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Chapter 17: Problem 3

Let \(\mathcal{H}\) be a Hilbert space, and \(\mathbf{T} \in \mathcal{L}(\mathcal{H})\) an isometry, i.e., a linear operator that does not change the norm of any vector. Show that \(\|\mathbf{T}\|=1\).

Short Answer

Expert verified
\( \|\mathbf{T}\| = 1 \)

Step by step solution

01

Definition of the operator norm

Recall the definition for the operator norm of a bounded linear operator \( \mathbf{T} \): \( \|\mathbf{T}\| = \sup_{\|\mathbf{x}\| \neq 0} \frac{\|\mathbf{T}(\mathbf{x})\|}{\|\mathbf{x}\|} \). The operator norm is the supremum (the least upper bound) of the set of numbers obtained by taking the ratio of the norm of \( \mathbf{T}(\mathbf{x}) \) to the norm \( \mathbf{x} \), where \( \mathbf{x} \) ranges over all nonzero vectors in the Hilbert space \( \mathcal{H} \).
02

Action of an isometry on a vector

Consider an arbitrary nonzero vector \( \mathbf{x} \in \mathcal{H} \). Because \( \mathbf{T} \) is an isometry, it does not change the norm of \( \mathbf{x} \). Therefore, we have \( \|\mathbf{T}(\mathbf{x})\| = \|\mathbf{x}\| \) for every \( \mathbf{x} \in \mathcal{H} \) such that \( \|\mathbf{x}\| \neq 0 \).
03

Application to operator norm definition

Using the fact that \( \|\mathbf{T}(\mathbf{x})\| = \|\mathbf{x}\| \) from Step 2, we substitute into the definition of the operator norm. This gives \( \|\mathbf{T}\| = \sup_{\|\mathbf{x}\| \neq 0} \frac{\|\mathbf{x}\|}{\|\mathbf{x}\|} \). Because the numerator and denominator are the same, this ratio is 1 for all \( \mathbf{x} \in \mathcal{H} \) such that \( \|\mathbf{x}\| \neq 0 \). As the supremum over all such 1s is also 1, we conclude that \( \|\mathbf{T}\| = 1 \).

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Key Concepts

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

Operator Norm
The operator norm is a fundamental concept in functional analysis when dealing with linear operators, especially in the context of Hilbert spaces. This norm essentially measures the "size" of an operator in terms of its action on vectors within the space. For any bounded linear operator \( \mathbf{T} \), the operator norm is defined as:
  • \( \|\mathbf{T}\| = \sup_{\|\mathbf{x}\| eq 0} \frac{\|\mathbf{T}(\mathbf{x})\|}{\|\mathbf{x}\|} \)
Here, the norm of \( \mathbf{T}(\mathbf{x}) \) is divided by the norm of \( \mathbf{x} \), and we take the supremum (or the least upper bound) of these values. What we are doing is essentially stretching the vector by \( \mathbf{T} \) and seeing by how much it can grow in terms of its norm.
When working with an isometry, which preserves the norm, the process simplifies. As seen in the problem, if \( \mathbf{T} \) is an isometry, then for any nonzero vector \( \mathbf{x} \), we have \( \|\mathbf{T}(\mathbf{x})\| = \|\mathbf{x}\| \). This property immediately indicates that the operator norm \( \|\mathbf{T}\| \) will equal 1, as it neither expands nor contracts any vector.
Isometry
An isometry is quite an interesting concept in mathematics, particularly within Hilbert spaces which deal extensively with distance and angle. An operator is called an isometry if it maintains the distance between any two vectors within the space. In simpler terms, applying an isometric operator does not change the length (or norm) of any vector it processes.
In formal terms, for an operator \( \mathbf{T} \) to be an isometry, for every vector \( \mathbf{x} \) in a Hilbert space \( \mathcal{H} \), the following condition must hold:
  • \( \|\mathbf{T}(\mathbf{x})\| = \|\mathbf{x}\| \)
This equality shows that \( \mathbf{T} \) maps each vector \( \mathbf{x} \) to another vector of the same length. This property is crucial because it implies that the operator norm \( \|\mathbf{T}\| \) is \( 1 \), as shown in our step-by-step exercise solution.
Isometries are significant because they preserve geometric properties of vector spaces, making them useful in various applications such as computer graphics and quantum computing, where the shape and size of figures need to be retained.
Linear Operator
Understanding linear operators is essential for mastering concepts in Hilbert spaces. A linear operator on a vector space is a function \( \mathbf{T} \) that satisfies two primary properties: additivity and homogeneity of degree 1.
  • Additivity: \( \mathbf{T}(\mathbf{x} + \mathbf{y}) = \mathbf{T}(\mathbf{x}) + \mathbf{T}(\mathbf{y}) \)
  • Homogeneity: \( \mathbf{T}(c\mathbf{x}) = c\mathbf{T}(\mathbf{x}) \), for any scalar \( c \)
By these two properties, any linear transformation retains the operations of vector addition and scalar multiplication, which is central in Hilbert space theory.
Linear operators are critical because they form the foundation for more complex operations and transformations in mathematical structures. They allow us to map one vector space to another while preserving linear relationships throughout the space. When such operators, like our isometry \( \mathbf{T} \) in the exercise, are also norm-preserving, it further indicates the robustness of linear operations in maintaining structural coherence in a space. Thus, a linear operator in a Hilbert space provides an elegant way to manipulate vectors while retaining their inherent properties.

