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Chapter 6: Problem 11

For a linear operator \(T\) on an inner product space \(V\), prove that \(T^{*} T=\) \(\mathrm{T}_{0}\) implies \(\mathrm{T}=\mathrm{T}_{0}\). Is the same result true if we assume that \(\mathrm{TT}^{*}=\mathrm{T}_{0}\) ?

Short Answer

Expert verified
In conclusion, if \(T^*T = T_0\) for a linear operator \(T\) on an inner product space \(V\), then \(T = T_0\). However, if \(TT^* = T_0\), we cannot conclude that \(T = T_0\). This is because when \(T^*T = T_0\), we have \(\langle Tv, T^*w \rangle = 0\) for all \(v, w \in V\), which implies that \(Tv\) is the zero vector for all \(v \in V\). On the other hand, when \(TT^* = T_0\), we only have \(\langle T^*v, Tw \rangle = 0\) for all \(v, w \in V\), which does not give us enough information to conclude that \(Tv\) is the zero vector for all \(v \in V\).

Step by step solution

01

Write down the definition of the adjoint of a linear operator

For linear operators, \(T: V \to V\) and \(T^* : V \to V\), we know that \[ \langle Tv, w \rangle = \langle v, T^*w \rangle \] for all \(v, w \in V\), where \(\langle \cdot, \cdot \rangle\) denotes the inner product on the vector space \(V\).
02

Show that if \(T^*T = T_0\), then \(T = T_0\)

Suppose that \(T^*T = T_0\). Then for any \(v, w \in V\), we have \[ \langle T^*Tv, w \rangle = \langle T_0v, w \rangle \] We substitute the definition of the adjoint in the left side of this equation: \[ \langle T^*Tv, w \rangle = \langle Tv, T^*w \rangle \] Since \(T^*T = T_0\), we get \(\langle Tv, T^*w \rangle = \langle 0, w \rangle\). By the properties of the inner product, this simplifies to: \[ \langle Tv, T^*w \rangle = 0 \] Thus, for all \(v, w \in V\), we have: \[ \langle Tv, T^*w \rangle = 0 \] Now consider any vector \(v \in V\) and take \(w = Tv\): \[ \langle Tv, T^* Tv \rangle = 0 \] Using the definition of the adjoint again, we get: \[ \langle Tv, 0 \rangle = 0 \] We know that the inner product of any vector with the zero vector is 0. This is true for all \(Tv\), which means that \(Tv\) must be the zero vector for all \(v \in V\). Thus, we have \(T = T_0\).
03

Investigate the case when \(TT^* = T_0\)

Suppose now that \(TT^* = T_0\). Then for any \(v, w \in V\), we have \[ \langle TT^*v, w \rangle = \langle T_0v, w \rangle \] We substitute the definition of the adjoint in the left side of this equation: \[ \langle TT^*v, w \rangle = \langle T^*v, Tw \rangle \] Since \(TT^* = T_0\), we get \(\langle T^*v, Tw \rangle = \langle 0, w \rangle\). By the properties of the inner product, this simplifies to: \[ \langle T^*v, Tw \rangle = 0 \] Thus, for all \(v, w \in V\), we have: \[ \langle T^*v, Tw \rangle = 0 \] Now consider any vector \(v \in V\) and take \(w = T^*v\): \[ \langle T^*v, TT^* v \rangle = 0 \] Again, using the property that the inner product of any vector with the zero vector is 0: \[ \langle T^*v, 0 \rangle = 0 \] However, this result only tells us information about \(T^*v\), not \(Tv\). Therefore, we cannot conclude that \(T = T_0\) in this case. In conclusion, the result \(T^*T = T_0\) implies \(T = T_0\), but the result \(TT^* = T_0\) does not imply \(T = T_0\).

