Show that $f(y)=0$ for every $y \in Y$. 
If $Y$ is a subspace of a vector space $X$ and $f$ is a linear functional on $X$ such that $f(Y)$ is not the whole scalar field of $X$, show that $f(y)=0$ for all $y \in Y$.

Suppose to the contrary, that there exists some $y_0 \in Y$ such that $f(y_0) = \alpha \neq 0$. Now notice that for any scalar $\beta$ is the underlying scalar field of $X$, we have that $$\beta = \frac{\beta}{\alpha}f(y_0) = f(\frac{\beta}{\alpha}y_0).$$ I am having trouble forming or seeing the contradiction though? Can someone please help point it out to me? 

You know that $f$ is linear, s
$$\text{dim}(\text{ker}(f))+\text{rg}(f)=\dim Y.$$
Because $\text{im}(f)$ is not the whole scalar field, and $\text{im}(f)$ is a vector space (because $f$ is linear) and the scalar field has dimension $1$ :
$$\text{rg}(f)=0.$$
So $\text{dim}(\text{ker}(f))=\dim Y$, and because $\ker(f)\subset Y$, you get
$$\ker(f)=Y$$.
This is exactly $$\forall y \in Y \quad f(y)=0.$$
