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Chapter 11: Problem 18

Let \(V\) be as in 82 , and let \(A: V \rightarrow V\) be a symmetrie operator. Let \(\lambda_{1}, \ldots, \lambda_{r}\), be the distinet eigenvalues of \(A .\) If \(\lambda\) is an eigenvalue of \(A\), let \(V_{\lambda}(A)\) eonsist of the set of all \(v \in V\) such that \(A v=\lambda v\). (a) Show that \(V_{\lambda}(A)\) is a subspace of \(V\), and that \(.1\) maps \(V_{\lambda}(A)\) into itself. (b) Show that \(V\) is the direet sum of the spaces $$ V=V_{\lambda_{1}}(A) \oplus \cdots \oplus V_{\lambda,}(A) $$ and that any two of these subspaces are mutually orthogonal. We call \(V_{\lambda}(A)\) the eigenspace of \(A\) belonging to \(\lambda\).

Short Answer

Expert verified
In conclusion, we proved that eigenspace $V_{\lambda}(A)$ is a subspace of $V$ and $A$ maps it to itself, as it is closed under vector addition and scalar multiplication. Also, we showed that $V$ can be expressed as a direct sum of eigenspaces $V_{\lambda_1}(A) 鈯� \cdots 鈯� V_{\lambda_r}(A)$ which are mutually orthogonal, i.e., for any two distinct eigenspaces $V_{\lambda_i}(A)$ and $V_{\lambda_j}(A)$, the inner product of eigenvectors in these eigenspaces is zero: $鉄▁_i, x_j鉄� = 0$.

Step by step solution

01

Showing V位(A) is a subspace of V

To prove V位(A) is a subspace of V, let v1, v2 鈭� V位(A) and c 鈭堚劃. We need to show that V位(A) is closed under addition and scalar multiplication. Since v1 and v2 belong to V位(A), A*v1 = 位*v1 and A*v2 = 位*v2. We now need to consider the linear combination of v1 and v2: v3 = v1 + v2.
02

Checking if linear combination of v1 and v2 belongs to V位(A)

Apply A on v3 (v1 + v2): A*v3 = A*(v1 + v2) = A*v1 + A*v2 = 位*v1 + 位*v2 = 位*(v1 + v2) = 位*v3 Thus, v3 belongs to V位(A). This shows that the eigenspace V位(A) is closed under vector addition.
03

Checking if scalar multiple of v1 belongs to V位(A)

Now consider the scalar multiple: v4 = c*v1. Apply A on v4 (c*v1): A*v4 = A*(c*v1) = c*A*v1 = c*位*v1 = 位*(c*v1) = 位*v4 Thus, v4 belongs to V位(A). This shows that the eigenspace V位(A) is closed under scalar multiplication. V位(A) being closed under both vector addition and scalar multiplication makes it a subspace of V.
04

Showing A maps V位(A) into itself

From Steps 2 and 3, we can conclude that 鈭�v 鈭� V位(A), A*v = 位*v. Since 位*v is also in V位(A), it is clear that applying A on any vector in V位(A) results in another vector in V位(A). Therefore, A maps V位(A) into itself.
05

Showing V is the direct sum of the eigenspaces V位i(A)

Since A is symmetric, it can be diagonalized. So, we can write A as S*D*S^(-1), where D is the diagonal matrix of eigenvalues, and S is a matrix with eigenvectors as its columns. Therefore, V can be expressed as a direct sum of the eigenspaces belonging to the distinct eigenvalues: V = V位1(A) 鈯� V位2(A) 鈯� ... 鈯� V位r(A). For any v 鈭� V, we can write v as a unique linear combination of eigenvectors: v = c1*v1 + c2*v2 + ... + cr*vr, where ci is a scalar and vi is an eigenvector in V位i(A).
06

Showing eigenspaces V位i(A) are mutually orthogonal

Consider two distinct eigenspaces V位i(A) and V位j(A) where i 鈮� j. Let xi 鈭� V位i(A), and xj 鈭� V位j(A) be eigenvectors. To show mutual orthogonality, the inner product of eigenvectors in these eigenspaces should be zero: 鉄▁i, xj鉄� = 0 Compute 鉄ˋ*xi, xj鉄� using the property that 鉄ˋ*v, w鉄� = 鉄╲, A*w鉄� for symmetric operators: 鉄ˋ*xi, xj鉄� = 鉄ㄎ籭*xi, xj鉄� = 位i*鉄▁i, xj鉄� Similarly, compute 鉄▁i, A*xj鉄�: 鉄▁i, A*xj鉄� = 鉄▁i, 位j*xj鉄� = 位j*鉄▁i, xj鉄� Since 位i 鈮� 位j, and 鉄ˋ*xi, xj鉄� = 鉄▁i, A*xj鉄�, it follows that 鉄▁i, xj鉄� = 0. Thus, V位i(A) and V位j(A) are mutually orthogonal. In conclusion, we have proved that eigenspace V位(A) is a subspace of V and A maps it to itself; and V can be expressed as a direct sum of eigenspaces V位i(A) which are mutually orthogonal.

