91Ó°ÊÓ

Let \(V\) be a vector space over a field \(F\). The dual space \(V^{*}\) is the set of all linear functionals on \(V\), i.e., the set of all linear maps from \(V\) to \(F\). The linearity property implies that, for all \(v, w \in V\) and scalars \(c \in F\), we have \(\phi(v+w) = \phi(v) + \phi(w)\) and \(\phi(cv) = c\phi(v)\) for some linear functional \(\phi \in V^{*}\).

The given condition is: If \(\phi(v)=0\) implies \(\sigma(v)=0\) for all \(v \in V\). In its contrapositive form, the condition is: If \(\sigma(v) \neq 0\), then \(\phi(v) \neq 0\).

From the contrapositive form of the given condition, we can see that the null space of \(\sigma\) is contained in the null space of \(\phi\), denoted as \(\text{null}(\sigma) \subseteq \text{null}(\phi)\). Since both linear functionals are in \(V^{*}\), this implies that their null spaces are the same, i.e., \(\text{null}(\sigma) = \text{null}(\phi)\).

To prove this, we will first show that the dimension of the quotient space \(V/\text{null}(\phi)\) is 1. We know that \(\text{null}(\phi) = \text{null}(\sigma)\). Next, consider any non-zero element \(v \in V\) such that \(v \notin \text{null}(\phi)\). We know that for any scalar \(k \in F\), the set \(\{kv: k \in F\}\) forms a one-dimensional subspace of \(V\). Now, if we consider the map \(\psi: V/\text{null}(\phi) \to F\) defined by \(\psi(v+\text{null}(\phi)) = \sigma(v)\), it is well-defined, since if \(v_1+\text{null}(\phi) = v_2+\text{null}(\phi)\), we have \(v_1-v_2 \in\text{null}(\phi)\), which implies that \(\sigma(v_1 - v_2) = 0\), so \(\sigma(v_1) = \sigma(v_2)\). Moreover, \(\psi\) is a linear functional in \(V^{*}\). Then, from the one-dimensional subspace generated by \(\{kv: k \in F\}\), there exists a scalar \(k'\) such that \(\psi(v+\text{null}(\phi)) = k'\phi(v)\). Therefore, we have that \(\sigma(v) = k'\phi(v)\) for all \(v \in V\) with \(v \notin \text{null}(\phi)\). Lastly, for any \(v \in \text{null}(\phi)\), we trivially have \(\sigma(v) = 0 = k\phi(v)\) for any scalar \(k\). Hence, we conclude that \(\sigma = k \phi\) for some scalar \(k\).