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Let $V$ be a vector space (not necessarily finite dimensional) over $\mathbb{K}$ ($\mathbb{R}$ or $\mathbb{C}$), and $\phi_1$ and $\phi_2$ be two independent elements of $V^*$. Consider the map $\Phi:V\longrightarrow \mathbb{K}^2$ given by $\Phi(x)=(\phi_1(x),\phi_2(x))$. The question is to show that $\Phi$ is surjective. Now, since each $\phi_i$ is non-zero, it follows that each is surjective, but how do we conclude that $\Phi$ is surjective?

This is what I have managed so far. We need to find $x_1$ and $x_2$ in $V$ such that $\phi_i(x_j)=\delta_{ij}$. If V were finite dimensional (or reflexive, more generally) this would following by taking the corresponding dual elements of $\phi_i$ in $V^{**}$. We can find $x_1$ and $x_2$ such that $\phi_i(x_i)=1$. But how do we ensure that $\phi_i(x_j)=0$ for $i\neq j$?

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Suppose $\Phi$ were not surjective. What can you say about $\operatorname{im}\Phi$ then? –  Daniel Fischer Oct 10 '13 at 10:39
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Im $\Phi$ would then be a proper non-zero subspace of $\mathbb{K}^2.$ So it would have to be a one dimensional subspace of $\mathbb{K}^2$ –  Arundhathi Oct 10 '13 at 10:47
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And if you then take a nonzero linear form on $\mathbb{K}^2$ that annihilates the image, that gives you ... what? –  Daniel Fischer Oct 10 '13 at 11:06
    
Since it is a proper closed subspace of $\mathbb{K}^2$ we get a non-zero linear form, say $\psi$ from $\mathbb{K}^2$ to $\mathbb{K}$, which must be surjective and whose kernel=Im $\phi$. So for a given $(y_1,y_2) \in \mathbb{K}^2$, consider $\psi((y_1,y_2))$. This must have a pre-image in $X$, say $x$. Then since $\psi \circ \phi (x)=\psi((y_1,y_2))$ and $\psi$ annihilates Im $\phi$, $\psi((y_1,y_2))=0$ whence $(y_1,y_2)$ is in Im $\phi$. Is that right? –  Arundhathi Oct 10 '13 at 11:25
    
You know the structure of linear functionals on $\mathbb{K}^2$. So $\psi((y_1,y_2)) = a\cdot y_1 + b\cdot y_2$ for some $a,b\in\mathbb{K}$, not both $0$. Then, how can you write $\psi\circ \Phi$? –  Daniel Fischer Oct 10 '13 at 11:35
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