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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
To prove that the restriction of a diagonalizable linear operator T to any nontrivial T-invariant subspace is also diagonalizable, we first consider a nontrivial T-invariant subspace W of a vector space V. We define the restricted operator \(T_W\) as the operator that acts on W, and show that the eigenspaces of T, when intersected with W, will form eigenspaces of \(T_W\). Then we calculate the sum of dimensions of the eigenspaces of the restricted operator and show that it equals the dimension of W. By using the result of Exercise 23, we can conclude that \(T_W\) is diagonalizable. Therefore, the restriction of a diagonalizable linear operator T to any nontrivial T-invariant subspace is also diagonalizable.

Step by step solution

01

Understanding diagonalizable linear operator

A linear operator T is called diagonalizable if there exists an ordered basis B of the vector space such that its matrix representation [T]_B is a diagonal matrix. In other words, T is diagonalizable if it can be represented by a matrix with non-zero diagonal entries and zeros everywhere else in the basis B.
02

Understanding T-invariant subspace

A subspace W of a vector space V is said to be T-invariant if for every vector v in W, T(v) is also in W. In other words, the action of the linear operator T keeps the subspace W within itself.
03

Recalling result of Exercise 23

Exercise 23 states that if T is a diagonalizable linear operator, then the sum of the dimensions of its eigenspaces is equal to the dimension of the entire vector space V, i.e. if T has distinct eigenvalues \(位_1, 位_2, ... , 位_k\), then \[\dim V = \dim E_{位_1} + \dim E_{位_2} + ... + \dim E_{位_k}\]
04

Consider a nontrivial T-invariant subspace W

Let W be a nontrivial T-invariant subspace of the vector space V. Our goal is to show that the restriction of T to W, (denoted \(T_W\)), is diagonalizable.
05

Construction of restricted operator and eigenspaces

Define the restricted operator \(T_W\) as the operator that acts on W, i.e., for any vector v in W, \(T_W(v) = T(v)\). Since W is T-invariant, \(T_W(v)\) will also be in W. Next, we will show that the eigenspaces \(E_{位_i}\) of T, when intersected with W, will form eigenspaces of \(T_W\). Let \(v 鈭� E_{位_i} 鈭� W\), then by definition, \(T(v) = 位_i v\). Since W is T-invariant, \(T_W(v) = T(v)\) and thus, \(T_W(v) = 位_i v\), which means v is an eigenvector of \(T_W\) with eigenvalue \(位_i\). This implies that \(E_{位_i} 鈭� W\) are eigenspaces of \(T_W\).
06

Calculate sum of dimensions of eigenspaces of the restricted operator

Since \(T_W\) has the same eigenvalues as T, we can calculate the sum of dimensions of its eigenspaces as follows: \[\dim W = \dim E_{位_1} 鈭� W + \dim E_{位_2} 鈭� W + ... + \dim E_{位_k} 鈭� W\]
07

Prove that \(T_W\) is diagonalizable

Now we can use the result of Exercise 23. If the sum of dimensions of the eigenspaces of the restricted operator equals the dimension of W, then \(T_W\) is diagonalizable. Since we have calculated the sum of dimensions of eigenspaces of \(T_W\) as \(\dim W\), we can conclude that \(T_W\) is indeed diagonalizable. Therefore, we have proved that the restriction of a diagonalizable linear operator T to any nontrivial T-invariant subspace is also diagonalizable.

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Most popular questions from this chapter

Prove: (a) The zero mapping \(0,\) defined by \(\mathbf{0}(v)=0 \in U\) for every \(v \in V,\) is the zero element of \(\operatorname{Hom}(V, U) .\) (b) The negative of \(F \in \operatorname{Hom}(V, U)\) is the mapping \((-1) F,\) that is, \(-F=(-1) F\)

Label the following statements as true or false. (a) Every linear operator on an \(n\)-dimensional vector space has \(n\) distinct eigenvalues. (b) If a real matrix has one eigenvector, then it has an infinite number of eigenvectors. (c) There exists a square matrix with no eigenvectors. (d) Eigenvalues must be nonzero scalars. (e) Any two eigenvectors are linearly independent. (f) The sum of two eigenvalues of a linear operator \(T\) is also an eigenvalue of \(T\). (g) Linear operators on infinite-dimensional vector spaces never have eigenvalues. (h) An \(n \times n\) matrix \(A\) with entries from a field \(F\) is similar to a diagonal matrix if and only if there is a basis for \(\mathrm{F}^{n}\) consisting of eigenvectors of \(A\). (i) Similar matrices always have the same eigenvalues. (j) Similar matrices always have the same eigenvectors. (k) The sum of two eigenvectors of an operator \(\mathrm{T}\) is always an eigenvector of \(T\).

Prove Theorem 5.6: Suppose \(V\) has finite dimension and \(F: V \rightarrow U\) is linear. Then \\[ \operatorname{dim} V=\operatorname{dim}(\operatorname{Ker} F)+\operatorname{dim}(\operatorname{Im} F)=\operatorname{nullity}(F)+\operatorname{rank}(F) \\]

Find the dimension \(d\) of \((a) \operatorname{Hom}\left(\mathbf{R}^{2}, \mathbf{R}^{8}\right),(b) \operatorname{Hom}\left(\mathbf{P}_{4}(t), \mathbf{R}^{3}\right),(\mathrm{c}) \operatorname{Hom}\left(\mathbf{M}_{2,4}, \mathbf{P}_{2}(t)\right)\)

Let \(V\) be of finite dimension and let \(T\) be a linear operator on \(V\) for which \(T R=I\), for some operator \(R\) on \(V\). (We call \(R\) a right inverse of \(T\).) (a) Show that \(T\) is invertible. (b) Show that \(R=T^{-1}\) (c) Give an example showing that the above need not hold if \(V\) is of infinite dimension.
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