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Chapter 2: Problem 15

Let \(X\) and \(Y\) be finite dimensional normed linear spaces with the same dimension, say \(n\). Let \(\left\\{u_{1}, \ldots, u_{n}\right\\}\) be a basis of \(X\) and \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) be a basis of \(Y\). For \(x=\sum_{j=1}^{n} \alpha_{j} u_{j} \in X\), define $$ T(x)=\sum_{j=1}^{n} \alpha_{j} v_{j} $$ Show that \(T: X \rightarrow Y\) is a linear homeomorphism from \(X\) onto \(Y\).

Short Answer

Expert verified
Since \(T\) is linear, bijective, continuous, and has a continuous inverse, it is a linear homeomorphism from \(X\) onto \(Y\).

Step by step solution

01

- Define Linearity

For the map to be linear, it must satisfy additivity and homogeneity. This means for any vectors \(x_1, x_2 \in X\) and any scalar \(c\), the following must hold: 1. \( T(x_1 + x_2) = T(x_1) + T(x_2) \) 2. \( T(cx_1) = cT(x_1) \)
02

- Prove Additivity

Let \( x_1 = \sum_{j=1}^{n} \alpha_{j1} u_{j} \) and \( x_2 = \sum_{j=1}^{n} \alpha_{j2} u_{j} \).Then, \(x_1 + x_2 = \sum_{j=1}^{n} (\alpha_{j1} + \alpha_{j2}) u_{j} \).Applying \(T\), \[ T(x_1 + x_2) = T\left(\sum_{j=1}^{n} (\alpha_{j1} + \alpha_{j2}) u_{j}\right) = \sum_{j=1}^{n} (\alpha_{j1} + \alpha_{j2}) v_{j} \]Simplifies to \[\sum_{j=1}^{n} \alpha_{j1} v_{j} + \sum_{j=1}^{n} \alpha_{j2} v_{j} = T(x_1) + T(x_2) \]Thus, \(T(x_1 + x_2) = T(x_1) + T(x_2)\) holds.
03

- Prove Homogeneity

Let \( x = \sum_{j=1}^{n} \alpha_{j} u_{j} \) and \( c \) be any scalar. Then, \[ T(cx) = T\left(c \sum_{j=1}^{n} \alpha_{j} u_{j}\right) = \sum_{j=1}^{n} (c \alpha_{j}) v_{j} = c \sum_{j=1}^{n} \alpha_{j} v_{j} = cT(x) \]Thus, \( T(cx) = cT(x) \) holds.
04

- Prove Bijectivity

To prove that \( T \) is a bijection, we need to show that \( T \) is both injective and surjective. First, for injectivity: Assume \( T(x) = T(x') \) for \( x, x' \in X \). Then, \[ \sum_{j=1}^{n} \alpha_{j} v_{j} = \sum_{j=1}^{n} \beta_{j} v_{j} \]Since \( \left\{v_{1}, \, \ldots, \, v_{n}\right\} \) is a basis of \( Y \), it follows that \( \alpha_{j} = \beta_{j} \) for all \( j \). Thus, \( x = x' \).For surjectivity: Let \( y \in Y \). Since \( \left\{v_{1}, \, \ldots, \, v_{n}\right\} \) is a basis of \( Y \), there exist scalars \( \beta_{j} \) such that \[ y = \sum_{j=1}^{n} \beta_{j} v_{j} \]Let \( x = \sum_{j=1}^{n} \beta_{j} u_{j} \). Then \( T(x) = y \), proving surjectivity.
05

- Prove Continuity and Open Mapping

A linear map between finite dimensional normed spaces is continuous if and only if it is bounded. Since \(Y\) is finite dimensional, the norm on \(Y\) is equivalent to any other norm. Thus, boundedness and continuity of \(T\) follow from the structure of norms on finite dimensional spaces. Because \(T\) is linear and bijective, for finite dimensional spaces, this also implies that the inverse map \(T^{-1}\) is linear and continuous. Thus, \(T\) is a homeomorphism.

