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Chapter 11: Problem 20

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 \(k\).

Short Answer

Expert verified
We are given that for all linear functionals \(\phi \in V^*\), if \(\phi(u) = 0\), then \(\phi(v) = 0\). We need to show that \(v=ku\) for some scalar \(k\). We construct a scalar \(k\) such that \(v = u - ku\), and then rewrite \(v\) as \(v = ku\). This proves that \(v=ku\) for some scalar \(k\).

Step by step solution

01

Linear functional definition

A linear functional \(\phi\) on a vector space \(V\) is a linear map from \(V\) to its scalar field (usually denoted as \(\mathbb{F}\)). That is, \(\phi: V \rightarrow \mathbb{F}\) such that for any \(x, y \in V\) and any scalar \(a\), we have: 1. \(\phi(x+y) = \phi(x) + \phi(y)\) 2. \(\phi(ax) = a\phi(x)\)
02

Using the given condition

Now, the given condition states that if \(\phi(u) = 0\), then \(\phi(v)=0\) for all linear functionals \(\phi \in V^*\). Let's consider an arbitrary linear functional \(\phi\in V^*\). Since \(\phi(u)=0\), the given condition implies that \(\phi(v)=0\).
03

Relating \(u\) and \(v\)

Since \(\phi(u) = \phi(v) = 0\), we can add the first property of linear functionals and write: \[\phi(u) - \phi(v) = \phi(u - v) = 0\] We can rewrite this equation as: \[\phi(u-v)=0\]
04

Consider an arbitrary linear functional \(\psi \in V^*\) such that \(\psi(u) \neq 0\)

Now let's consider another arbitrary linear functional \(\psi\in V^*\) such that \(\psi(u) \neq 0\). Note that the existence of such \(\psi\) does not contradict the given condition in the problem. Since the given condition implies that \(\psi(v) = 0\), we can apply the property of linear functionals and write: \[\psi(u) - \psi(v) = \psi(u - v) \neq 0\] Comparing this result with our earlier result that \(\phi(u-v) = 0\), we have that: \[\phi(u-v) = 0 \text{ and } \psi(u-v) \neq 0\]
05

Construct the scalar \(k\)

Now since \(\phi(u-v) = 0\), let's define a scalar \(k\) such that \(v = u - ku\). We can replace this expression of \(k\) back to the equation \(\phi(u-v) = 0\) to get: \[\phi(u - ku) = 0\]
06

Concluding the result

So, we have found a scalar \(k\) such that \(v = u - ku\). Thus, we can rewrite \(v\) as: \[v = ku\] This completes our proof that \(v = ku\) for some scalar \(k\).

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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 \(V\) be the vector space of polynomials over \(K\). For \(a \in K\), define \(\phi_{a}: V \rightarrow K\) by \(\phi_{a}(f(t))=f(a) .\) Show that \((\mathrm{a}) \phi_{a}\) is linear; \((\mathrm{b})\) if \(a \neq b,\) then \(\phi_{a} \neq \phi_{b}\).

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 \(\phi_{2}=\mathbf{0}\).

Let \(U\) and \(W\) be subspaces of \(V\). Prove that \((U+W)^{0}=U^{0} \cap W^{0}\).

Prove Theorem 11.2: Let \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) be a basis of \(V\) and let \(\left\\{\phi_{1}, \ldots, \phi_{n}\right\\}\) be the dual basis in \(V^{*} .\) For any \(u \in V\) and any \(\sigma \in V^{*},(\mathrm{i}) u=\sum_{i} \phi_{i}(u) v_{i}\) (ii) \(\sigma=\sum_{i} \phi\left(v_{i}\right) \phi_{i}\)
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