I have the following question here.
A certain linear transformation $\varphi:P_3 \rightarrow\mathbb{R}^2$
maps the polynomial $x^3+3x^2+5x+7$ to $\begin{pmatrix}0 \\ 0\\
\end{pmatrix}$ and $x^3+2x^2+3x+4$ to $\begin{pmatrix}1 \\ 0\\\end{pmatrix}$. What can we conclude aboout $\varphi$?a) It is injective and surjective.
b) It is injective but not surjective.
c) It is surjective but not injective.
d) It is neither injective nor surjective.
e) It is impossible to decide whether it is surjective, but we know it is not injective.
I'm tempted to say neither. For the transformation to be surjective, $\ker(\varphi)$ must be the zero polynomial but I can't really say that's the case here. I can't even conclude if the transformation is $1-1$ so I don't think it's injective.
For surjectivity, the codomain of the linear transformation (The image) has to describe all of $\mathbb{R}^2$ but I don't think I can really conclude this either.
So as such, my answer choice is D but I don't think this is right.
Can someone offer some guidance?
EDIT: I changed $\mathbb{R}^3$ to $\mathbb{R}^2$.
Best Answer
Firstly, I think you mean in your first paragraph that for the transformation to be injective, $\operatorname{ker} \varphi$ must be the zero polynomial. This is a true statement, so the fact that $\varphi$ maps $x^3+3x^2+5x+7$ to $(0,0)$ tells us that $\varphi$ is not injective.
Secondly, for surjectivity, the image of the linear transformation has to be all of $\mathbb R^2$. The only information we have here is that $\varphi$ maps $x^3+2x^2+3x+4$ to $(1,0)$, so your reasoning is correct that we can't necessarily conclude that $\varphi$ is surjective.
Now can you conclude?