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Chapter 2: Problem 30

$$ \text { Prove that the subspaces }\\{0\\}, V, R(T) \text {, and } N(T) \text { are all } T \text {-invariant. } $$

Short Answer

Expert verified
In summary, all four subspaces {}, V, R(T), and N(T) are T-invariant because: 1. The empty subspace {} only contains the zero vector, which is always mapped to itself by any linear transformation T. 2. The entire vector space V, by definition, contains all vectors and their images under the linear transformation T. 3. The range of the linear transformation R(T) consists of all vectors in V that are images of some vector under T, making it T-invariant. 4. The nullspace of the linear transformation N(T) consists of all vectors in V with T(v) = 0, and applying T to these vectors results in the zero vector, which is also in N(T). Hence, all subspaces are T-invariant under the given linear transformation.

Step by step solution

01

1. The empty subspace {}

Since the empty subspace only contains the zero vector, any linear transformation maps the zero vector to itself. Therefore, for any linear transformation T, the image of the zero vector under T is also in the empty subspace {}. So, the empty subspace {} is T-invariant.
02

2. The entire vector space V

As V is the entire vector space and T: V 鈫� V is a linear transformation, any vector v in V will have its image T(v) also in V. Hence, the entire vector space V is T-invariant.
03

3. The range of the linear transformation R(T)

By definition, R(T) consists of all vectors w in V such that w = T(v) for some v in V. Now, let w be any vector in R(T), then there exists some v in V such that w = T(v). If we apply the linear transformation T to w, we get T(w) = T(T(v)). As T is a linear transformation, T(T(v)) is also an element of R(T) since it is also the image of a vector in V (specifically, the image of T(v) in V). Therefore, R(T) is T-invariant.
04

4. The nullspace of the linear transformation N(T)

By definition, N(T) consists of all vectors v in V such that T(v) = 0. Now, let v be any vector in N(T) such that T(v) = 0. Applying the linear transformation T to the zero vector, we get T(0) = 0. Since T is a linear transformation, T(v) has to be in N(T) as well. Thus, N(T) is T-invariant.

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Most popular questions from this chapter

Recall the definition of \(\mathrm{P}(R)\) on page 11. Define $$ \mathrm{T}: \mathrm{P}(R) \rightarrow \mathrm{P}(R) \quad \text { by } \quad \mathrm{T}(f(x))=\int_{0}^{x} f(t) d t . $$ Prove that \(\mathrm{T}\) linear and one-to-one, but not onto.

Let \(\beta=\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) be a basis for a vector space \(\mathrm{V}\) and \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{V}\) be a linear transformation. Prove that \([\mathrm{T}]_{\beta}\) is upper triangular if and only if \(\mathrm{T}\left(v_{j}\right) \in \operatorname{span}\left(\left\\{v_{1}, v_{2}, \ldots, v_{j}\right\\}\right)\) for \(j=1,2, \ldots, n\). Visit goo.gl/k9ZrQb for a solution.

Label the following statements as true or false. In each part, \(V\) and \(W\) are finite-dimensional vector spaces (over \(F\) ), and \(\mathrm{T}\) is a function from \(\mathrm{V}\) to \(\mathrm{W}\). (a) If \(\mathrm{T}\) is linear, then \(\mathrm{T}\) preserves sums and scalar products. (b) If \(\mathrm{T}(x+y)=\mathrm{T}(x)+\mathrm{T}(y)\), then \(\mathrm{T}\) is linear. (c) \(\mathrm{T}\) is one-to-one if and only if the only vector \(x\) such that \(\mathrm{T}(x)=0\) is \(x=0\). (d) If \(T\) is linear, then \(T\left(0_{v}\right)=0_{w}\). (e) If \(T\) is linear, then nullity \((T)+\operatorname{rank}(T)=\operatorname{dim}(W)\). (f) If \(\mathrm{T}\) is linear, then \(\mathrm{T}\) carries linearly independent subsets of \(\mathrm{V}\) onto linearly independent subsets of W. (g) If \(\mathrm{T}, \mathrm{U}: \mathrm{V} \rightarrow \mathrm{W}\) are both linear and agree on a basis for \(\mathrm{V}\), then \(T=U\). (h) Given \(x_{1}, x_{2} \in \mathrm{V}\) and \(y_{1}, y_{2} \in \mathrm{W}\), there exists a linear transformation \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) such that \(\mathrm{T}\left(x_{1}\right)=y_{1}\) and \(\mathrm{T}\left(x_{2}\right)=y_{2}\). For Exercises 2 through 6, prove that \(\mathrm{T}\) is a linear transformation, and find bases for both \(N(T)\) and \(R(T)\). Then compute the nullity and rank of \(T\), and verify the dimension theorem. Finally, use the appropriate theorems in this section to determine whether \(\mathrm{T}\) is one-to-one or onto.

In \(\mathrm{R}^{2}\), let \(L\) be the line \(y=m x\), where \(m \neq 0\). Find an expression for \(\mathrm{T}(x, y)\), where (a) \(\mathrm{T}\) is the reflection of \(\mathrm{R}^{2}\) about \(L\). (b) \(\mathrm{T}\) is the projection on \(L\) along the line perpendicular to \(L\). (See the definition of projection in the exercises of Section 2.1.)

Let \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) be linear, \(b \in \mathrm{W}\), and \(K=\\{x \in \mathrm{V}: \mathrm{T}(x)=b\\}\) be nonempty. Prove that if \(s \in K\), then \(K=\\{s\\}+\mathrm{N}(\mathrm{T})\). (See page 22 for the definition of the sum of subsets.) The following definition is used in Exercises 25-28 and in Exercise 31 . Definition. Let \(\mathrm{V}\) be a vector space and \(\mathrm{W}_{1}\) and \(\mathrm{W}_{2}\) be subspaces of \(\mathrm{V}\) such that \(\mathrm{V}=\mathrm{W}_{1} \oplus \mathrm{W}_{2}\). (Recall the definition of direct sum given on page 22.) The function \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{V}\) defined by \(\mathrm{T}(x)=x_{1}\) where \(x=x_{1}+x_{2}\) with \(x_{1} \in \mathrm{W}_{1}\) and \(x_{2} \in \mathrm{W}_{2}\), is called the projection of \(\mathrm{V}\) on \(\mathrm{W}_{1}\) or the projection on \(\mathrm{W}_{1}\) along \(\mathrm{W}_{2}\).
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