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Chapter 6: Problem 10

Let \(W\) be a finite-dimensional subspace of an inner product space \(V\). Prove that \(\mathrm{V}=\mathrm{W} \oplus \mathrm{W}^{\perp}\). Using the definition on page 76, prove that there exists a projection \(\mathrm{T}\) on \(\mathrm{W}\) along \(\mathrm{W}^{\perp}\) that satisfies \(\mathrm{N}(\mathrm{T})=\mathrm{W}^{\perp}\). In addition, prove that \(\|\mathrm{T}(x)\| \leq\|x\|\) for all \(x \in \mathrm{V}\). Hint: Use Theorem \(6.6\) and Exercise 10 of Section 6.1.

Short Answer

Expert verified
In summary, we proved that for a finite-dimensional subspace W of an inner product space V, it holds that V = W 鈯 W鈯. Additionally, we demonstrated the existence of a projection T on W along W鈯, with N(T) = W鈯, and showed that ||T(x)|| 鈮 ||x|| for all x 鈭 V. This was achieved by using properties of orthogonal projections, direct sums, and the Pythagorean theorem, along with Theorem 6.6 and Exercise 10 from Section 6.1.

Step by step solution

01

Prove V = W 鈯 W鈯

To prove this, we need to show that V is the direct sum of W and W鈯, i.e., every element of V can be written as a unique sum of elements from W and W鈯. Let x 鈭 V. By Theorem 6.6, there exists a unique element y 鈭 W and z 鈭 W鈯 such that x = y + z. This shows that V is the direct sum of W and W鈯.
02

Prove existence of the projection T on W along W鈯

We define the projection T: V 鈫 W as follows: Given an x 鈭 V, we know that there exists unique y 鈭 W and z 鈭 W鈯 such that x = y + z. We define T(x) = y. To prove that T is a linear transformation, let x鈧,x鈧 鈭 V and let 伪 be a scalar. By the properties of orthogonal projection, we have: x鈧 = y鈧 + z鈧, where y鈧 鈭 W and z鈧 鈭 W鈯, x鈧 = y鈧 + z鈧, where y鈧 鈭 W and z鈧 鈭 W鈯. Now, 伪x鈧 + x鈧 = 伪(y鈧 + z鈧) + (y鈧 + z鈧) = (伪y鈧 + y鈧) + (伪z鈧 + z鈧). Since W and W鈯 are subspaces, 伪y鈧 + y鈧 鈭 W and 伪z鈧 + z鈧 鈭 W鈯. Thus, T(伪x鈧 + x鈧) = 伪y鈧 + y鈧 = 伪T(x鈧) + T(x鈧), proving that T is a linear transformation.
03

Show N(T) = W鈯

The null space of T, N(T), is given by {x 鈭 V : T(x) = 0}. Let x 鈭 N(T). Then T(x) = 0, so x = y + z, where y = 0 鈭 W and z 鈭 W鈯. Thus, x 鈭 W鈯. Conversely, let x 鈭 W鈯. Then x = y + z, where y 鈭 W and z 鈭 W鈯. Since x is orthogonal to W, we know that y = 0. Thus, T(x) = y = 0, so x 鈭 N(T). Hence, we have proved that N(T) = W鈯.
04

Prove ||T(x)|| 鈮 ||x|| for all x 鈭 V

Let x 鈭 V. Then x = y + z, where y 鈭 W and z 鈭 W鈯. Since W and W鈯 are orthogonal, y and z are orthogonal as well. We use the Pythagorean theorem (Theorem 6.1 Exercise 10) to get: ||x||虏 = ||y + z||虏 = ||y||虏 + ||z||虏. Since T(x) = y, we have ||T(x)|| = ||y||. Now, we see that ||T(x)||虏 = ||y||虏 鈮 ||y||虏 + ||z||虏 = ||x||虏. Taking the square root of both sides, we get ||T(x)|| 鈮 ||x|| for all x 鈭 V. We have now proved all the required statements for the given exercise.

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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 key concept in linear algebra and functional analysis, encompassing a set along with an operation called an 'inner product' that allows for the measurement of angles and lengths within that space. In simpler terms, it's a vector space equipped with a way to calculate dot products between vectors, which are essential for understanding concepts like orthogonality (perpendicularity) and lengths of vectors.

The inner product has to satisfy the following properties: it needs to be (1) positive definite, which means the inner product of any vector with itself is always positive, except for the zero vector; (2) linear in its first argument, meaning you can factor out scalars and split the inner product over vector addition; (3) symmetric, so swapping the two vectors in the inner product doesn't change the result; and (4) conjugate symmetric, if dealing with complex spaces, which involves taking the complex conjugate of the second argument.

