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Chapter 6: Problem 8

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 \(\left(\mathrm{T}^{*}\right)^{-1}=\left(\mathrm{T}^{-1}\right)^{*} .\)

Short Answer

Expert verified
In a finite-dimensional inner product space V, if T is an invertible linear operator, then its adjoint, T*, is also invertible with the inverse \(((T^*)^{-1}\)=(T^{-1})^* .\) We proved this by first finding a candidate for the inverse of T* and then verifying that it is indeed the left and right inverse of T*. This confirmed the invertibility of T* and the equality \(((T^*)^{-1}\)=(T^{-1})^* .\)

Step by step solution

01

Define the given terms

We are given a finite-dimensional inner product space V, a linear operator T on V, and we know that T is invertible. We are also given the adjoint of T, denoted by T*. Our goal is to show that T* is invertible and \(((T^*)^{-1}\)=(T^{-1})^* .\)
02

Show that T* is invertible by finding a candidate for its inverse

Since T is invertible, there exists an inverse operator, denoted by \( T^{-1} \). We will now show that the adjoint of \(T^{-1}\), denoted by \((T^{-1})^*\), is a candidate for the inverse of T*. To start, let's define the identity operator I on V, such that I(x) = x for all x in V. Then, we can check whether \((T^{-1})^*\) is a left and right inverse of T* using the following equalities: 1. \((T^{-1})^* T* = I\) 2. \(T* (T^{-1})^* = I\)
03

Prove that \((T^{-1})^*\) is a left inverse of T*

We want to show that the following equality holds true: \[((T^{-1})^* T*) (x) = I(x) \quad \forall x \in V\] Let y = T*(x). Since T* is a linear operator, it also has an adjoint operator denoted by \((T^*)^{-1}\), such that \(T^*(y) = x\). Now, we evaluate \((T^{-1})^* T*(x)\): \[((T^{-1})^* T*) (x) = ((T^{-1})^* T*) T^*(y)= ((T^{-1})^*) (T T^*)(y)\] Now, using the definition of adjoint, we know that \((T T^*)(y) = (T^* T)(y)\). Therefore, we have: \[((T^{-1})^*) (T^* T)(y) = I(y)\] By definition of the adjoint, \((T^{-1})^*\) is the inverse of T, thus: \[((T^{-1})^*) (T^* T)(y) = T^{-1}(T(y)) = I(y)\] So, we have shown that \((T^{-1})^* T* = I\).
04

Prove that \((T^{-1})^*\) is a right inverse of T*

We want to show that the following equality holds true: \[T* ((T^{-1})^*) (x) = I(x) \quad \forall x \in V\] Let y = T(x). Now, we evaluate \(T* ((T^{-1})^*) (x)\): \[T* ((T^{-1})^*) (x) = T* ((T^{-1})^*) T(y)= T (T^* (T^{-1})(y))\] Now, using the definition of adjoint and the inverse, we have: \[T (T^* (T^{-1})(y)) = T (I(y)) = I(x)\] So, we have shown that \(T* (T^{-1})^* = I\).
05

Conclude that T* is invertible and \(((T^*)^{-1}\)=(T^{-1})^* .\)

Since we have shown that \((T^{-1})^*\) is both the left and right inverse of T*, we can conclude that T* is invertible and its inverse is \((T^{-1})^*\). Thus, we have proven that if T is invertible, then T* is invertible and \(((T^*)^{-1}\)=(T^{-1})^* .\) This completes the proof.

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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
Understanding the structure of a finite-dimensional inner product space is crucial when delving into more complex concepts in linear algebra such as linear operators and their adjoints. An inner product space is a vector space equipped with an additional structure called an inner product. This inner product allows for the generalization of geometrical notions like length and angles to vector spaces.

Adjoint of a Linear Operator
In the realm of linear algebra, the adjoint of a linear operator plays a significant role, especially in the context of inner product spaces. For a given linear operator \( T \) on a finite-dimensional inner product space \( V \), its adjoint \( T^* \) is a unique operator that satisfies \( \langle T(x), y \rangle = \langle x, T^*(y) \rangle \) for all vectors \( x, y \) in \( V \).

Proof Techniques in Linear Algebra
Linear algebra is a field rich with methods for proving various properties of vectors and matrices. Proof techniques in linear algebra often involve verifying that certain operations on vectors or matrices lead to expected outcomes, such as showing the existence of an inverse or demonstrating that certain properties hold for all vectors within the vector space.