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Key Concepts

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

Adjoint of a Linear Operator
In the world of linear algebra, an adjoint of a linear operator is a fundamental concept that enhances our understanding of transformations in inner product spaces. Simply put, if you have a linear operator \(T\) acting on a vector space \(V\), there's another operator, denoted as \(T^*\), that is associated with it. The mathematical relationship that defines an adjoint can be written as:\[\langle Tv, w \rangle = \langle v, T^*w \rangle\]for all vectors \(v, w \in V\). This equation tells us that the inner product of \(Tv\) with \(w\) is the same as the inner product of \(v\) with \(T^*w\). Here is what makes this interesting:
  • An adjoint essentially "reverses" the action of \(T\) under the inner product.
  • The operator \(T^*\) exists and is unique for each linear operator \(T\).
  • The adjoint plays a critical role in self-adjoint operators, unitary operators, and normal operators, among other concepts.
Understanding adjoints is key, especially in expanding the dynamics between linear operators and the geometrical structure of vector spaces.
Inner Product Space
An inner product space is a vector space equipped with an additional structure called the inner product. This structure allows you to compute "dot products" between vectors, providing a measure of angle and length. Let's break it down some more:
  • The inner product, denoted as \(\langle v, w \rangle\), is a form of operation that combines pairs of vectors \(v\) and \(w\) to produce scalars.
  • One key result of an inner product space is that it enables the definition of orthogonality鈥攖wo vectors \(v\) and \(w\) are orthogonal if \(\langle v, w \rangle = 0\).
  • This space not only brings geometric intuition to vectors, like angles and lengths, but also ensures a solid foundation for more complex vector operations.
Think of an inner product space as a setting where vectors "live" with rules that allow us to understand their orientation and magnitude. Building problem-solving solutions within this framework is critically important. It connects the abstract world of vector spaces to tangible concepts like geometry.
Properties of Inner Products
The properties of inner products are rules that govern how vectors interact in inner product spaces. These properties include several familiar and useful facts:
  • Conjugate Symmetry: For vectors \(v\) and \(w\), \(\langle v, w \rangle = \overline{\langle w, v \rangle}\), where the overline denotes the complex conjugate.
  • Linearity: The inner product is linear in its first argument, meaning \(\langle av + bw, u \rangle = a\langle v, u \rangle + b\langle w, u \rangle\) for scalars \(a\) and \(b\).
  • Positive Definiteness: The inner product of a vector with itself is always non-negative, i.e., \(\langle v, v \rangle \geq 0\), and it is zero if and only if the vector is the zero vector.
These tangible, easy-to-check properties significantly simplify calculations and offer predictable results when manipulating vectors. In practical terms, these properties allow operations such as projections, decompositions, and transformations to be precisely defined and effortlessly executed.

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Most popular questions from this chapter

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 \((\mathrm{T}+i \mathrm{I})^{-1}\).

Let \(\mathrm{T}\) be a normal operator on a finite-dimensional complex inner product space \(\mathrm{V}\), and let \(\mathrm{W}\) be a subspace of \(\mathrm{V}\). Prove that if \(\mathrm{W}\) is T-invariant, then \(\mathrm{W}\) is also \(\mathrm{T}^{*}\)-invariant. Hint: Use Exercise 24 of Section \(5.4\).

Let \(\mathrm{V}\) be a finite-dimensional inner product space, and let \(\mathrm{T}\) be a linear operator on \(\mathrm{V}\). Prove that if \(\mathrm{T}\) is invertible, then \(\mathrm{T}^{*}\) is invertible and \(\left(\mathrm{T}^{*}\right)^{-1}=\left(\mathrm{T}^{-1}\right)^{*} .\)

Assume that \(\mathrm{T}\) is a linear operator on a complex (not necessarily finitedimensional) inner product space \(\mathrm{V}\) with an adjoint \(\mathrm{T}^{*}\). Prove the following results. (a) If \(\mathrm{T}\) is self-adjoint, then \(\langle\mathrm{T}(x), x\rangle\) is real for all \(x \in \mathrm{V}\). (b) If \(\mathrm{T}\) satisfies \(\langle\mathrm{T}(x), x\rangle=0\) for all \(x \in \mathrm{V}\), then \(\mathrm{T}=\mathrm{T}_{0}\). Hint: Replace \(x\) by \(x+y\) and then by \(x+i y\), and expand the resulting inner products. (c) If \(\langle\mathrm{T}(x), x\rangle\) is real for all \(x \in \mathrm{V}\), then \(\mathrm{T}\) is self-adjoint.

Let \(V\) be a finite-dimensional inner product space over \(F\). (a) Parseval's Identity. Let \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) be an orthonormal basis for \(\mathrm{V}\). For any \(x, y \in \mathrm{V}\) prove that $$ \langle x, y\rangle=\sum_{i=1}^{n}\left\langle x, v_{i}\right\rangle \overline{\left\langle y, v_{i}\right\rangle} . $$ (b) Use (a) to prove that if \(\beta\) is an orthonormal basis for \(V\) with inner product \(\langle\cdot, \cdot\rangle\), then for any \(x, y \in \mathrm{V}\) $$ \left\langle\phi_{\beta}(x), \phi_{\beta}(y)\right\rangle^{\prime}=\left\langle[x]_{\beta},[y]_{\beta}\right\rangle^{\prime}=\langle x, y\rangle, $$ where \(\langle\cdot, \cdot\rangle^{\prime}\) is the standard inner product on \(\mathrm{F}^{n}\).
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