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Chapter 6: Problem 10

Let \(\mathrm{T}\) be a self-adjoint operator on a finite-dimensional inner product space \(\mathrm{V}\). Prove that for all \(x \in \mathrm{V}\) $$ \|\mathrm{T}(x) \pm i x\|^{2}=\|\mathrm{T}(x)\|^{2}+\|x\|^{2} . $$ Deduce that \(\mathrm{T}-i \mathrm{l}\) is invertible and that the adjoint of \((\mathrm{T}-i \mathrm{I})^{-1}\) is \((\mathrm{T}+i \mathrm{I})^{-1}\).

Short Answer

Expert verified
In summary, we have proved the equality \(\|\mathrm{T}(x) \pm i x\|^{2} = \|\mathrm{T}(x)\|^{2}+\|x\|^{2}\) for a self-adjoint operator \(T\) acting on a finite-dimensional inner product space \(V\). We have also deduced that the operator \(T - iI\) is invertible, and its adjoint of its inverse is given by (\(T + iI\))^{-1}.

Step by step solution

01

Expanding the norm

We want to show that for all \(x \in \mathrm{V}\), \(\|\mathrm{T}(x) \pm i x\|^{2}=\|\mathrm{T}(x)\|^{2}+\|x\|^{2}\). Using the definition of the norm and the properties of the inner product, we can rewrite the left-hand side as follows: \(\|\mathrm{T}(x) \pm i x\|^{2} = \langle\mathrm{T}(x) \pm i x, \mathrm{T}(x) \pm i x\rangle\)
02

Using the property of self-adjointness

Recall that \(T\) is self-adjoint, which means that \(\langle\mathrm{T}(x),y\rangle = \langle x,\mathrm{T}(y)\rangle\) for all \(x, y \in V\). Using this property and expanding the inner product, we obtain: \(\langle\mathrm{T}(x) \pm i x, \mathrm{T}(x) \pm i x\rangle = \langle\mathrm{T}(x),\mathrm{T}(x)\rangle \pm i\langle\mathrm{T}(x),x\rangle \mp i\langle x,\mathrm{T}(x)\rangle + \langle x,x\rangle\) But we can use the property of self-adjointness once again to simplify this expression: \(\langle\mathrm{T}(x),\mathrm{T}(x)\rangle \pm i\langle\mathrm{T}(x),x\rangle \mp i\langle x,\mathrm{T}(x)\rangle + \langle x,x\rangle = \langle\mathrm{T}(x),\mathrm{T}(x)\rangle + \langle x,x\rangle\)
03

Relating the expanded expression to the given equality

Now, using the definition of the norm, we can rewrite the expanded expression as: \(\|\mathrm{T}(x)\|^{2}+\|x\|^{2}\) This shows that for all \(x \in V\): \(\|\mathrm{T}(x) \pm i x\|^{2} = \|\mathrm{T}(x)\|^{2}+\|x\|^{2}\) Now that we have proved the given equality, let's deduce the properties of the operator \(T - iI\).
04

Proving the invertibility of T - i I

Now, suppose we have an operator \(S = T - iI\). Note that: \[\|S(x)\|^2 = \|T(x) - i x\|^2 = \|\mathrm{T}(x)\|^{2}+\|x\|^{2}\] Since \(\|S(x)\|^2 = \|\mathrm{T}(x)\|^{2}+\|x\|^{2} > 0\) for all nonzero \(x\), we can conclude that \(S\) is an injective operator. For finite-dimensional spaces, injectivity implies invertibility. Therefore, \(S = T - iI\) is invertible.
05

Finding the adjoint of the inverse of T - i I

Let \(S^{-1}\) be the inverse of \(S = T - iI\). To find the adjoint of \(S^{-1}\), we start by finding the adjoint of \(S\). We know that: \(S^*\) = (\(T - iI\))^* = \(T^* + iI^*\) Since \(T\) is self-adjoint, \(T^* = T\). Also, the adjoint of the identity operator is also the identity operator. Therefore: \(S^*\) = \(T + iI\) Now, using the property that \(S^* = (S^{-1})^{-1}\), we get: \((S^{-1})^* = (S^*)^{-1} = (T + iI)^{-1}\) Thus, the adjoint of (\(T - iI\))^{-1} is (\(T + iI\))^{-1}.

