/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Let \(\mathrm{V}\) be an inner p... [FREE SOLUTION] | 91影视

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Let \(\mathrm{V}\) be an inner product space, and let \(y, z \in \mathrm{V}\). Define \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{V}\) by \(\mathrm{T}(x)=\langle x, y\rangle z\) for all \(x \in \mathrm{V}\). First prove that \(\mathrm{T}\) is linear. Then show that \(\mathrm{T}^{*}\) exists, and find an explicit expression for it. The following definition is used in Exercises 15-17 and is an extension of the definition of the adjoint of a linear operator. Definition. Let \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) be a linear transformation, where \(\mathrm{V}\) and \(\mathrm{W}\) are finite-dimensional inner product spaces with inner products \(\langle\cdot, \cdot\rangle_{1}\) and \(\langle\cdot, \cdot\rangle_{2}\), respectively. A function \(\mathrm{T}^{*}: \mathrm{W} \rightarrow \mathrm{V}\) is called an adjoint of \(\mathrm{T}\) if \(\langle\mathrm{T}(x), y\rangle_{2}=\left\langle x, \mathrm{~T}^{*}(y)\right\rangle_{1}\) for all \(x \in \mathrm{V}\) and \(y \in \mathrm{W}\).

Short Answer

Expert verified
In order to prove that the transformation $\mathrm{T}(x)=\langle x, y\rangle z$ is linear, we first showed that it satisfies additivity and homogeneity. Then, to find the adjoint $\mathrm{T}^*(y)$, we used the definition of the adjoint of a linear operator: $\langle\mathrm{T}(x),y\rangle_2=\langle x, \mathrm{T}^*(y)\rangle_1$. By comparing the expressions, we found that the adjoint $\mathrm{T}^*(y)$ is given by: \[ \mathrm{T}^*(y) = \langle z,y \rangle_2 y \]

Step by step solution

01

Proving T is linear

To prove T is linear, we will demonstrate that T satisfies additivity and homogeneity. 1. Additivity: Show that T(x+y) = T(x) + T(y) for all x and y in V T(x+y) = 鉄▁+y, y鉄﹝ = (鉄▁, y鉄 + 鉄▂, y鉄)z (by linearity of inner product) = 鉄▁, y鉄﹝ + 鉄▂, y鉄﹝ = T(x) + T(y) 2. Homogeneity: Show that T(cx) = cT(x) for all x in V and scalar c T(cx) = 鉄╟x, y鉄﹝ = c鉄▁, y鉄﹝ (by linearity of inner product) = cT(x) Since both additivity and homogeneity are satisfied, T is linear.
02

Finding the adjoint of T

To find the adjoint T*, we will use the definition of the adjoint of a linear operator: 鉄═(x),y鉄┾倐=鉄▁, T*(y)鉄┾倎 for all x in V and y in W. We know that T(x) = 鉄▁, y鉄﹝. Now, we need an expression for 鉄═(x),y鉄┾倐. 鉄═(x),y鉄┾倐 = 鉄ㄢ煥x, y鉄﹝, y鉄┾倐 = 鉄▁, y鉄 鉄▃, y鉄┾倐 (by linearity of inner product) The right side of the definition of the adjoint is given by: 鉄▁, T*(y)鉄┾倎. We are to find a function T* such that the right side of the definition equals the left side we just found. Thus, for all x in V, we want: 鉄▁, T*(y)鉄┾倎 = 鉄▁, y鉄 鉄▃, y鉄┾倐 Notice that if T*(y) = 鉄▃,y鉄┾倐 y, then 鉄▁, T*(y)鉄┾倎 becomes: 鉄▁, 鉄▃,y鉄┾倐 y鉄┾倎 = 鉄▁, y鉄 鉄▃, y鉄┾倐 Now we have found a function T* that satisfies the definition of the adjoint for all x in V. Therefore, the adjoint T*(y) of T is given by: T*(y) = 鉄▃,y鉄┾倐 y

