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Chapter 11: Problem 30

Let \(\mathcal{B}=\left\\{v_{1}, \ldots, v_{n}\right\\}\) be a basis for \(V\). Prove that \(r \in \mathcal{L}(V, W)\) is an isometry if and only if it is bijective and \(\left\langle\tau v_{i}, \tau v_{j}\right\rangle=\left\langle v_{i,} v_{j}\right\rangle\) for all \(i, j\).

Short Answer

Expert verified
To prove that a linear map \(r: V \rightarrow W\) is an isometry if and only if it is bijective and \(\langle r(v_i), r(v_j) \rangle_W = \langle v_i, v_j \rangle_V\), we consider two directions. (Forward) If r is an isometry, it preserves the norm and thus it is injective. Additionally, dimensions of V and W are equal, making r bijective. For the basis elements, we have \(\langle r(v_i), r(v_j) \rangle_W = \langle v_i, v_j \rangle_V\). (Converse) Given r is bijective and preserves the inner product for basis elements, we need to show r preserves the inner product for all elements in V. Using the linearity of r and bilinearity of the inner product, we find that \(\langle r(x), r(y) \rangle_W = \langle x, y \rangle_V\), proving r is an isometry. Therefore, r is an isometry if and only if it is bijective and preserves the inner product for the basis elements.

Step by step solution

01

Definition of Isometry

A linear map \(r: V \rightarrow W\) is called an isometry if it preserves the norm, i.e., for all \(x,y \in V\), we have \(\langle r(x), r(y) \rangle_W = \langle x, y \rangle_V\).
02

Forward Direction: Suppose r is an isometry.

If \(r: V \rightarrow W\) is an isometry, we have that for all \(x, y \in V\), \(\langle r(x), r(y) \rangle_W = \langle x, y \rangle_V\). Now let's check the two conditions given in the problem: 1. Bijective: Since r is an isometry, it must be injective (one-to-one). This is because if we have \(r(x) = r(y)\), we get \(\langle r(x), r(y) \rangle_W = \langle x, y \rangle_V\), which implies \(x=y\). Moreover, since \(V\) and \(W\) are finite-dimensional and have the same dimensions, r must also be surjective (onto), and therefore r is bijective. 2. Preservation of inner product: for a basis \(\mathcal{B}\) of \(V\), we have \(\langle r(v_i), r(v_j) \rangle_W = \langle v_i, v_j \rangle_V\) for all \(i, j\), which was given in the problem statement. Thus, if r is an isometry, it satisfies both the conditions given in the problem.
03

Converse Direction: Suppose r is bijective and preserves the inner product for the basis elements.

If r is bijective and preserves the inner product for the basis elements, we need to show that r is an isometry. In other words, we need to show that r preserves the inner product for all elements in V. Let \(x, y \in V\). Then we can write \(x = \sum_{i=1}^{n} a_{i}v_{i}\) and \(y = \sum_{i=1}^{n} b_{i}v_{i}\), where \(a_{i}\) and \(b_{i}\) are scalar coefficients and \(v_{i}, v_{j} \in \mathcal{B}\). Now, consider the inner product \(\langle r(x), r(y) \rangle_W\): \(\langle r(x), r(y) \rangle_W = \langle r(\sum_{i=1}^{n} a_{i}v_{i}), r(\sum_{i=1}^{n} b_{i}v_{i}) \rangle_W\) Since r is linear, we can rewrite: \(\langle r(x), r(y) \rangle_W = \langle \sum_{i=1}^{n} a_{i}r(v_{i}), \sum_{i=1}^{n} b_{i}r(v_{i}) \rangle_W\) Using bilinearity of the inner product, we can expand the terms: \(\langle r(x), r(y) \rangle_W = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i}b_{j}\langle r(v_{i}), r(v_{j}) \rangle_W\) Since r preserves the inner product for basis elements: \(\langle r(x), r(y) \rangle_W = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i}b_{j}\langle v_{i}, v_{j} \rangle_V\) This simplifies to: \(\langle r(x), r(y) \rangle_W = \langle x, y \rangle_V\) Thus, r preserves the inner product for all elements in V, and so r is an isometry. In conclusion, a linear map r is an isometry if and only if it satisfies the given criteria of being bijective and preserving the inner product for the basis elements.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Map
A linear map is a function between two vector spaces that respects both vector addition and scalar multiplication. If you have two vector spaces, say \( V \) and \( W \), a function \( r: V \rightarrow W \) is a linear map if for any vectors \( x, y \in V \) and any scalar \( c \), the following conditions hold:

  • Linearity of Addition: \( r(x + y) = r(x) + r(y) \)
  • Linearity of Scalar Multiplication: \( r(cx) = c r(x) \)
These properties ensure that the function maintains the structure and operations of the original space while mapping it onto the new space. Linear maps are central to linear algebra as they help in simplifying problems related to vector transformations and changes of basis.
Inner Product Spaces
Inner product spaces are a type of vector space that possess a way to "multiply" vectors to yield a scalar, known as the inner product. This peculiar multiplication allows us to define angles and lengths in the space, turning it into a geometrical entity.