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Key Concepts

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

Finite-dimensional Inner Product Space
A finite-dimensional inner product space is a mathematical structure that combines two important concepts: finiteness and inner products. It is a vector space equipped with an inner product, which is a function that assigns a number to each pair of vectors. This number, called the inner product, helps us understand angles and lengths within the space.

Imagine this space as a setting where you have a limited number of dimensions, much like 3D space we are familiar with. However, in mathematics, the number of dimensions isn't restricted to just three. The term 'finite-dimensional' indicates that the space, unlike infinite-dimensional spaces, has a limited number of dimensions. This makes computations and various mathematical operations more manageable.

To visualize, think of vectors in a space with dimensions, say, 2 or 3. The inner product allows us to easily compute things like the length of a vector (using the formula for magnitude) and the angle between two vectors (using the cosine formula). These calculations are fundamental in many areas of mathematics and physics, making finite-dimensional inner product spaces highly useful.
Injective Operator
An injective operator is a function between two spaces that efficiently captures the concept of being one-to-one. This principle ensures that no two different inputs produce the same output. In simpler terms, imagine a function where each value maps uniquely to a distinct image, so no overlaps or duplications occur.

In mathematical formalism, if an operator \( f \) is called injective, it means that for any two distinct elements \( a \) and \( b \) in the domain, \( f(a) eq f(b) \). This property is crucial in fields such as algebra and calculus, as ensuring no data is "lost" or "replicated" during transformation helps retain the entire set's structure.

For finite-dimensional spaces, an injective operator also implies that the operator is invertible. Invertibility means you can "reverse" the operation, recovering original inputs from outputs. It provides a strong implication, signifying deeper control over the relationships between the spaces involved.
Invertibility
In mathematics, invertibility of an operator suggests that the operator can be reversed - meaning if you apply the operator and then apply its inverse, you return to where you started. This concept is central to solving equations and understanding function behavior.

An operator is invertible if there exists another operator that "undoes" its action. If you visualize an operator as a machine transforming inputs, then an invertible operator's inverse is like a machine that reverses these transformations.

Within finite-dimensional spaces, an operator is invertible if and only if it is both injective and surjective. Surjective means that every possible output is covered by some input. Therefore, when both injectivity and surjectivity are satisfied, the input-output relationship is both one-to-one and onto, ensuring a perfect mapping in both directions.

Understanding and proving invertibility is a common task in linear algebra, pivotal in ensuring the stability and solvability of systems of linear equations.
Adjoint Operator
An adjoint operator is a concept from linear algebra and functional analysis, providing a way to relate a given operator to another operator closely. For any operator \( T \) in an inner product space, its adjoint, denoted \( T^* \), is defined such that for any vectors \( x \) and \( y \), the inner product \( \langle T(x), y \rangle = \langle x, T^*(y) \rangle \).

This relationship between the operator and its adjoint captures the symmetry in transformations, similar to how reflection works. For self-adjoint operators, which are their own adjoints, this symmetry is perfect, meaning the transformation is entirely "mirrored" in a particular sense.

The adjoint is particularly useful when studying properties such as orthogonality and self-adjointness, which play a vital role in many areas of mathematics, including quantum mechanics and differential equations. When dealing with operators on a finite-dimensional space, understanding the adjoint becomes a powerful way to simplify complex problems, providing insights into the underlying structures involved.

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Most popular questions from this chapter

Let \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) be a linear transformation, where \(\mathrm{V}\) and \(\mathrm{W}\) are finitedimensional inner product spaces. Prove the following results. (a) \(\mathrm{T}^{*} \mathrm{~T}\) and \(\mathrm{TT}^{*}\) are positive semidefinite. (See Exercise 15 of Section 6.3.) (b) \(\operatorname{rank}\left(\mathrm{T}^{*} \mathrm{~T}\right)=\operatorname{rank}\left(\mathrm{TT}^{*}\right)=\operatorname{rank}(\mathrm{T})\).

