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

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?

• Looks great! Well done
– Rob
Commented Jan 13, 2022 at 5:24
• That's the easier part, but you should first prove that $T$ is a linear transformation.
– dxiv
Commented Jan 13, 2022 at 5:26
• Surely this is the hard part? It follows from linearity of the derivative of that $T$ is linear. Commented Jan 13, 2022 at 5:30
• @AlfredYerger I meant linearity is the easier part, but the wording didn't come out quite right.
– dxiv
Commented Jan 13, 2022 at 5:32
• Careful: $T(c_0+c_1x+c_2x^2+\cdots+c_nx^n) = (c_0+c_1)+(c_1+2c_2)x+\cdots+(c_{n-1}+nc_n)x^{n-1}+c_nx^n$. Commented Jan 13, 2022 at 5:33

1 Answer

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.