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Key Concepts

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

Finite Dimensional Normed Spaces
A finite dimensional normed space is a vector space equipped with a norm, and it has a finite number of basis vectors. For example, in the space \( \textbf{R}^n \), the standard basis has exactly \( n \) vectors. Finite dimensional spaces have significant properties, such as every norm being equivalent. This means that any two norms on such a space are comparable; if a sequence converges in one norm, it will converge in another.
Basis
A basis is a set of vectors in a vector space that are linearly independent and span the entire space. In the given exercise, the set \( \left\{u_1, u_2, \, ..., \, u_n \right\} \) forms a basis for \( X \), and the set \( \left\{v_1, v_2, \, ..., \, v_n \right\} \) forms a basis for \( Y \). Any vector in \( X \) or \( Y \) can be uniquely expressed as a linear combination of their respective basis vectors. This property is fundamental in defining the linear map \( T \).
Linearity
Linearity is a crucial property of the map \( T \). A function is linear if it satisfies two conditions: additivity and homogeneity. Additivity means \( T(u + v) = T(u) + T(v) \) for any vectors \( u \) and \( v \). Homogeneity means \( T(cu) = cT(u) \) for any scalar \( c \). These properties ensure that the structure of the vector space is preserved under the map. The exercise shows that \( T \) fulfills both conditions, verifying its linearity.
Injectivity
Injectivity, or one-to-one property, ensures that distinct inputs map to distinct outputs. To prove \( T \) is injective, we assume \( T(x) = T(x') \) and show that this implies \( x = x' \). In the exercise, since \( T \) maps \( x = \sum_{j=1}^{n} \alpha_{j} u_{j} \) to \( \sum_{j=1}^{n} \alpha_{j} v_{j} \) and \( \left\{v_1, v_2, \, ..., \, v_n \right\} \) is a basis for \( Y \), it follows that coefficients must match, proving \( x = x' \).
Surjectivity
Surjectivity, or onto property, requires that every element in the codomain \( Y \) is the image of at least one element in the domain \( X \). The exercise demonstrates this by showing for any \( y \in Y \), there is an \( x \in X \) such that \( T(x) = y \). Because \( \left\{v_1, v_2, ..., v_n \right\} \) spans \( Y \), for any \( y \), we can find coefficients such that \( y = \sum_{j=1}^{n} \beta_{j} v_{j} \). This proves the surjectivity of \( T \).
Continuity
Continuity in finite dimensional normed spaces means that small changes in input result in small changes in output. A linear map between finite dimensional normed spaces is continuous if it is bounded. The exercise mentions that due to the finite dimensional nature, the norm equivalences ensure that \( T \) is bounded and thus continuous. This continuity extends to its inverse \( T^{-1} \), confirming that \( T \) is more than a linear bijective map鈥攊t's a linear homeomorphism.
Open Mapping
The Open Mapping Theorem states that a surjective continuous linear map between Banach spaces (complete normed spaces) is an open map, meaning it maps open sets to open sets. Because \( X \) and \( Y \) are finite dimensional, they are complete, making \( T \) and its inverse \( T^{-1} \) continuous. Hence, \( T \) is an open mapping. These properties combined confirm that \( T \) is a linear homeomorphism, a map that is linear, bijective, and bi-continuous.

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Most popular questions from this chapter

Let \(\mathcal{N}\) be the set of all norms on a linear space \(X\) such that \(\sup _{\nu \in \mathcal{N}} \nu(x)<\infty\) for every \(x \in X\). Show that \(x \mapsto \sup _{\nu \in \mathcal{N}} \nu(x)\) is a norm on \(X\).

For every linear space \(X\), does there exist a norm on \(X\) ?

Let \(t_{1}, \ldots, t_{n}\) be distinct points in \([a, b] .\) For \(x \in C[a, b]\), let $$ \nu_{p}=\left\\{\begin{array}{ll} \left(\sum_{j=1}^{n}\left|x\left(t_{j}\right)\right|^{p}\right)^{1 / p} & \text { if } 1 \leq p<\infty \\ \max \left\\{\left|x\left(t_{j}\right)\right|: j=1, \ldots, n\right\\} & \text { if } p=\infty \end{array}\right. $$

Let \(X\) be an inner product space and \(\left(x_{n}\right)\) be a sequence in \(X\). For \(x \in X\), show that \(\left\|x_{n}-x\right\| \rightarrow 0\) as \(n \rightarrow \infty\) whenever \(\left\|x_{n}\right\| \rightarrow\|x\|\) and \(\left\langle x_{n}, x\right\rangle \rightarrow\|x\|^{2}\) as \(n \rightarrow \infty\).

Give an example of a sequence in \(C[a, b]\) to illustrate that there does not exist \(c>0\) such that $$ \|x\|_{\infty} \leq c\|x\|_{1} \quad \forall x \in C[a, b] . $$
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