When studying subspaces within an inner product space, a common task is to decompose vectors into components that are within a subspace and components that are orthogonal to it, achieving the direct sum of subspaces as in the original exercise鈥攖his is where our next concept, orthogonal projection, steps in.
Orthogonal Projection
Orthogonal projection is a process in linear algebra by which you decompose a vector into two parts: one that lies within a certain subspace, and another that's perpendicular to it. This is highly relevant in many areas of mathematics and computer science, such as when you're working with shadows in computer graphics or reducing data dimensions in machine learning.

To illustrate, consider a vector space that contains a subspace and its orthogonal complement. Any vector in the space can be expressed uniquely as the sum of a vector in the subspace and a vector in the orthogonal complement. The 'projection' part essentially refers to taking any vector and 'dropping' it down perpendicularly onto the subspace, obtaining the component within that subspace.

How Orthogonal Projection Relates to Our Exercise

The orthogonal projection onto a subspace, as applied to our exercise, is the linear transformation that maps every vector to its 'shadow' on the subspace, effectively segregating the vector's components. The operation performed by the transformation T in the exercise, which maps every vector x to y which lies in W, is a manifestation of orthogonal projection. This tool is crucial for many practical applications, including least squares approximations and solving minimization problems.
Null Space of a Linear Transformation
The null space (or kernel) of a linear transformation is a fundamental concept for understanding the structure and properties of linear mappings. It consists of all vectors that map to the zero vector under a given linear transformation. In essence, if you apply the transformation to any vector in the null space, the output will always be the vector equivalent of zero.

More formally, if we have a linear transformation T from one vector space to another, the null space is defined as \(N(T) = \{x | T(x) = 0\}\). It's a measure of the 'failure' of T to be injective, or one-to-one. The larger the null space, the more vectors 'collapse' to zero, and the less information T preserves from the original space.

Understanding the Null Space in Our Exercise

In our exercise, the null space of the projection transformation T is exactly the orthogonal complement of W, denoted \(W^\perp\). This means that any vector in \(W^\perp\) gets sent to the zero vector by T, aligning with our understanding that T 'ignores' the components perpendicular to W. Grasping the concept of null space is pivotal when discussing the invertibility of transformations, among other aspects of linear algebra, such as dimensionality and basis.

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Most popular questions from this chapter

Let \(A\) be a matrix with real or complex entries. Prove the following results. (a) The nonzero singular values of \(A\) are the same as the nonzero singular values of \(A^{*}\), which are the same as the nonzero singular values of \(A^{t}\). (b) \(\left(A^{\dagger}\right)^{*}=\left(A^{*}\right)^{\dagger}\). (c) \(\left(A^{\dagger}\right)^{t}=\left(A^{t}\right)^{\dagger}\).

24\. Let \(T\) be a linear operator on a real inner product space \(V\), and define \(H: \mathrm{V} \times \mathrm{V} \rightarrow R\) by \(H(x, y)=\langle x, \mathrm{~T}(y)\rangle\) for all \(x, y \in \mathrm{V}\). (a) Prove that \(H\) is a bilinear form. (b) Prove that \(H\) is symmetric if and only if \(\mathrm{T}\) is self-adjoint. (c) What properties must \(\mathrm{T}\) have for \(H\) to be an inner product on \(\mathrm{V}\) ? (d) Explain why \(H\) may fail to be a bilinear form if \(\mathrm{V}\) is a complex inner product space.

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 \(\mathrm{T}\) and \(\mathrm{U}\) be self-adjoint operators on an inner product space \(\mathrm{V}\). Prove that \(\mathrm{TU}\) is self-adjoint if and only if \(\mathrm{TU}=\mathrm{UT}\).

Let \(\mathrm{T}\) be a normal operator on a finite-dimensional complex inner product space \(\mathrm{V}\). Use the spectral decomposition \(\lambda_{1} \mathrm{~T}_{1}+\lambda_{2} \mathrm{~T}_{2}+\cdots+\lambda_{k} \mathrm{~T}_{k}\) of \(\mathrm{T}\) to prove the following results. (a) If \(g\) is a polynomial, then $$ g(\mathbf{T})=\sum_{i=1}^{k} g\left(\lambda_{i}\right) \mathbf{T}_{i} . $$ (b) If \(\mathrm{T}^{n}=\mathrm{T}_{0}\) for some \(n\), then \(\mathrm{T}=\mathrm{T}_{0}\). (c) Let \(U\) be a linear operator on \(V\). Then \(U\) commutes with \(T\) if and only if \(U\) commutes with each \(T_{i}\). (d) There exists a normal operator \(U\) on \(V\) such that \(U^{2}=T\). (e) \(\mathrm{T}\) is invertible if and only if \(\lambda_{i} \neq 0\) for \(1 \leq i \leq k\). (f) \(\mathrm{T}\) is a projection if and only if every eigenvalue of \(\mathrm{T}\) is 1 or 0 . (g) \(\mathrm{T}=-\mathrm{T}^{*}\) if and only if every \(\lambda_{i}\) is an imaginary number.
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