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Key Concepts

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

Inner Product Space
An inner product space is a fundamental notion in linear algebra. It's essentially a vector space equipped with an additional structure called the inner product. The inner product allows us to measure angles and lengths, making it an indispensable tool in geometry and physics.
The inner product is a function that takes two vectors and returns a scalar. Mathematically, if \( V \) is a vector space, the inner product is denoted by \( \langle \cdot, \cdot \rangle \), where for vectors \( x, y \in V \), \( \langle x, y \rangle \) represents the inner product.
Key properties of inner products include:
  • Conjugate symmetry: \( \langle x, y \rangle = \overline{\langle y, x \rangle} \), meaning swapping vectors conjugates the result.
  • Linearity in the first argument: \( \langle ax + by, z \rangle = a \langle x, z \rangle + b \langle y, z \rangle \), where \( a \) and \( b \) are scalars.
  • Positive-definiteness: \( \langle x, x \rangle \geq 0 \) with equality if and only if \( x \) is the zero vector.
Utilizing these properties, inner products help determine orthogonality (perpendicularity) and norm (length) of vectors. Their existence empowers a variety of mathematical tools like orthogonal projections and Gram-Schmidt process, all fundamental to vector calculus and functional analysis.
Linear Transformation
Linear transformations are functions that map vectors from one vector space to another while preserving vector addition and scalar multiplication. If \( T: V \rightarrow W \) is a linear transformation from vector space \( V \) to another \( W \), it satisfies:
  • Additivity: \( T(x+y) = T(x) + T(y) \) for all \( x, y \in V \).
  • Homogeneity: \( T(cx) = cT(x) \) for every scalar \( c \) and all \( x \in V \).
Linear transformations simplify complex problems by reducing them to simpler vector operations, supporting both computational and theoretical work. These transformations are essential in virtually every field in mathematics, physics, and engineering.
There are well-known types of linear transformations such as:
  • Rotations and translations: Operations preserving the vector magnitudes.
  • Projections: Transformations reducing the space's dimension.
Interestingly, the inner product spaces provide an extra layer of structure for linear transformations, enabling concepts like transposition and adjoints. This leads to defining operators that can "flip" a transformation around a standard line or plane, portraying symmetry and duality in various mathematical contexts.
Additivity and Homogeneity
The principles of additivity and homogeneity form the cornerstone of linear transformations. These properties define how transformations respond to vector addition and scalar multiplication鈥攃ore operations in vector spaces.
Additivity states that if a transformation \( T \) is applied to the sum of two vectors, the result is equivalent to applying \( T \) to both vectors separately and adding the results: \[ T(x+y) = T(x) + T(y) \] Here, no matter how vectors \( x \) and \( y \) are combined within their space, \( T \) splits the work evenly among both.
Homogeneity, on the other hand, asserts that scaling a vector by some scalar \( c \) before applying \( T \) is identical to scaling the transformation\( T(x)\) by \( c \): \[ T(cx) = cT(x) \] This shows that scalar multiplication outside the transformation yields the same result as doing it inside. Bridging these laws, any linear transformation can be understood and visualized as both disentangling vector combinations and stretching/shrinking them appropriately within the space's structure.
Understanding these fundamental properties can help interpret how various changes affect vector spaces in practice, whether in simple mappings or intricate physical systems.

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Most popular questions from this chapter

Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.

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 \(U\) be a unitary operator on an inner product space \(V\), and let \(W\) be a finite-dimensional U-invariant subspace of V. Prove that (a) \(U(W)=W\); (b) \(\mathrm{W}^{\perp}\) is U-invariant. Contrast (b) with Exercise 16 .

Let \(\mathrm{T}\) and \(\mathrm{U}\) be self-adjoint linear operators on an \(n\)-dimensional inner product space \(\mathrm{V}\), and let \(A=[\mathrm{T}]_{\beta}\), where \(\beta\) is an orthonormal basis for V. Prove the following results. (a) \(\mathrm{T}\) is positive definite [semidefinite] if and only if all of its eigenvalues are positive [nonnegative]. (b) \(\mathrm{T}\) is positive definite if and only if $$ \sum_{i, j} A_{i j} a_{j} \bar{a}_{i}>0 \text { for all nonzero } n \text {-tuples }\left(a_{1}, a_{2}, \ldots, a_{n}\right) \text {. } $$ (c) \(\mathrm{T}\) is positive semidefinite if and only if \(A=B^{*} B\) for some square matrix \(B\). (d) If \(T\) and \(U\) are positive semidefinite operators such that \(T^{2}=U^{2}\), then \(\mathrm{T}=\mathrm{U}\). (e) If \(T\) and \(U\) are positive definite operators such that \(T U=U T\), then TU is positive definite. (f) \(\mathrm{T}\) is positive definite [semidefinite] if and only if \(A\) is positive definite [semidefinite]. Because of (f), results analogous to items (a) through (d) hold for matrices as well as operators.

Let \(H: \mathrm{R}^{2} \times \mathrm{R}^{2} \rightarrow R\) be the function defined by $$ H\left(\left(\begin{array}{l} a_{1} \\ a_{2} \end{array}\right),\left(\begin{array}{l} b_{1} \\ b_{2} \end{array}\right)\right)=a_{1} b_{2}+a_{2} b_{1} \quad \text { for }\left(\begin{array}{l} a_{1} \\ a_{2} \end{array}\right),\left(\begin{array}{l} b_{1} \\ b_{2} \end{array}\right) \in \mathrm{R}^{2} $$ (a) Prove that \(H\) is a bilinear form. (b) Find the \(2 \times 2\) matrix \(A\) such that \(H(x, y)=x^{t} A y\) for all \(x, y \in \mathrm{R}^{2} .\)

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