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(a) Let \(\mathrm{T}\) be a linear operator on a vector space \(\mathrm{V}\), and let \(x\) be an eigenvector of \(\mathrm{T}\) corresponding to the eigenvalue \(\lambda\). For any positive integer \(m\), prove that \(x\) is an eigenvector of \(\mathrm{T}^{m}\) corresponding to the eigenvalue \(\lambda^{m}\). (b) State and prove the analogous result for matrices.

Short Answer

Expert verified
In summary, for any linear operator T and a corresponding eigenvector x with eigenvalue λ, we have proven that x is an eigenvector of \(T^m\) with eigenvalue \(\lambda^m\). Similarly, for any matrix A and a corresponding eigenvector x with eigenvalue λ, we have proven that x is an eigenvector of \(A^m\) with eigenvalue \(\lambda^m\).

Step by step solution

01

(a) Apply the definition of eigenvector to T(x)

We know x is an eigenvector of T with eigenvalue λ, meaning that \(T(x)=\lambda x\).
02

(a) Apply T repeatedly to both sides of the equation

We can apply T m times to both sides of the equation: \(T^m(x) = \lambda^m x\) This can be shown by applying T repeatedly (m times), and using the fact that λ is an eigenvalue: \(T(T^{m-1}(x)) = \lambda T^{m-1}(x)\) \(T(\lambda^{m-1}x) = \lambda(\lambda^{m-1}x)\) Since \(T(\lambda^{m-1}x) = \lambda^{m-1}T(x)\), we get: \(\lambda^{m-1}T(x) = \lambda^m x\)
03

(a) Conclude the proof

We showed that \(T^m(x) = \lambda^m x\), meaning that x is an eigenvector of \(T^m\) with eigenvalue \(\lambda^m\). Thus, the statement is proven.
04

(b) Apply the definition of eigenvector to Ax

Now let A be a matrix and x be an eigenvector of A with eigenvalue λ, meaning that \(Ax=\lambda x\).
05

(b) Apply A repeatedly to both sides of the equation

Similarly to part (a), we can apply A m times to both sides of the equation: \(A^m x = \lambda^m x\) This can be shown by applying A repeatedly (m times), and using the fact that λ is an eigenvalue: \(A(A^{m-1}x) = \lambda A^{m-1}(x)\) \(A(\lambda^{m-1}x) = \lambda(\lambda^{m-1}x)\) Since \(A(\lambda^{m-1}x) = \lambda^{m-1}Ax\), we get: \(\lambda^{m-1}Ax = \lambda^{m}x\)
06

(b) Conclude the proof

We’ve shown that \(A^m x = \lambda^m x\), meaning that x is an eigenvector of \(A^m\) with eigenvalue \(\lambda^m\). Thus, the analogous result for matrices is proven.

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Key Concepts

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

Linear Transformations
A linear transformation is a function between two vector spaces that preserves the operations of addition and scalar multiplication. This means, if you have a vector space \(V\) and another vector space \(W\), a linear transformation \(T\) from \(V\) to \(W\) (denoted as \(T: V \rightarrow W\)) will satisfy two main properties for all vectors \(u\) and \(v\) in \(V\), and all scalars \(c\):
  • Additivity: \(T(u + v) = T(u) + T(v)\)
  • Homogeneity: \(T(cu) = cT(u)\)

Linear transformations give rise to important concepts in linear algebra, such as eigenvectors and eigenvalues. When a vector \(x\) is transformed by \(T\) and only changes by a scalar factor, it is called an eigenvector of \(T\), with the corresponding scalar being the eigenvalue. This relationship is expressed as \(T(x) = \lambda x\), where \(\lambda\) is the eigenvalue. In the original exercise, we explore this relationship and extend it to powers of the linear operator, demonstrating that if \(x\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), then it is also an eigenvector of \(T^m\), corresponding to \(\lambda^m\).
Vector Spaces
Vector spaces are essential structures in linear algebra that consist of a set of vectors, where vector addition and scalar multiplication are defined and satisfy certain axioms. Some of these axioms include:
  • Closure under addition and scalar multiplication
  • Existence of an additive identity (zero vector)
  • Existence of an additive inverse
  • Distributive properties for vectors and scalars

