(-1\).

Proof

TODO