/*! 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 6 Suppose \(v \in V, v \neq 0,\) a... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 6

Suppose \(v \in V, v \neq 0,\) and \(\operatorname{dim} V=n .\) Show that there exists \(\phi \in V^{*}\) such that \(\phi(v) \neq 0\).

Short Answer

Expert verified
Given a vector space \(V\) with \(\operatorname{dim}V = n\), and a non-zero vector \(v \in V\), we need to show that there exists a linear functional \(\phi \in V^*\) such that \(\phi(v) \neq 0\). To prove this, we extend \(v\) to a basis for \(V\), \(\{v_n = v, v_1, v_2, \dots, v_{n-1}\}\), and consider the dual basis \(\{\varphi_1, \varphi_2, \dots, \varphi_n\}\) of the dual space \(V^*\). Then, choose \(\phi = \varphi_n\). Evaluating \(\phi(v) = \varphi_n(v) = \varphi_n(v_n) = \delta_{nn} = 1\), we find that \(\phi(v) \neq 0\), as required.

Step by step solution

01

1. Choose a basis for V

Since the dimension of the vector space V is n, we can choose a basis \(\{v_1, v_2, \dots, v_n\}\). First, we want to include the vector v in this basis. For this, extend v to a basis for V. We can do this by adding linearly independent vectors to the set containing v until we have n vectors in total. Finally, let's relabel these vectors to form a new basis: \(\{v_n = v, v_1, v_2, \dots, v_{n-1}\}\).
02

2. Define the linear functional 蠁

Now, consider the dual space of V, denoted by \(V^*\). Recall that the dual space is the set of all linear functionals that map from V to the field F, i.e., functions 蠁: V 鈫� F that satisfy linearity: \(\phi(\alpha u + \beta v) = \alpha \phi(u) + \beta \phi(v)\) for all vectors u, v in V and all scalars 伪, 尾 in F. As the basis for V is \(\{v_n, v_1, v_2, \dots, v_{n-1}\}\), we can define a linear functional 蠁 in \(V^*\) using this basis' dual basis, denoted by \(\{\varphi_1, \varphi_2, \dots, \varphi_n\}\), where \(\varphi_i: V 鈫� F\) such that \(\varphi_i(v_j) = \delta_{ij}\) (the Kronecker delta, which is 1 if i = j and 0 otherwise).
03

3. Show that 蠁(v) 鈮� 0

Now, we need to show that there exists a linear functional 蠁 in \(V^*\) such that 蠁(v) 鈮� 0. Choose this 蠁 to be the last element of the dual basis, that is, \(\phi = \varphi_n\). Let's evaluate 蠁(v), which is the same as evaluating \(\varphi_n(v)\): \[\varphi_n(v) = \varphi_n(v_n) = \delta_{nn} = 1\] Since \(\varphi_n(v) = 1\), we have found a linear functional 蠁 in \(V^*\) such that 蠁(v) 鈮� 0. This completes the proof.

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Most popular questions from this chapter

Show that if \(\phi \in V^{*}\) annihilates a subset \(S\) of \(V\), then \(\phi\) annihilates the linear span \(L(S)\) of \(S\). Hence, \(S^{0}=[\operatorname{span}(S)]^{0}\).

Let \(W\) be the subspace of \(\mathbf{R}^{4}\) spanned by \((1,2,-3,4), \quad(1,3,-2,6), \quad(1,4,-1,8) .\) Find a basis of the annihilator of \(W\).

Let \(V=\\{a+b t: a, b \in \mathbf{R}\\},\) the vector space of real polynomials of degree \(\leq 1 .\) Find the basis \(\left\\{v_{1}, v_{2}\right\\}\) of \(V\) that is dual to the basis \(\left\\{\phi_{1}, \phi_{2}\right\\}\) of \(V^{*}\) defined by \\[ \phi_{1}(f(t))=\int_{0}^{1} f(t) d t \quad \text { and } \quad \phi_{2}(f(t))=\int_{0}^{2} f(t) d t \\]

Let \(V\) be a vector space over \(\mathbf{R}\). Let \(\phi_{1}, \phi_{2} \in V^{*}\) and suppose \(\sigma: V \rightarrow \mathbf{R},\) defined by \(\sigma(v)=\phi_{1}(v) \phi_{2}(v)\) also belongs to \(V^{*}\). Show that either \(\phi_{1}=\mathbf{0}\) or \(\phi_{2}=\mathbf{0}\).

Suppose \(u, v \in V\) and that \(\phi(u)=0\) implies \(\phi(v)=0\) for all \(\phi \in V^{*} .\) Show that \(v=k u\) for some scalar \(k\).
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