Show $T: \mathcal{P}_n \rightarrow \mathcal{P}_n$ defined by $T(p(x)) = p(x) + p'(x)$ is an isomorphism

Question

Show $T: \mathcal{P}_n \rightarrow \mathcal{P}_n$ defined by $T(p(x)) = p(x) + p'(x)$ is an isomorphism

Let $c_0 + c_1x+c_2x^2 + ... + c_nx^n \in \mathcal{P}_n$

Suppose $T(c_0 + c_1x+c_2x^2 + ... + c_nx^n) = 0$

$\Longrightarrow(c_0 + c_1) + (c_1 + c_2)x + ... + (c_{n-1} + c_n)x^{n-1} + c_nx^n = 0$

$\Longrightarrow c_n = 0$

$\Longrightarrow c_1 = c_2 = ... = c_n = 0$

Thus $kerT = \{\mathbb{0}\}$

And so $T$ is injective.

It follows from injectivity and equal dimensions of domain and codomain, that $T$ must be surjective.

Therefore $T$ is an isomorphism.

Is the above proof valid?

Answer 1

Assuming we already know that these maps are linear transformations, your proof is valid; here's another one in case you're interested. The inverse transformation is $$ T^{-1}(q(x)) = q(x) - q'(x) + q''(x) - \cdots + (-1)^n q^{(n)}(x), $$ as is easy to check once it's written down.