TITLE: Does that fact that the dual map is zero imply that the map is zero?
QUESTION [3 upvotes]: Let $X$ and $Y$ be vector spaces over a field $\mathbb{F}$ and let $f:X\to Y$ be a linear map. If the dual map $f^*: Y^*\to X^∗:y^*\mapsto y^*\circ f$ is the zero map, is  the original map $f$ the zero map too?

REPLY [2 votes]: Yes. If $x\in X$ and $f(x)\ne 0$, then let $\phi\in Y^*$ be a linear form for which $\phi(f(x))\ne 0$ (to see that this is possible, pick a basis of $Y$ that contains $f(x)$). Then $f^*(\phi)=f\circ \phi\ne 0$ (because it maps $x$ to something nonozero).

REPLY [0 votes]: Suppose $f$ is not the zero map, so $f(x)=y\neq 0$ say. Let $g$ be a linear functional on $Y$ that maps $y$ to $1$ (you can construct $g$ by extending $\{ y\}$ to a basis of $Y$ and taking the dual basis). Then $(f^*(g))(x)=g(f(x))=g(y)=1$, so $f^*(g)$ is not the zero functional, so $f^*$ is not the zero map.
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