The inner product of two vectors \( x \) and \( y \) in an inner product space \( V \) is denoted by \( \langle x, y \rangle \). Two important properties of an inner product are:

  • Bilinear: it satisfies \( \langle ax + by, z \rangle = a \langle x, z \rangle + b \langle y, z \rangle \) for all vectors and scalars.
  • Symmetric and Positive-definite: \( \langle x, y \rangle = \langle y, x \rangle \) and \( \langle x, x \rangle > 0 \) for all non-zero vectors \( x \).
Understanding inner product spaces is vital because they allow us to extend geometric concepts like orthogonality (perpendicularity) and projection in arbitrary vector spaces.
Bijective Linear Transformation
A bijective linear transformation is a linear map that is both injective (one-to-one) and surjective (onto). Let's dig into these terms:

  • Injective means that different inputs lead to different outputs, ensuring that the map doesn't "collapse" the space too much. In mathematical terms, \( r(x) = r(y) \) implies that \( x = y \).
  • Surjective implies that every possible output in \( W \) is hit by some input in \( V \). This ensures that the map is comprehensive, covering the whole space \( W \).
For a map \( r: V \rightarrow W \) to be bijective, \( V \) and \( W \) generally must have the same dimension. A bijective linear transformation not only confirms a perfect "fit" between two spaces but also allows you to find an inverse transformation, denoted \( r^{-1} \), mapping \( W \) back to \( V \). Understanding bijective transformations is essential in fields such as linear algebra and differential equations, where solving systems and transforming functions often require such precise mappings.

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Most popular questions from this chapter

Let \(V\) be a nonsingular symplectic geometry and let \(T_{v, n}\) be a symplectic transvection. Prove that a) \(T_{\Sigma, a} T_{z, h}=T_{r, a+b}\) b) For any symplectic transformation \(\sigma\), $$ \sigma T_{v, a} \sigma^{-1}=T_{\sigma v, a} $$ c) For \(b \in F^{*}\), $$ T_{k, a}=\tau_{x, a t^{2}} $$ d) For a fixed \(v \neq 0\), the map \(a \mapsto \tau_{v, a}\) is an isomorphism from the additive group of \(F\) onto the group \(\left\\{T_{v, a} \mid a \in F\right\\} \subseteq \operatorname{Sp}(V)\).

Let \(\operatorname{dim}(V)=\operatorname{dim}(W)\). Prove that \(V / \operatorname{rad}(V) \approx W / \operatorname{rad}(W)\) implies \(V \approx W\).

Let \(V\) be a nonsingular orthogonal geometry and let \(\tau \in \mathcal{L}(V)\) be an isometry. a) Show that \(\operatorname{dim}(\operatorname{ker}(\iota-\tau))=\operatorname{dim}\left(\operatorname{im}(\iota-\tau)^{\perp}\right)\). b) Show that \(\operatorname{ker}(t-\tau)=\operatorname{im}(t-\tau)^{\perp}\). How would you describe \(\operatorname{ker}(t-\tau)\) in words? c) If \(\tau\) is a symmetry, what is \(\operatorname{dim}(\operatorname{ker}(t-\tau))\) ? d) Can you characterize symmetries by means of \(\operatorname{dim}(\operatorname{ker}(\iota-\tau))\) ?

Let \(U, W\) be subspaces of a metric vector space \(V\). Show that a) \((U+W)^{\perp}=U^{\perp} \cap W^{\perp}\) b) \((U \cap W)^{\perp}=U^{\perp}+W^{\perp}\)

Let \(V=N \odot S\), where \(N\) is a totally degenerate space. a) Prove that \(N=\operatorname{rad}(V)\) if and only if \(S\) is nonsingular. b) If \(S\) is nonsingular, prove that \(S \approx V / \mathrm{rad}(V)\).
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