A previous exam problem says:
If possible, construct a linear transformation $T$ of $P_2(\mathbb{R})$ (Polynomials of degree less than or equal to $2$) in $\mathbb{R^3}$ such that $T(x^2+1)=(1, -1, 0)$ and $T(x+1)=(0, 2, 1)$.
That was the full question. Now, I think it isn't possible because they should provide me with one more linearly independent vector from $P_2$ and its respective transformation to $\mathbb{R^3}$, so that I could find a correspondence rule to the transformation through that basis. That's because $P_2$ is of dimension $3$, the same as $\mathbb{R^3}$, so I'd need a basis of three vectors from $P_2$ and its corresponding transformed vectors on $\mathbb{R^3}$.
That's my reasoning, am I wrong? Is there any possible way to build such a transformation?
Yes, you are wrong. In other words, yes, it is possible to define such a function. For instance, take$$T(a+bx+cx^2)=(c,2b-c,b).$$It works. I obtained this function as the only linear map $T\colon P_2(\mathbb R)\longrightarrow\mathbb R^3$ such that:
Of course, this $(0,0,0)$ is an arbitrary choice. If I had chosen another vector, I would have got another answer.