For $p(x)\in\mathcal P_2(\mathbb R)$, define $L:\mathcal P_2(\mathbb R)\to\mathbb R^3$ by $$ L(p(x))=[p(0) \space\space p'(0) \space\space p''(0)]^T. $$ Here, $\mathcal P_2(\mathbb R)$ is the set of all real polynomials with degree at most $2$.
I'm asked to verify that $L$ is an isomorphism.
I begin by noticing that only the $p(0)$ element has the possibility of having a nonzero value (the constant term). We can write $$ \mathrm{range}(L)=\mathrm{span} \{[1 \space\space 0 \space\space 0]^T\}, $$ which implies $\mathrm{dim}(\mathrm{range}(L))=1$. However, $\mathrm{dim}(\mathcal P_2(\mathbb R))=3$ so $L$ cannot be surjective and therefore cannot be an isomorphism.
Is my reasoning correct here? This question is part of a larger one and this verification step seems like something that should be simple to show. The fact that I'm getting a different result raises doubts. Thanks!
We know that if $ B $ is a basis of $ \mathcal{P}_2(\mathbb{R}) $, then $ Range(T) = Span(T(B)) $.
Let's take the standard basis $ B = \{1, x, x^2\} $.
You can check that:
$ T(1) = [1, 0, 0]^T $
$ T(x) = [0, 1, 0]^T $
$ T(x^2) = [0, 0, 2]^T $
The set $ C=\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix}\} $ is linearly independent (check!) and has exactly $ 3 $ elements. $ \dim(\mathbb{R}^3)=3$, which means $ C $ is a basis for $ \mathbb{R}^3 $, and therefore $ Range(T)= \mathbb{R}^3 $.
There is a theorem stating that if $ V $ and $ W $ are two finite-dimensional vector spaces of dimension $ n $ and there exists a linear transformation $ S : V \to W $ such that $ Range(S) = W $, then $ S $ is an isomorphism.
$ \dim(\mathcal{P}_2(\mathbb{R}))= \dim(\mathbb{R}^3) $ and therefore, by the theorem above, we know that $ T $ is indeed an isomorphism.