- #1

mathmari

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Let $\mathbb{K}$ be a field, $V,W$ finite dimensional $\mathbb{K}$-vector spaces and $\Phi:V\rightarrow W$ a linear map.

I want to show that the following two propositions are equivalent:

- $\Phi$ is surjective
- For each linear form $\beta:W\rightarrow \mathbb{K}$ it holds:

Is $\beta\circ\Phi:V\rightarrow \mathbb{K}$ the zero map, then $\beta$ is the zero map.

$(1)\Rightarrow (2)$

The map $\Phi$ is surjective.

Let $\beta : W\rightarrow \mathbb{K}$ be a linear form.

If $\beta\circ \Phi:V\rightarrow \mathbb{K}$ is the zero map, then it holds that $(\beta \circ \Phi)(v)=0$ for all $v\in V$.

We assume that $\beta$ is not the zero map. Then there is a $w\in W$ such that $\beta (w)\neq 0$.

Since $\Phi$ is surjective, there is a $v\in V$ with $\Phi (v)=w$, for this $v$ it holds \begin{equation*}\beta (\Phi (v))=\beta (w) \Rightarrow (\beta \circ \Phi )(v)=\beta (w)\Rightarrow 0=\beta (w)\neq 0\end{equation*} a contradiction.

So the assumption is wrong, therefore $\beta$ must be the zero map.

Is this direction correct? Could we improve something? (Wondering)

$(2)\Rightarrow (1)$

Is $\beta\circ\Phi:V\rightarrow \mathbb{K}$ the zero map, then $\beta$ is the zero map.

We assume that $\Phi$ is not surjective.

Could you give me a hint for that direction? (Wondering)