/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Let \(T: V \rightarrow V\) be li... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 15

Let \(T: V \rightarrow V\) be linear. Let \(U=\) Ker \(T^{i}\) and \(W=\) Ker \(T^{i+1}\). Show that (a) \(U \subseteq W\) (b) \(T(W) \subseteq U\). (a) Suppose \(u \in U=\operatorname{Ker} T^{i}\). Then \(T^{i}(u)=0\) and so \(T^{i+1}(u)=T\left(T^{i}(u)\right)=T(0)=0\). Thus, \(u \in \operatorname{Ker} T^{i+1}=W\). But this is true for every \(u \in U\); hence, \(U \subseteq W\) (b) Similarly, if \(w \in W=\operatorname{Ker} T^{i+1}\), then \(T^{i+1}(w)=0\). Thus, \(T^{i+1}(w)=T^{i}(T(w))=T^{i}(0)=0\) and so \(T(W) \subseteq U .\)

Short Answer

Expert verified
(a) Since \(T^i(u) = 0\) for all \(u \in U\), we have \(T^{i+1}(u) = T(0) = 0\). Thus, \(u \in W\), and we conclude that \(U \subseteq W\). (b) For all \(w \in W\), we have \(T^{i+1}(w) = 0\), so \(T^i(T(w)) = 0\). Thus, \(T(w) \in U\), and we conclude that \(T(W) \subseteq U\).

Step by step solution

01

(a) Prove that \(U \subseteq W\)

Let \(u \in U\). By definition, \(U = \operatorname{Ker}(T^i)\), which means that \(T^i(u) = 0\). Now we want to show that \(u \in W\). By definition, \(W = \operatorname{Ker}(T^{i+1})\). So we need to show that \(T^{i+1}(u) = 0\). Notice that \(T^{i+1}(u) = T\left(T^i(u)\right)\). Since we know that \(T^i(u) = 0\), this expression becomes \(T^{i+1}(u) = T(0)\). By the properties of linear transformations, we know that \(T(0) = 0\). Thus, \(T^{i+1}(u) = 0\), which means \(u \in W\). Since this is true for every \(u \in U\), we have proved that \(U \subseteq W\).
02

(b) Prove that \(T(W) \subseteq U\)

Let \(w \in W\). By definition, \(W = \operatorname{Ker}(T^{i+1})\), which means that \(T^{i+1}(w) = 0\). Now we want to show that \(T(w) \in U\). By definition, \(U = \operatorname{Ker}(T^i)\). So we need to show that \(T^i(T(w)) = 0\). Notice that \(T^{i+1}(w) = T^i(T(w))\). Since we know that \(T^{i+1}(w) = 0\), this expression becomes \(0 = T^i(T(w))\). This means that \(T(w) \in \operatorname{Ker}(T^i) = U\). Since this is true for every \(w \in W\), we have proved that \(T(W) \subseteq U\).

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Most popular questions from this chapter

Let \(W\) be a subspace of a vector space \(V .\) Suppose \(\left\\{w_{1}, \ldots, w_{r}\right\\}\) is a basis of \(W\) and the set of cosets \(\left\\{\bar{v}_{1}, \ldots, \bar{v}_{s}\right\\}\), where \(\bar{v}_{j}=v_{j}+W\), is a basis of the quotient space. Show that the set of vectors \(B=\left\\{v_{1}, \ldots, v_{s}, w_{1}, \ldots, w_{r}\right\\}\) is a basis of \(V .\) Thus, \(\operatorname{dim} V=\operatorname{dim} W+\operatorname{dim}(V / W)\) Suppose \(u \in V\). Because \(\left\\{\bar{v}_{j}\right\\}\) is a basis of \(V / W\) $$ \bar{u}=u+W=a_{1} \bar{v}_{1}+a_{2} \bar{v}_{2}+\cdots+a_{s} \bar{v}_{s} $$ Hence, \(u=a_{1} v_{1}+\cdots+a_{s} v_{s}+w\), where \(w \in W\). Since \(\left\\{w_{i}\right\\}\) is a basis of \(W\), $$ u=a_{1} v_{1}+\cdots+a_{s} v_{s}+b_{1} w_{1}+\cdots+b_{r} w_{r} $$Accordingly, \(B\) spans \(V\). We now show that \(B\) is linearly independent. Suppose $$ \begin{gathered} c_{1} v_{1}+\cdots+c_{s} v_{s}+d_{1} w_{1}+\cdots+d_{r} w_{r}=0 \\ c_{1} \bar{v}_{1}+\cdots+c_{s} \bar{v}_{s}=\overline{0}=W \end{gathered} $$ Because \(\left\\{\bar{v}_{j}\right\\}\) is independent, the \(c\) 's are all 0 . Substituting into (1), we find \(d_{1} w_{1}+\cdots+d_{r} w_{r}=0 .\) Because \(\left\\{w_{i}\right\\}\) is independent, the \(d\) 's are all \(0 .\) Thus, \(B\) is linearly independent and therefore a basis of \(V .\)

