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Chapter 5: Problem 17

5-17. (a) Let \(T: \mathbf{R}^{n} \rightarrow \mathbf{R}^{n}\) be self-adjoint with matrix \(A-\left(a_{i j}\right)\), so that \(a_{i j}=a_{j i} . \quad\) If \(f(x)=\langle T x, x\rangle=\Sigma a_{i j} x^{i} x^{j}\), show that \(D_{k} f(x)=\) \(2 \sum_{j-1}^{n} a_{k j} x^{j} .\) By considering the maximum of \(\langle T x, x\rangle\) on \(S^{n-1}\) show that there is \(x \in S^{n-1}\) and \(\lambda \in \mathbf{R}\) with \(T x=\lambda x\). (b) If \(V=\left\\{y \in \mathbf{R}^{n}:\langle x, y\rangle=0\right\\}\), show that \(T(V) \subset V\) and \(T: V \rightarrow V\) is self-adjoint. (c) Show that \(T\) has a basis of eigenvectors.

Short Answer

Expert verified
\( D_k f(x) = 2 \sum_{j=1}^{n} a_{k j} x^{j} \). There exists \( x \in S^{n-1} \) and \( \lambda \in \mathbf{R} \) with \( T x = \lambda x \). The operator \( T \) on \( V \) remains self-adjoint and \( T \) has a basis of eigenvectors.

Step by step solution

01

Understand the Concept of Self-Adjoint Operators and Their Matrix Representation

A self-adjoint operator, or symmetric operator, in linear algebra is one where its matrix representation is equal to its transpose. For operator \(T\) with matrix \(A = (a_{ij})\), this means \(a_{ij} = a_{ji}\).
02

Define the Function \( f(x) \)

Given \( f(x) = \langle T x, x \rangle = \sum a_{ij} x^i x^j \), where \( \langle T x, x \rangle \) is the inner product of \( T x \) and \( x \). Rewrite the inner product using matrix representation to understand its structure.
03

Compute the Partial Derivative \( D_k f(x) \)

To find \( D_k f(x) \), compute the partial derivative of \( f(x) \) with respect to \( x^k \): \( \frac{\partial f(x)}{\partial x^k} = \frac{\partial}{\partial x^k} \left( \sum_{i,j} a_{ij} x^i x^j \right) \). Since \( a_{ij} = a_{ji} \), we get \( D_k f(x) = 2 \sum_{j=1}^{n} a_{kj} x^j \).
04

Consider the Maximum of \( \langle T x, x \rangle \) on \( S^{n-1} \)

Maximize \( \langle T x, x \rangle \) on the unit sphere \( S^{n-1} \) in \( \mathbf{R}^{n} \). Show that there exists a vector \( x \in S^{n-1} \) and a scalar \( \lambda \in \mathbf{R} \) such that \( T x = \lambda x \). This means \( x \) is an eigenvector and \( \lambda \) is the corresponding eigenvalue.
05

Prove that \( T(V) \subset V \) and Self-Adjoint Property on \( V \)

If \( V = \left\{ y \in \mathbf{R}^n : \langle x, y \rangle = 0 \right\} \), show that applying \( T \) to any vector in \( V \) keeps the resulting vector in \( V \). Also, demonstrate that within this subspace, the operator \( T \) remains self-adjoint.
06

Show that \( T \) Has a Basis of Eigenvectors

Using previously established results, show that the entire space \( \mathbf{R}^n \) can be spanned by eigenvectors of the operator \( T \). This completes the proof that \( T \) has a basis of eigenvectors.

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Key Concepts

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

Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are fundamental in understanding transformations in linear algebra.
An eigenvector of a linear transformation is a vector that does not change its direction when the transformation is applied.
It only gets scaled by a certain factor, known as the eigenvalue.
Mathematically, if \(T\) is a transformation and \(x\) is an eigenvector, then \(T(x) = \lambda x\) where \(\lambda\) is the eigenvalue.
Eigenvalues and eigenvectors help in simplifying complex matrix operations.
They are used in many applications like stability analysis, vibrations, and quantum mechanics.
When dealing with self-adjoint operators, the eigenvectors can be chosen to be orthogonal.
Inner Product
The inner product is a way of measuring angles and lengths in vector spaces.
For vectors \(x\) and \(y\) in \(\mathbf{R}^n\), the inner product is denoted as \(\langle x, y \rangle\) and computed as \(\sum x^i y^i\).
This operation results in a scalar.
It helps in defining concepts like orthogonality (when the inner product is zero).
In the context of self-adjoint operators, we deal with expressions like \(\langle T x, x \rangle\), representing how the transformation \(T\) interacts with vectors in the space.
This becomes crucial while analyzing the properties such as maximum and minimum values over unit spheres.
Matrix Representation
A matrix represents a linear transformation in numerical form.
Each element \(a_{ij}\) in the matrix corresponds to how the \(j\)-th component of the input vector affects the \(i\)-th component of the output vector.
For a self-adjoint (or symmetric) matrix, \(a_{ij} = a_{ji}\).
This symmetry simplifies many calculations.
For instance, when computing the derivative \(D_k f(x)\), the symmetry ensures that certain cross terms match up, making the derivative expression more elegant.
Additionally, a self-adjoint matrix has real eigenvalues and its eigenvectors can be chosen to be orthonormal, which is helpful in various proofs and applications.
Subspaces
A subspace is a subset of a vector space that is also a vector space itself.
It must be closed under vector addition and scalar multiplication.
For example, the set of all vectors orthogonal to a given vector form a subspace.
In the given problem, we considered \(V = \{ y \in \mathbf{R}^n : \langle x, y \rangle = 0 \}\), which means \(V\) consists of vectors orthogonal to \(x\).
It's important to show that applying the transformation \(T\) to any vector in \(V\) keeps the resulting vector in \(V\).
This maintains the integrity of the subspace under \(T\).
Proving that \(T\) is self-adjoint within the subspace further extends the properties of self-adjointness to subdomains of our space, making our mathematical frameworks robust and consistent.

