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Chapter 10: Problem 1

Suppose \(T: V \rightarrow V\) is linear. Show that each of the following is invariant under \(T\) : (a) \(\\{0\\}\), (b) \(V\), (c) kernel of \(T\), (d) image of \(T\). (a) We have \(T(0)=0 \in\\{0\\}\); hence, \(\\{0\\}\) is invariant under \(T\). (b) For every \(v \in V, T(v) \in V\); hence, \(V\) is invariant under \(T\). (c) Let \(u \in \operatorname{Ker} T\). Then \(T(u)=0 \in\) Ker \(T\) because the kernel of \(T\) is a subspace of \(V\). Thus, Ker \(T\) is invariant under \(T\). (d) Because \(T(v) \in \operatorname{Im} T\) for every \(v \in V\), it is certainly true when \(v \in \operatorname{Im} T\). Hence, the image of \(T\) is invariant under \(T\).

Short Answer

Expert verified
(a) \(T(0) = 0 \in\{0\}\), so \(\{0\}\) is invariant under \(T\). (b) \(\forall v \in V, T(v) \in V\), so \(V\) is invariant under \(T\). (c) \(\forall u \in \operatorname{Ker} T, T(u) = 0 \in\operatorname{Ker} T\), so \( \operatorname{Ker} T\) is invariant under \(T\). (d) \(\forall v \in \operatorname{Im} T, v = T(u) \text{ for some } u \in V \Rightarrow T(v) \in \operatorname{Im} T\). Hence, \(\operatorname{Im} T\) is invariant under \(T\).

Step by step solution

01

Part (a) - Invariance of {0}

To show that the set containing only the zero vector, \({0}\), is invariant under \(T\), we need to demonstrate that when we apply the transformation \(T\) to the zero vector, the result is still in \({0}\). Since \(T\) is a linear transformation, we know that \(T(0) = 0\). Thus, \({0}\) is invariant under \(T\).
02

Part (b) - Invariance of V

To show that the entire vector space \(V\) is invariant under \(T\), we need to demonstrate that for every vector \(v \in V\), the result of applying the transformation \(T(v)\) is also in \(V\). Since \(T\) is a linear transformation mapping vectors in \(V\) to other vectors in \(V\), this is always true. Therefore, \(V\) is invariant under \(T\).
03

Part (c) - Invariance of the Kernel of T

To show that the kernel of \(T\) (denoted as \(\operatorname{Ker} T\)) is invariant under \(T\), we need to demonstrate that when a vector \(u \in \operatorname{Ker} T\) is acted upon by the transformation, the result is still in \(\operatorname{Ker} T\). By definition, the kernel of a linear transformation is the set of vectors that map to the zero vector: \(\operatorname{Ker} T = \{u \in V : T(u) = 0\}\). Since \(T(u) = 0\) for all \(u \in \operatorname{Ker} T\), by definition, the kernel is invariant under \(T\).
04

Part (d) - Invariance of the Image of T

To show that the image of \(T\) (denoted as \(\operatorname{Im} T\)) is invariant under \(T\), we need to demonstrate that when a vector \(v \in \operatorname{Im} T\) is acted upon by the transformation, the result is still in \(\operatorname{Im} T\). By definition, the image of a linear transformation is the set of vectors that can be obtained by applying the transformation to some vector \(u \in V\): \(\operatorname{Im} T = \{v \in V : v = T(u) \text{ for some } u \in V\}\). Since every vector in \(V\) is mapped to a vector in the image when acted upon by \(T\), the image must be invariant under \(T\).

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Most popular questions from this chapter

Suppose dim \(V=n\). Show that \(T: V \rightarrow V\) has a triangular matrix representation if and only if there exist \(T\) -invariant subspaces \(W_{1} \subset W_{2} \subset \cdots \subset W_{n}=V\) for which \(\operatorname{dim} W_{k}=k, k=1, \ldots, n\).

Suppose \(E: V \rightarrow V\) is a projection. Prove (i) \(I-E\) is a projection and \(V=\operatorname{Im} E \oplus \operatorname{Im}(I-E)\), (ii) \(I+E\) is invertible (if \(1+1 \neq 0\) ).

Prove Theorem 10.9: A linear operator \(T: V \rightarrow V\) has a diagonal matrix representation if and only if its minimal polynomal \(m(t)\) is a product of distinct linear polynomials.Suppose \(m(t)\) is a product of distinct linear polynomials, say, $$ m(t)=\left(t-\lambda_{1}\right)\left(t-\lambda_{2}\right) \cdots\left(t-\lambda_{r}\right) $$ where the \(\lambda_{i}\) are distinct scalars. By the Primary Decomposition Theorem, \(V\) is the direct sum of subspaces \(W_{1}, \ldots, W_{r}\), where \(W_{i}=\operatorname{Ker}\left(T-\lambda_{i} I\right) .\) Thus, if \(v \in W_{i}\), then \(\left(T-\lambda_{i} I\right)(v)=0\) or \(T(v)=\lambda_{i} v .\) In other words, every vector in \(W_{i}\) is an eigenvector belonging to the eigenvalue \(\lambda_{i} .\) By Theorem \(10.4\), the union of bases for \(W_{1}, \ldots, W_{r}\) is a basis of \(V .\) This basis consists of eigenvectors, and so \(T\) is diagonalizable. Conversely, suppose \(T\) is diagonalizable (i.e., \(V\) has a basis consisting of eigenvectors of \(T\) ). Let \(\lambda_{1}, \ldots, \lambda_{s}\) be the distinct eigenvalues of \(T\). Then the operator $$ f(T)=\left(T-\lambda_{1} I\right)\left(T-\lambda_{2} I\right) \cdots\left(T-\lambda_{s} I\right) $$ maps each basis vector into \(0 .\) Thus, \(f(T)=0\), and hence, the minimal polynomial \(m(t)\) of \(T\) divides the polynomial $$ f(t)=\left(t-\lambda_{1}\right)\left(t-\lambda_{2}\right) \cdots\left(t-\lambda_{s} I\right) $$ Accordingly, \(m(t)\) is a product of distinct linear polynomials.

Let \(V\) be a vector space and \(W\) a subspace of \(V\). Show that the natural map \(\eta: V \rightarrow V / W\), defined by \(\eta(v)=v+W,\) is linear.

Find all invariant subspaces of \(A=\left[\begin{array}{ll}2 & -5 \\ 1 & -2\end{array}\right]\) viewed as an operator on \(\mathbf{R}^{2}\).
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