Understanding vector spaces is crucial for grasping the concept of eigenvectors and eigenvalues. In our context, a vector space \(V\) provides the setting where a linear operator \(T\) acts. When the operator \(T\) is applied to a vector \(x\) in \(V\) and the result is a scaled version of \(x\), \(x\) is recognized as an eigenvector. The vector space offers the structure to study transformations, allowing us to comprehend how these transformations behave through repeated application, as observed with matrix powers. These properties were utilized in the exercise to show that repeated application continues to satisfy the eigenvector conditions provided the conditions hold initially.
Matrix Powers
Matrix powers involve multiplying a square matrix by itself a certain number of times. If \(A\) is a square matrix, then \(A^m\) denotes the matrix multiplied by itself \(m\) times. This is particularly useful for analyzing linear transformations within vector spaces, as matrices can represent linear operators. When exploring eigenvectors and eigenvalues, examining matrix powers allows us to predict the behavior of transformations over multiple steps.

In the exercise, the concept of matrix powers was used to illustrate how an eigenvector \(x\) of a matrix \(A\) with eigenvalue \(\lambda\) remains an eigenvector of \(A^m\) with eigenvalue \(\lambda^m\). The step-by-step solution demonstrated how repeated applications of a matrix (or linear transformation \(T\)) to its eigenvector not only satisfy the original eigenvector equation, but retain their meaning under these reiterated operations. This concept is pivotal in simplifying calculations in various applications, like solving linear differential equations or determining powers of matrices used in computer graphics and systems theory.

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Most popular questions from this chapter

(a) Prove that similar matrices have the same characteristic polynomial. (b) Show that the definition of the characteristic polynomial of a linear operator on a finite-dimensional vector space \(V\) is independent of the choice of basis for \(\mathrm{V}\).

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\). Prove that \(T\) is diagonalizable if and only if \(V\) is the direct sum of one- dimensional T-invariant subspaces.

Prove that if \(A \in \mathbf{M}_{n \times n}(C)\) is diagonalizable and \(L=\lim _{m \rightarrow \infty} A^{m}\) exists, then either \(L=I_{n}\) or \(\operatorname{rank}(L)

Let \(\mathcal{C}\) be a collection of diagonalizable linear operators on a finitedimensional vector space V. Prove that there is an ordered basis \(\beta\) such that \([T]_{\beta}\) is a diagonal matrix for all \(T \in \mathcal{C}\) if and only if the operators of \(\mathcal{C}\) commute under composition. (This is an extension of Exercise 25.) Hints for the case that the operators commute: The result is trivial if each operator has only one eigenvalue. Otherwise, establish the general result by mathematical induction on \(\operatorname{dim}(\mathrm{V})\), using the fact that \(V\) is the direct sum of the eigenspaces of some operator in \(\mathcal{C}\) that has more than one eigenvalue.

For each of the following matrices \(A \in \mathrm{M}_{n \times n}(F)\), (i) Determine all the eigenvalues of \(A\). (ii) For each eigenvalue \(\lambda\) of \(A\), find the set of eigenvectors corresponding to \(\lambda\). (iii) If possible, find a basis for \(\mathrm{F}^{n}\) consisting of eigenvectors of \(A\). (iv) If successful in finding such a basis, determine an invertible matrix \(Q\) and a diagonal matrix \(D\) such that \(Q^{-1} A Q=D\).258 Chap. 5 Diagonalization (a) \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 2\end{array}\right) \quad\) for \(F=R\) (b) \(A=\left(\begin{array}{rrr}0 & -2 & -3 \\ -1 & 1 & -1 \\ 2 & 2 & 5\end{array}\right) \quad\) for \(F=R\) (c) \(A=\left(\begin{array}{rr}i & 1 \\ 2 & -i\end{array}\right) \quad\) for \(F=C\) (d) \(A=\left(\begin{array}{ccc}2 & 0 & -1 \\ 4 & 1 & -4 \\ 2 & 0 & -1\end{array}\right) \quad\) for \(F=R\)

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