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Most popular questions from this chapter

Find new coordinates \(x^{\prime}, y^{\prime}\) so that the following quadratic forms can be written as \(\lambda_{1}\left(x^{\prime}\right)^{2}+\lambda_{2}\left(y^{\prime}\right)^{2}\). (a) \(x^{2}+4 x y+y^{2}\) (b) \(2 x^{2}+2 x y+2 y^{2}\) (c) \(x^{2}-12 x y-4 y^{2}\) (d) \(3 x^{2}+2 x y+3 y^{2}\) (e) \(x^{2}-2 x y+y^{2}\)

Let \(W\) be a finite-dimensional subspace of an inner product space \(V\). Show that if \(\mathrm{T}\) is the orthogonal projection of \(\mathrm{V}\) on \(\mathrm{W}\), then \(\mathrm{I}-\mathrm{T}\) is the orthogonal projection of \(\mathrm{V}\) on \(\mathrm{W}^{\perp}\).

Let \(\mathrm{W}=\operatorname{span}(\\{(i, 0,1)\\})\) in \(\mathrm{C}^{3}\). Find orthonormal bases for \(\mathrm{W}\) and \(\mathrm{W}^{\perp}\).

Let \(V\) be a finite-dimensional inner product space, and let \(T\) and \(U\) be self-adjoint operators on \(V\) such that \(T\) is positive definite. Prove that both TU and UT are diagonalizable linear operators that have only real eigenvalues. Hint: Show that UT is self-adjoint with respect to the inner product \(\langle x, y\rangle^{\prime}=\langle\mathrm{T}(x), y\rangle\). To show that TU is self-adjoint, repeat the argument with \(\mathrm{T}^{-1}\) in place of \(\mathrm{T}\).

Recall the moving space vehicle considered in the study of time contraction. Suppose that the vehicle is moving toward a fixed star located on the \(x\)-axis of \(S\) at a distance \(b\) units from the origin of \(S\). If the space vehicle moves toward the star at velocity \(v\), Earthlings (who remain "almost" stationary relative to \(S\) ) compute the time it takes for the vehicle to reach the star as \(t=b / v\). Due to the phenomenon of time contraction, the astronaut perceives a time span of \(t^{\prime}=t \sqrt{1-v^{2}}=(b / v) \sqrt{1-v^{2}}\). A paradox appears in that the astronaut perceives a time span inconsistent with a distance of \(b\) and a velocity of \(v\). The paradox is resolved by observing that the distance from the solar system to the star as measured by the astronaut is less than \(b\). (a) At time \(t\) (as measured on \(C\) ), the space-time coordinates of the star relative to \(S\) and \(C\) are $$ \left(\begin{array}{l} b \\ 0 \\ t \end{array}\right) \text {. } $$ (b) At time \(t\) (as measured on \(C\) ), the space-time coordinates of the star relative to \(S^{\prime}\) and \(C^{\prime}\) are $$ \left(\begin{array}{c} \frac{b-v t}{\sqrt{1-v^{2}}} \\ 0 \\ \frac{t-b v}{\sqrt{1-v^{2}}} \end{array}\right) $$ (c) For $$ x^{\prime}=\frac{b-t v}{\sqrt{1-v^{2}}} \quad \text { and } \quad t^{\prime}=\frac{t-b v}{\sqrt{1-v^{2}}}, $$ we have \(x^{\prime}=b \sqrt{1-v^{2}}-t^{\prime} v\). This result may be interpreted to mean that at time \(t^{\prime}\) as measured by the astronaut, the distance from the astronaut to the star, as measured by the astronaut, (see Figure 6.9) is $$ b \sqrt{1-v^{2}}-t^{\prime} v . $$ Assuming that the coordinate systems \(S\) and \(S^{\prime}\) and clocks \(C\) and \(C^{\prime}\) are as in the discussion of time contraction, prove the following results. (d) Conclude from the preceding equation that (1) the speed of the space vehicle relative to the star, as measured by the astronaut, is \(v\); (2) the distance from Earth to the star, as measured by the astronaut, is \(b \sqrt{1-v^{2}}\). Thus distances along the line of motion of the space vehicle appear to be contracted by a factor of \(\sqrt{1-v^{2}}\).
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