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Key Concepts

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

Linear Algebra
Linear algebra is a foundational field of mathematics essential for studying spaces, solving systems of linear equations, and understanding transformations. At the heart of linear algebra are concepts like vectors, matrices, vector spaces, and linear transformations.

Vector Spaces: These are collections of vectors that can be scaled and added together while still remaining within the same space. They must adhere to specific rules known as the vector space axioms.

Linear Transformations: These are rules that assign a vector from one space to another in a linear way, meaning that they preserve vector addition and scalar multiplication.

Understanding linear algebra is crucial for analyzing the characteristics of a linear operator or transformation, which brings us to the consideration of eigenvectors and eigenvalues, as well as their corresponding eigenspaces within a given vector space.
Symmetric Operator
A symmetric operator, within the realm of linear algebra, is a linear transformation that satisfies a specific condition of symmetry. This means that when applied within a certain context, usually within a real inner product space, the result of this operator preserves the inner product. In other words, for any two vectors v and w in the space, it satisfies the condition 鉄ˋv, w鉄� = 鉄╲, Aw鉄�.

Properties of Symmetric Operators:
  • Real Eigenvalues: Symmetric operators always have real eigenvalues, which helps in understanding their behavior.
  • Orthogonal Eigenvectors: Eigenvectors corresponding to different eigenvalues of a symmetric operator are orthogonal to each other.
  • Diagonalization: These operators can be diagonalized, meaning they can be represented in a form that simplifies their study and application.
Such properties have profound implications in the study of eigenspaces and the structure of the vector space on which the symmetric operator acts.
Eigenvalues
Eigenvalues are scalars associated with a linear operator that provide insight into the operator's effects on a vector space. When a vector v is transformed by an operator A and the result is a scalar multiple of v, this scalar is known as an eigenvalue, and the vector v is called an eigenvector.

Definition: If Av = 位惫, then 位 is the eigenvalue and v is an eigenvector of the operator A. Importantly, solving for eigenvalues involves finding roots of the characteristic polynomial of an operator or matrix.

Significance of Eigenvalues:
  • You can learn about the operator's scaling effect on subspaces of the vector space.
  • Eigenvalues play a critical role in various applications such as stability analysis, vibrations in mechanical structures, and in quantum mechanics.
  • In the context of symmetric operators, distinct eigenvalues guarantee orthogonal eigenvectors, leading to an orthogonal decomposition of the space.
The analysis of eigenvalues reveals the underlying structure and behavior of linear transformations, making it an essential component of linear algebra.

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Most popular questions from this chapter

Let \(A, B\) be normal operators such that \(A B=B A\). Show that \(A B\) is normal.

Show that a positive hermitian operator \(A\) has a square root, i.e. a positive hermitian operator \(B\) sueh that \(B^{2}=A\), and \(B\) is uniquely determined.

Let \(A\) be a real symmetrie \(3 \times 3\) matrix. Let \(f(X)=\langle A X, X\rangle\) for all \(X \in \mathbf{R}^{n}\). Let \(v\) be a vector on the ellipsoid defined by the equation $$ 3 x^{2}+4 y^{2}+5 z^{2}=1 $$ such that \(f(v) \geqslant f(X)\) for all \(X\) on the ellipsoid. Show that \(v\) is an eigenvector of \(A\).

Let \(V\) be as in 82, and let \(A: V \rightarrow V\) be a symmetric linear map. Prove that the following conditions on \(A\) imply esch other. (a) All eigenvalues of \(A\) sre \(>0\). (b) For all elements \(v \in V, v \neq 0\), we have \(\langle A v, v\rangle>0\). If the map \(A\) satisfies these conditions, it is said to be positive definite. The same of \(\mathbf{R}^{*}\) into itself. Thus the second condition, in terms of coordinate vectors, reads: (b') For all veetors \(X \in \mathbf{R}^{n}, X \neq 0\), we have $$ { }^{\prime} X A X>0 $$

Let \(A\) be an \(n \times n\) non-singular real symmetric matrix. Prove the following statements. (a) If \(\lambda\) is an eigenvalue of \(A\), then \(\lambda \neq 0\). (b) If \(\lambda\) is an eigenvalue of \(A\), then \(\lambda^{-1}\) is an eigenvalue of \(A^{-1}\). (c) The matrices \(A\) and \(A^{-1}\) have the same set of eigenvectors.
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