Let \(T: V\) be linear. Let \(X=\operatorname{Ker} T^{i-2}, Y=\operatorname{Ker} T^{i-1}, Z=\operatorname{Ker} T^{i} .\) Therefore (Problem 10.14), \(X \subseteq Y \subseteq Z .\) Suppose $$ \left\\{u_{1}, \ldots, u_{r}\right\\}, \quad\left\\{u_{1}, \ldots, u_{r}, v_{1}, \ldots, v_{s}\right\\}, \quad\left\\{u_{1}, \ldots, u_{r}, v_{1}, \ldots, v_{s}, w_{1}, \ldots, w_{t}\right\\} $$ are bases of \(X, Y, Z\), respectively. Show that $$ S=\left\\{u_{1}, \ldots, u_{r}, T\left(w_{1}\right), \ldots, T\left(w_{t}\right)\right\\} $$ is contained in \(Y\) and is linearly independent. By Problem 10.14, \(T(Z) \subseteq Y\), and hence \(S \subseteq Y\). Now suppose \(S\) is linearly dependent. Then there exists a relation $$ a_{1} u_{1}+\cdots+a_{r} u_{r}+b_{1} T\left(w_{1}\right)+\cdots+b_{t} T\left(w_{t}\right)=0 $$ where at least one coefficient is not zero. Furthermore, because \(\left\\{u_{i}\right\\}\) is independent, at least one of the \(b_{k}\) must be nonzero. Transposing, we find $$ \begin{gathered} b_{1} T\left(w_{1}\right)+\cdots+b_{t} T\left(w_{t}\right)=-a_{1} u_{1}-\cdots-a_{r} u_{r} \in X=\operatorname{Ker} T^{i-2} \\ T^{i-2}\left(b_{1} T\left(w_{1}\right)+\cdots+b_{t} T\left(w_{t}\right)\right)=0 \end{gathered} $$ Thus, \(\quad T^{i-1}\left(b_{1} w_{1}+\cdots+b_{t} w_{t}\right)=0, \quad\) and so \(\quad b_{1} w_{1}+\cdots+b_{t} w_{t} \in Y=\operatorname{Ker} T^{i-1}\) Because \(\left\\{u_{i}, v_{j}\right\\}\) generates \(Y\), we obtain a relation among the \(u_{i}, v_{j}, w_{k}\) where one of the coefficients (i.e., one of the \(b_{k}\) ) is not zero. This contradicts the fact that \(\left\\{u_{i}, v_{j}, w_{k}\right\\}\) is independent. Hence, \(S\) must also be independent.

Let \(W\) be invariant under \(T_{1}: V \rightarrow V\) and \(T_{2}: V \rightarrow V\). Prove \(W\) is also invariant under \(T_{1}+T_{2}\) and \(T_{1} T_{2}\).

Prove the following: The cosets of \(W\) in \(V\) partition \(V\) into mutually disjoint sets. That is, (a) Any two cosets \(u+W\) and \(v+W\) are either identical or disjoint. (b) Each \(v \in V\) belongs to a coset; in fact, \(v \in v+W\) Furthermore, \(u+W=v+W\) if and only if \(u-v \in W\), and so \((v+w)+W=v+W\) for any \(w \in W\) Let \(v \in V\). Because \(0 \in W\), we have \(v=v+0 \in v+W\), which proves (b). Now suppose the cosets \(u+W\) and \(v+W\) are not disjoint; say, the vector \(x\) belongs to both \(u+W\) and \(v+W\). Then \(u-x \in W\) and \(x-v \in W\). The proof of (a) is complete if we show that \(u+W=v+W\). Let \(u+w_{0}\) be any element in the coset \(u+W\). Because \(u-x, x-v, w_{0}\) belongs to \(W\), $$ \left(u+w_{0}\right)-v=(u-x)+(x-v)+w_{0} \in W $$ Thus, \(u+w_{0} \in v+W\), and hence the cost \(u+W\) is contained in the coset \(v+W\). Similarly, \(v+W\) is contained in \(u+W\), and so \(u+W=v+W\). The last statement follows from the fact that \(u+W=v+W\) if and only if \(u \in v+W\), and, by Problem \(10.21\), this is equivalent to \(u-v \in W\)

Suppose \(V=U \oplus W\) and suppose \(T: V \rightarrow V\) is linear. Show that \(U\) and \(W\) are both \(T\) -invariant if and only if \(T E=E T,\) where \(E\) is the projection of \(V\) into \(U\).
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