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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

Let \(\beta\) be an ordered basis for a finite-dimensional vector space \(\mathrm{V}\), and let \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{V}\) be linear. Prove that, for any nonnegative integer \(k\), \(\left[\mathrm{T}^{k}\right]_{\beta}=\left([\mathrm{T}]_{\beta}\right)^{k} .\)

Let \(\mathrm{T}: \mathrm{P}(R) \rightarrow \mathrm{P}(R)\) be defined by \(\mathrm{T}(f(x))=f^{\prime}(x)\). Recall that \(\mathrm{T}\) is linear. Prove that \(\mathrm{T}\) is onto, but not one-to-one.

Using the notation in the definition above, assume that \(T: V \rightarrow V\) is the projection on \(\mathrm{W}_{1}\) along \(\mathrm{W}_{2}\). (a) Prove that \(\mathrm{T}\) is linear and \(\mathrm{W}_{1}=\\{x \in \mathrm{V}: \mathrm{T}(x)=x\\}\). (b) Prove that \(\mathrm{W}_{1}=\mathrm{R}(\mathrm{T})\) and \(\mathrm{W}_{2}=\mathrm{N}(\mathrm{T})\). (c) Describe \(\mathrm{T}\) if \(\mathrm{W}_{1}=\mathrm{V}\). (d) Describe \(T\) if \(W_{1}\) is the zero subspace.

Let \(\mathrm{V}\) be an \(n\)-dimensional vector space, and let \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{V}\) be a linear transformation. Suppose that \(\mathrm{W}\) is a T-invariant subspace of \(\mathrm{V}\) (see the exercises of Section 2.1) having dimension \(k\). Show that there is a basis \(\beta\) for \(\mathrm{V}\) such that \([\mathrm{T}]_{\beta}\) has the form $$ \left(\begin{array}{ll} A & B \\ O & C \end{array}\right), $$ where \(A\) is a \(k \times k\) matrix and \(O\) is the \((n-k) \times k\) zero matrix.

Let \(A\) and \(B\) be matrices for which the product matrix \(A B\) is defined, and let \(u_{j}\) and \(v_{j}\) denote the \(j\) th columns of \(A B\) and \(B\), respectively. If \(v_{p}=c_{1} v_{j_{1}}+c_{2} v_{j_{2}}+\cdots+c_{k} v_{j_{k}}\) for some scalars \(c_{1}, c_{2}, \ldots c_{k}\), prove that \(u_{p}=c_{1} u_{j_{1}}+c_{2} u_{j_{2}}+\cdots+c_{k} u_{j_{k}} .\) Visit goo.gl/sRpves for a solution.
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