Exercise 3.F.6

Proof. _(a)_ Suppose $v_1,\dots ,v_m$ spans $V$. Let $\phi ,\psi \in V^{\prime }$, such that $\Gamma (\phi )=\Gamma (\psi )$. We need to prove that $\phi =\psi$. By definition of $\Gamma$, we have

Hence $\phi (v_j)=\psi (v_j)$ for $j=1,\dots ,m$. By 2.31, it follows that $v_1,\dots ,v_m$ contains a basis and then, by the uniqueness part in 3.5, $\phi$ and $\psi$ are equal. Thus $\Gamma$ is injective.

Conversely, suppose $\Gamma$ is injective. Let $U=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m)$ and suppose by contradiction that $U\ne V$. By _Exercise 4_, there is a _non-zero_ linear functional $\phi$ in $V^{\prime }$ such that $\phi (u)=0$ for every $u\in U$. We have $\Gamma (\phi )=0$, hence $\phi \in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\Gamma$. But $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\Gamma =0$ (since $\Gamma$ is injective), we get a contradiction.

_(b)_ Suppose $v_1,\dots ,v_m$ is linearly independent. We can extend it to a basis $v_1,\dots ,v_m,u_1,\dots ,u_n$ of $V$. Let $x=(x_1,\dots ,x_m)$ be any vector in ${𝔽}^{\text{⋗}}$. Define $\phi \in V^{\prime }$ by

By 3.5 $\phi$ is a well defined linear map. And we have

Since $x$ was arbitrary, it follows that $\Gamma$ is surjective.

Conversely, suppose $\Gamma$ is surjective. Let $U=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_m)$. We need to prove that $\dim <!--\; FUNCTION\; APPLICATION\; -->U=m$. We have

Where the first equation follows from 3.106, the second from 3.95, the third from the Fundamental Theorem of Linear Maps and the fourth because $\Gamma$ is surjective.

Now it suffices to show that $\dim <!--\; FUNCTION\; APPLICATION\; -->U^0=\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\Gamma$.

Suppose $\phi \in U^0$. Clearly $\phi \in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\Gamma$, thus $U^0\subset \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\Gamma$. To show the inclusion in the other direction, suppose $\phi \in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\Gamma$. Let $u\in U$. There are scalars $a_1,\dots ,a_m$ such that $u=a_1{v}_1+\cdots +a_m{v}_m$. Thus,

Since $u$ was arbitrary, we have $\phi \in U^0$. Hence $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\Gamma \subset U^0$. Therefore $U^0=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\Gamma$ and $\dim <!--\; FUNCTION\; APPLICATION\; -->U^0=\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\Gamma$ as desired. □