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Chapter 5: Problem 24

Prove that the restriction of a diagonalizable linear operator \(\mathrm{T}\) to any nontrivial \(\mathrm{T}\)-invariant subspace is also diagonalizable. Hint: Use the result of Exercise \(23 .\)

Short Answer

Expert verified
In conclusion, to prove that the restriction of a diagonalizable linear operator T to any nontrivial T-invariant subspace is also diagonalizable, we used the result of Exercise 23 to show the existence of complementary T-invariant subspaces. By constructing a basis for the given T-invariant subspace W consisting of eigenvectors of T'|W (the restriction of T on W), we showed that the matrix representing T'|W with respect to this basis will be diagonal, thus proving that T'|W is diagonalizable.

Step by step solution

01

Understand the definition of T-invariant subspace

A subspace W 鈯� V is T-invariant if T(w) 鈭� W for every w 鈭� W. In other words, the action of T on a vector of the subspace W doesn't take the vector outside the subspace W.
02

Recall the definition of diagonalizable linear operator and its properties

A linear operator T on a finite-dimensional vector space V is called diagonalizable if there exists a basis of V consisting of eigenvectors of T. In matrix form, if A is a linear operator and there exists an invertible matrix P and a diagonal matrix D such that \(A = PDP^{-1}\), then A is diagonalizable.
03

Use Exercise 23

Exercise 23 is an important result that says that if we have a diagonalizable linear operator with a T-invariant subspace, then there exists a complementary T-invariant subspace such that V = W 鈯� W'. Here, W' is another T-invariant subspace and 鈯� represents the direct sum.
04

Consider T restricted to the T-invariant subspace W

Let's call the restriction of T on subspace W as T'|W. Since T is diagonalizable, W is a T-invariant subspace, and W' is also T-invariant by the result of Exercise 23. We know that there exists a basis of eigenvectors of T for the entire vector space V. Let's denote this basis as {v1, v2, ..., vn}.
05

Construct a basis for the subspace W

From the complementary T-invariant subspaces (W and W'), pick a basis for W consisting of eigenvectors of T'|W (since these eigenvectors are also eigenvectors of T), and call this basis {w1, w2, ..., wk}. Note that k 鈮� n, due to the hierarchical structure of T-invariant subspace and complementary space.
06

Prove the diagonalizable property of T'|W

Since we have a basis for the subspace W consisting of eigenvectors of T'|W, it implies that T'|W is diagonalizable because the matrix representing T'|W, with respect to this basis, will be a diagonal matrix with its eigenvalues as the diagonal elements. In conclusion, we have shown that a diagonalizable linear operator T restricted to a T-invariant subspace W is also diagonalizable, as we have found a basis for W consisting of eigenvectors corresponding to T'|W.

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Key Concepts

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

T-invariant subspace
A T-invariant subspace is a crucial concept in linear algebra, especially when working with linear operators like transformations or matrices. Imagine you have a vector space, and you want to see how a particular linear operation affects a part of that space. Well, a subspace is called T-invariant if when you apply the operator T to any vector in this subspace, the result is still within the subspace.

Essentially, it means that the transformation 鈥渟tays within the lines鈥� of that subspace. This property is vital because it allows us to observe and study the behavior of the operator in a more confined environment without losing generality.

When dealing with diagonalizable linear operators, which are simpler to handle as they involve eigenvectors and eigenvalues neatly lined up, T-invariant subspaces allow for breaking problems into smaller pieces while maintaining their characteristics.
Eigenvectors
Eigenvectors come into play when we want to understand how a matrix or linear operator affects particular directions in space. An eigenvector of a linear transformation T is a nonzero vector that doesn't change direction when T is applied to it. Think of it as a vector that simply gets stretched or squished, but not turned, by the transformation.

Mathematically, if T is a linear operator and v is a nonzero vector such that \(T(v) = \lambda v\), then v is the eigenvector of T, and \(\lambda\) is its corresponding eigenvalue. Having a set of eigenvectors is extremely useful because they form the backbone of the diagonalization process.

By expressing an operator in terms of its eigenvectors, we can transform a complex matrix into a diagonal form, which simplifies many calculations and theoretical deductions. This process is especially efficient if the entire vector space can be built from eigenvectors, which is the case for diagonalizable operators.
Basis
Understanding the concept of a basis is fundamental when working with vector spaces. A basis is essentially a set of vectors that can uniquely represent every vector in the space through a linear combination. These vectors must be linearly independent (no one vector in the set can be composed as a linear combination of the others), and they must span the entire vector space.