For each of the following matrices \(A\) with entries from \(R\), find a diagonal matrix \(D\) and an invertible matrix \(Q\) such that \(Q^{t} A Q=D\). (a) \(\left(\begin{array}{ll}1 & 3 \\ 3 & 2\end{array}\right)\) (b) \(\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\) (c) \(\left(\begin{array}{rrr}3 & 1 & 2 \\ 1 & 4 & 0 \\ 2 & 0 & -1\end{array}\right)\) Hint for \((b)\) : Use an elementary operation other than interchanging columns.

Let \(V\) be the vector space defined in Example 5 of Section 1.2, the space of all sequences \(\sigma\) in \(F\) (where \(F=R\) or \(F=C\) ) such that \(\sigma(n) \neq 0\) for only finitely many positive integers \(n\). For \(\sigma, \mu \in \mathrm{V}\), we define \(\langle\sigma, \mu\rangle=\sum_{n=1}^{\infty} \sigma(n) \overline{\mu(n)}\). Since all but a finite number of terms of the series are zero, the series converges. (a) Prove that \(\langle\cdot, \cdot\rangle\) is an inner product on \(\mathrm{V}\), and hence \(\mathrm{V}\) is an inner product space. (b) For each positive integer \(n\), let \(e_{n}\) be the sequence defined by \(e_{n}(k)=\delta_{n k}\), where \(\delta_{n k}\) is the Kronecker delta. Prove that \(\left\\{e_{1}, e_{2}, \ldots\right\\}\) is an orthonormal basis for \(\mathrm{V}\). (c) Let \(\sigma_{n}=e_{1}+e_{n}\) and \(\mathrm{W}=\operatorname{span}\left(\left\\{\sigma_{n}: n \geq 2\right\\}\right.\). (i) Prove that \(e_{1} \notin \mathrm{W}\), so \(\mathrm{W} \neq \mathrm{V}\). (ii) Prove that \(\mathrm{W}^{\perp}=\\{0\\}\), and conclude that \(\mathrm{W} \neq\left(\mathrm{W}^{\perp}\right)^{\perp}\). Thus the assumption in Exercise 13(c) that \(\mathrm{W}\) is finite-dimensional is essential.

Consider three coordinate systems \(S, S^{\prime}\), and \(S^{\prime \prime}\) with the corresponding axes \(\left(x, x^{\prime}, x^{\prime \prime} ; y, y^{\prime}, y^{\prime \prime} ;\right.\) and \(\left.z, z^{\prime}, z^{\prime \prime}\right)\) parallel and such that the \(x-, x^{\prime}-\), and \(x^{\prime \prime}\)-axes coincide. Suppose that \(S^{\prime}\) is moving past \(S\) at a velocity \(v_{1}>0\) (as measured on \(S\) ), \(S^{\prime \prime}\) is moving past \(S^{\prime}\) at a velocity \(v_{2}>0\) (as measured on \(S^{\prime}\) ), and \(S^{\prime \prime}\) is moving past \(S\) at a velocity \(v_{3}>0\) (as measured on \(S\) ), and that there are three clocks \(C, C^{\prime}\), and \(C^{\prime \prime}\) such that \(C\) is stationary relative to \(S, C^{\prime}\) is stationary relative to \(S^{\prime}\), and \(C^{\prime \prime}\) is stationary relative to \(S^{\prime \prime}\). Suppose that when measured on any of the three clocks, all the origins of \(S, S^{\prime}\), and \(S^{\prime \prime}\) coincide at time 0 . Assuming that \(\mathrm{T}_{v_{3}}=\mathrm{T}_{v_{2}} \mathrm{~T}_{v_{1}}\) (i.e., \(B_{v_{3}}=B_{v_{2}} B_{v_{1}}\) ), prove that $$ v_{3}=\frac{v_{1}+v_{2}}{1+v_{1} v_{2}} . $$ Note that substituting \(v_{2}=1\) in this equation yields \(v_{3}=1\). This tells us that the speed of light as measured in \(S\) or \(S^{\prime}\) is the same. Why would we be surprised if this were not the case?

Let \(B\) be a symmetric matrix. Prove that \(\min _{x \neq 0} R(x)\) equals the smallest eigenvalue of \(B\).
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