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Most popular questions from this chapter

5-12. (a) If \(F\) is a differentiable vector field on \(M \subset \mathbf{R}^{n}\), show that there is an open set \(A \supset M\) and a differentiable vector field \(\tilde{F}\) on \(A\) with \(\tilde{F}(x)=F(x)\) for \(x \in M .\) Hint: Io this locally and use partitions of unity. (b) If \(M\) is closed, show that we can choose \(A=\mathbf{R}^{n}\).

5-31. Consider the 2 -form \(\omega\) defined on \(\mathbf{R}^{3}-0\) by $$ \omega=\frac{x d y \wedge d z+y d z \wedge d x+z d x \wedge d y}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}} $$ (a) Show that \(\omega\) is closed. (b) Show that $$ \omega(p)\left(v_{p}, w_{p}\right)=\frac{\langle v \times w, p\rangle}{|p|^{3}} $$ For \(r>0\) let \(S^{2}(r)=\left\\{x \in \mathbf{R}^{3}:|x|=r\right\\} . \quad\) Show that \(\omega\) restricted to the tangent space of \(S^{2}(r)\) is \(1 / r^{2}\) times the volume element, and that \(\int S^{2}(r) \omega=4 \pi . \quad\) Conclude that \(\omega\) is not exact. Nevertheless we denote \(\omega\) by \(d \Theta\) since, as we shall see, \(d \Theta\) is the analogue of the 1 -form \(d \theta\) on \(\mathbf{R}^{2}-0\) (c) If \(v_{p}\) is a tangent vector such that \(v=\lambda p\) for some \(\lambda \in \mathbf{R}\) show that \(d \Theta(p)\left(v_{p}, w_{p}\right)=0\) for all \(w_{p}\). If a two-dimensional manifold \(M\) in \(\mathbf{R}^{3}\) is part of a generalized cone, that is, \(M\) is the union of segments of rays through the origin, show that \(\int_{M} d \theta=0\) (d) Let \(M \subset \mathbf{R}^{3}-0\) be a compact two-dimensional manifoldwith-boundary such that every ray through 0 intersects \(M\) at most once (Figure 5-10). The union of those rays through 0 which intersect. \(M\), is a solid cone \(C(M) .\) The solid angle subtended by \(M\) is defined as the area of \(C(M) \cap S^{2}\), or equivalently as \(1 / r^{2}\) times the area of \(C(M) \cap S^{2}(r)\) for \(r>0\). Prove that the solid angle subtended by \(M\) is \(\left|\int_{M} d \Theta\right| .\) Hint: Choose \(r\) small enough so that there is a three-dimensional manifold-with-boundary \(N\) (as in Figure \(5-10)\) such that \(\partial N\) is the union of \(M\) and \(C(M) \cap S^{2}(r)\), and a part of a generalized cone. (Actually, \(N\) will be a manifold-

5-30. (a) If \(f:[0,1] \rightarrow \mathbf{R}\) is differentiable and \(c:[0,1] \rightarrow \mathbf{R}^{2}\) is defined by \(c(x)=(x, f(x))\), show that \(c([0,1])\) has length \(\int_{0}^{1} \sqrt{1+\left(f^{\prime}\right)^{2}} .\) (b) Show that this length is the least upper bound of lengths of inscribed broken lines. Hint: If \(0=t_{0} \leq t_{1} \leq \cdots \leq t_{n}=1\), then $$ \begin{aligned} \left|c\left(t_{i}\right)-c\left(t_{i-1}\right)\right| &=\sqrt{\left(t_{i}-t_{i-1}\right)^{2}+\left(f\left(t_{i}\right)-f\left(t_{i-1}\right)\right)^{2}} \\\ &=\sqrt{\left(t_{i}-t_{i-1}\right)^{2}+f^{\prime}\left(s_{i}\right)^{2}\left(t_{i}-t_{i-1}\right)^{2}} \end{aligned} $$ for some \(s_{i} \in\left[t_{i-1}, t_{i}\right]\).

5-22. If \(M_{1} \subset \mathbf{R}^{n}\) is an \(n\)-dimensional manifold-with- boundary and \(M_{2} \subset M_{1}-\partial M_{1}\) is an \(n\)-dimensional manifold-with-boundary, and \(M_{1}, M_{2}\) are compact, prove that $$ \int_{\partial M_{1}} \omega=\int_{\lambda M_{2}} \omega $$ where \(\omega\) is an \((n-1)\)-form on \(M_{1}\), and \(\partial M_{1}\) and \(\partial M_{2}\) have the orientations induced by the usual orientations of \(M_{1}\) and \(M_{2}\). Hint: Find a manifold-with-boundary \(M\) such that \(\partial M=\partial M_{1} \cup \partial M_{2}\) and such that the induced orientation on \(\partial M\) agrees with that for \(\partial M_{1}\) on \(\partial M_{1}\) and is the negative of that for \(\partial M_{2}\) on \(\partial M_{2}\)

5-10. Suppose \(e\) is a collection of coordinate systems for \(M\) such that (1) For each \(x \in M\) there is \(f \in \mathcal{e}\) which is a coordinate system around \(x\); (2) if \(f, g \in e\), then det \(\left(f^{-1} \circ g\right)^{\prime}>0\). Show that there is a unique orientation of \(M\) such that \(f\) is orientation-preserving if \(f \in e\)
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