Think of the basis as the "coordinate system" for your vector space. In the context of diagonalization, when we say that a basis consists of eigenvectors of a linear operator, it means that each vector in the space can be expressed in terms of these eigenvectors. This arrangement is what makes an operator diagonalizable, allowing the transformation matrix to be expressed in simpler, diagonal form.

By choosing a basis appropriately, often as the eigenvectors of the operator, it becomes possible to view and manipulate complex problems in a much more manageable and elegant manner.

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Most popular questions from this chapter

Definitions. Two linear operators \(T\) and \(U\) on a finite-dimensional vector space \(\mathrm{V}\) are called simultaneously diagonalizable if there exists an ordered basis \(\beta\) for \(V\) such that both \([T]_{\beta}\) and \([U]_{\beta}\) are diagonal matrices. Similarly, \(A, B \in \mathrm{M}_{n \times n}(F)\) are called simultaneously diagonalizable if there exists an invertible matrix \(Q \in \mathrm{M}_{n \times n}(F)\) such that both \(Q^{-1} A Q\) and \(Q^{-1} B Q\) are diagonal matrices. 18\. (a) Prove that if \(T\) and \(U\) are simultaneously diagonalizable linear operators on a finite-dimensional vector space \(\mathrm{V}\), then the matrices \([T]_{\beta}\) and \([U]_{\beta}\) are simultaneously diagonalizable for any ordered basis \(\beta\). (b) Prove that if \(A\) and \(B\) are simultaneously diagonalizable matrices, then \(L_{A}\) and \(L_{B}\) are simultaneously diagonalizable linear operators.

Each of the matrices that follow is a regular transition matrix for a three- state Markov chain. In all cases, the initial probability vector is $$ P=\left(\begin{array}{l} 0.3 \\ 0.3 \\ 0.4 \end{array}\right) $$ For each transition matrix, compute the proportions of objects in each state after two stages and the eventual proportions of objects in each state by determining the fixed probability vector. (a) \(\left(\begin{array}{rrr}0.6 & 0.1 & 0.1 \\ 0.1 & 0.9 & 0.2 \\ 0.3 & 0 & 0.7\end{array}\right)\) (b) \(\left(\begin{array}{lll}0.8 & 0.1 & 0.2 \\ 0.1 & 0.8 & 0.2 \\ 0.1 & 0.1 & 0.6\end{array}\right)\) (c) \(\left(\begin{array}{rrr}0.9 & 0.1 & 0.1 \\ 0.1 & 0.6 & 0.1 \\ 0 & 0.3 & 0.8\end{array}\right)\) (d) \(\left(\begin{array}{lll}0.4 & 0.2 & 0.2 \\ 0.1 & 0.7 & 0.2 \\ 0.5 & 0.1 & 0.6\end{array}\right)\) (e) \(\left(\begin{array}{lll}0.5 & 0.3 & 0.2 \\ 0.2 & 0.5 & 0.3 \\ 0.3 & 0.2 & 0.5\end{array}\right)\) (f) \(\left(\begin{array}{rrr}0.6 & 0 & 0.4 \\ 0.2 & 0.8 & 0.2 \\ 0.2 & 0.2 & 0.4\end{array}\right)\)

(a) Prove that if \(\mathrm{T}\) and \(\mathrm{U}\) are simultaneously diagonalizable operators, then \(T\) and \(U\) commute (i.e., \(T U=U T\) ). (b) Show that if \(A\) and \(B\) are simultaneously diagonalizable matrices, then \(A\) and \(B\) commute. The converses of (a) and (b) are established in Exercise 25 of Section \(5.4\).

Let \(A=\left(\begin{array}{rrr}1 & 1 & -3 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right)\), let \(\mathrm{T}=\mathrm{L}_{A}\), and let \(\mathrm{W}\) be the cyclic subspace of \(\mathrm{R}^{3}\) generated by \(e_{1}\). (a) Use Theorem \(5.21\) to compute the characteristic polynomial of \(\mathrm{T}_{\mathrm{W}}\). (b) Show that \(\left\\{e_{2}+\mathrm{W}\right\\}\) is a basis for \(\mathrm{R}^{3} / \mathrm{W}\), and use this fact to compute the characteristic polynomial of \(\overline{\mathrm{T}}\). (c) Use the results of (a) and (b) to find the characteristic polynomial of \(A\).

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)
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