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Most popular questions from this chapter

(a) Take the inner product of \(|u\rangle=(\mathbf{A}-\lambda \mathbf{1})|v\rangle\) with \(|v\rangle\) and show that for a hermitian \(\mathbf{A}, \operatorname{Im}\langle v \mid u\rangle=-(\operatorname{Im} \lambda)\|v\|^{2}\). Now use the Schwarz inequality to obtain $$\|v\| \leq \frac{\|u\|}{|\operatorname{Im} \lambda|} \Rightarrow \| \mathbf{R}_{\lambda}(\mathbf{A})|u\rangle \| \leq \frac{\|u\|}{|\operatorname{Im} \lambda|} .$$ (b) Use this result to show that $$\|\left(\lambda-\lambda_{j}\right) \mathbf{R}_{\lambda}(\mathbf{A})|u\rangle\left\|\leq\left(1+\left|\frac{\operatorname{Re}\left(\lambda-\lambda_{j}\right)}{\operatorname{Im}\left(\lambda-\lambda_{j}\right)}\right|\right)\right\| u\|=(1+|\cot \theta|)\| u \|$$ where \(\theta\) is the angle that \(\lambda-\lambda_{j}\) makes with the real axis and \(\lambda\) is chosen to have an imaginary part. From this result conclude that \(\mathbf{R}_{\lambda}(\mathbf{A})\) has a simple pole when \(\mathbf{A}\) is hermitian.

Show that for any hermitian operator \(\mathbf{H}\), we have $$\begin{aligned} 4\langle\mathbf{H} x \mid y\rangle=&\langle\mathbf{H}(x+y) \mid x+y\rangle-\langle\mathbf{H}(x-y) \mid x-y\rangle \\\&+i[\langle\mathbf{H}(x+i y) \mid x+i y\rangle-\langle\mathbf{H}(x-i y) \mid x-i y\rangle] . \end{aligned}$$ Now let \(|x\rangle=\lambda|z\rangle\) and \(|y\rangle=|\mathbf{H} z\rangle / \lambda\), where \(\lambda=(\|\mathbf{H} z\| /\|z\|)^{1 / 2}\), and show that $$\|\mathbf{H} z\|^{2}=\langle\mathbf{H} x \mid y\rangle \leq M\|z\|\|\mathbf{H} z\|,$$ where \(M=\max \left\\{|\langle\mathbf{H} z \mid z\rangle| / \mid\|z\|^{2}\right\\}\). Now conclude that \(\|\mathbf{H}\| \leq M\).

Suppose that \(\mathbf{S}\) is a bounded operator, \(\mathbf{T}\) an invertible operator, and that $$\|\mathbf{T}-\mathbf{S}\|<\frac{1}{\left\|\mathbf{T}^{-1}\right\|}$$ Show that \(\mathbf{S}\) is invertible. Hint: Show that \(\mathbf{T}^{-1} \mathbf{S}\) is invertible. Thus, an operator that is "sufficiently close" to an invertible operator is invertible.

Let \(V\) and \(\mathcal{W}\) be finite-dimensional vector spaces. Show that \(\mathbf{T} \in\) \(\mathcal{L}(\mathcal{V}, \mathcal{W})\) is necessarily bounded.

Show that (a) the unit operator is not compact, and that (b) the inverse of a compact operator cannot be bounded.
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