TITLE: Surjective/injective linear function
QUESTION [0 upvotes]: Assuming $U = \Bbb R^3$, $V = \Bbb R^3$ with a canonical structure of $\Bbb R$-vector spaces. If a linear application $f:U\to\Bbb R$ has $\dim(\mathrm{im}(f)) = 3$, then $f$ is surjective and/or injective? Proof.

REPLY [0 votes]: Proof of Surjectivity.
The image of $f$ is a subspace. But $\Bbb R^3$ has no proper 3-dimensional subspaces. So the image is all of $\Bbb R^3$. $\square$
Proof of Injectivity. 
If $f(x)=f(y)$ for $x\not=y$ then $f(e_1)=0$ for $e_1:=x-y\not=0$. If $\{e_1,e_2,e_3\}$ is a basis of $\Bbb R^3$, then a vector $v= v_1e_1+v_2e_2+v_3e_3$ gets mapped to
$$f(v)=v_1f(e_1)+v_2f(e_2)+v_3f(e_3)=0+v_2f(e_2)+v_3f(e_3).$$
So the image is 2-dimensional. Therefore $f$ must be injective. $\square$
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