Question

\(V\) denotes a finite dimensional inner product space. Show that the following are equivalent for a linear transformation \(S: V \rightarrow V\) where \(V\) is finite dimensional and \(S \neq 0\) : $$ \text { 1. }\langle S(\mathbf{v}), S(\mathbf{w})\rangle=0 \text { whenever }\langle\mathbf{v}, \mathbf{w}\rangle=0 $$ 2\. \(S=a T\) for some isometry \(T: V \rightarrow V\) and some \(a \neq 0\) in \(\mathbb{R}\) 3\. \(S\) is an isomorphism and preserves angles between nonzero vectors. [Hint: Given (1), show that \(\|S(\mathbf{e})\|=\|S(\mathbf{f})\|\) for all unit vectors \(\mathbf{e}\) and \(\mathbf{f}\) in \(V\).]

Short answer

The three statements are equivalent: 1 implies 3, 3 implies 2, and 2 implies 1.

Step-by-step solution

Understand the Problem

To prove the equivalence of the three statements for a linear transformation \( S: V \rightarrow V \), we need to demonstrate that each condition implies the others. Start by considering what each statement entails in the context of a finite-dimensional inner product space \( V \).

Show 1 implies 3

Given statement (1), we have \( \langle S(\mathbf{v}), S(\mathbf{w}) \rangle = 0 \) if \( \langle \mathbf{v}, \mathbf{w} \rangle = 0 \). Consider two unit vectors \( \mathbf{e} \) and \( \mathbf{f} \) that are orthogonal. This implies that \( \|S(\mathbf{e})\| = \|S(\mathbf{f})\| \). Extend this idea to conclude that \( S \) preserves angles, as \( \|S(\mathbf{v})\| = \|S(\mathbf{w})\| \) must hold for all \( \mathbf{v}, \mathbf{w} \). S is injective and preserves angles, making it an isomorphism, which is condition (3).

Show 3 implies 2

Statement (3) tells us \( S \) is an isomorphism and preserves angles. An isomorphism \( S \) that preserves angles must scale vectors by some nonzero scalar \( a \) uniformly. Therefore, there exists an isometry \( T \) such that \( S = aT \), which corresponds to condition (2).

Show 2 implies 1

Statement (2) says \( S = aT \) for some isometry \( T \) and \( a eq 0 \). Since \( T \) is an isometry, it preserves inner products: \( \langle T(\mathbf{v}), T(\mathbf{w}) \rangle = \langle \mathbf{v}, \mathbf{w} \rangle \). Scaling by \( a \) does not affect orthogonality: if \( \langle \mathbf{v}, \mathbf{w} \rangle = 0 \), then \( \langle S(\mathbf{v}), S(\mathbf{w}) \rangle = a^2 \langle T(\mathbf{v}), T(\mathbf{w}) \rangle = 0 \). Thus, condition (1) holds.

Key concepts

Inner Product Space

An inner product space is a vector space equipped with an additional structure called an inner product. This concept is pivotal in studying linear transformations. The inner product is a way of multiplying two vectors to get a scalar. It is represented by the expression \(\langle \mathbf{v}, \mathbf{w} \rangle\).
This operation must satisfy:

• Linearity: \(\langle a\mathbf{v} + b\mathbf{w}, \mathbf{u} \rangle = a\langle \mathbf{v}, \mathbf{u} \rangle + b\langle \mathbf{w}, \mathbf{u} \rangle\)
• Symmetry: \(\langle \mathbf{v}, \mathbf{w} \rangle = \langle \mathbf{w}, \mathbf{v} \rangle\)
• Positive Definiteness: \(\langle \mathbf{v}, \mathbf{v} \rangle > 0\) for any nonzero vector \(\mathbf{v}\)

These properties help in defining concepts like orthogonality (when the inner product is zero). This foundation supports deeper investigations into linear transformations, such as the one in our problem, ensuring the transformations maintain certain properties of the space.

Isometries

Isometries are transformations that preserve distances and, consequently, inner products between vectors. In the context of inner product spaces, an isometry \( T: V \rightarrow V \) satisfies \( \langle T(\mathbf{v}), T(\mathbf{w}) \rangle = \langle \mathbf{v}, \mathbf{w} \rangle \) for all vectors \( \mathbf{v} \) and \( \mathbf{w} \). This property makes isometries crucial for understanding how shapes and angles behave under transformations.
Since isometries do not alter the inner product values, any orthogonality (perpendicularity) in the space is preserved. Thus, these transformations are powerful tools in geometrical and analytical studies, explaining why they play a major role in finding solutions to problems involving linear transformations, like the one tackled in the exercise.

Isomorphism

An isomorphism in linear algebra is a transformation that preserves the overall structure of the vector space. It implies that there is a one-to-one mapping that is both injective (every vector in the image comes from a unique vector in the domain) and surjective (every vector in the codomain is an image of some vector in the domain).
In simpler terms, an isomorphism allows us to "translate" vectors from one vector space into another without losing any structural information. The transformation \( S \) is an isomorphism if it maintains the properties of addition and scalar multiplication. In our exercise, proving that \( S \) is an isomorphism further guarantees it functions uniformly across vector relations and dimensions in the space \( V \), underlining its equivalence to other described properties.

Angle Preservation

Angle preservation is a property of certain linear transformations where the angle between any two vectors does not change under the transformation. This means that if \( \mathbf{v} \) and \( \mathbf{w} \) make an angle \( \theta \), and if \( S \) is the linear transformation, then the angle between \( S(\mathbf{v}) \) and \( S(\mathbf{w}) \) is also \( \theta \).
For a transformation to preserve angles, it must scale vectors in the same proportion – commonly, this implies it is an isometry, potentially multiplied by a nonzero scalar. Because angle preservation connects directly with the equality \( \langle S(\mathbf{v}), S(\mathbf{w}) \rangle = a^2 \langle T(\mathbf{v}), T(\mathbf{w}) \rangle \), it is an integral characteristic when examining if transformations such as \( S \) are equivalent to having isometric properties. Understanding angle preservation is key, as it relates closely to the idea of maintaining the geometric structure of a vector space under transformation.