# Constructing a linear transformation $T: \mathbb F[x] \to \mathbb F[x]$ satisfying $R(T) = N(T)$

Let $$T : V \to V$$ be linear. Is there a general way to find $$T$$ such that $$N(T)= R(T)$$ given any vector space $$V$$?

I know that $$N(T) = R(T)$$ implies that $$T^2(x) = 0$$ for all $$x\in\mathbb{F}[x]$$ and that $$R(T)$$ and $$N(T)$$ should have an infinite dimension and that $$T^2 = 0\Rightarrow R(T) \subset N(T).$$

I know that a linear mapping from $$\mathbb{R}^2$$ to $$\mathbb{R}^2$$ that satisfies this constraint is $$(a,b) \mapsto (b, 0),$$ but I can't find a general way to find $$T$$ such that $$N(T) = R(T)$$ for the vector space $$\mathbb{R}^{[0,1]},$$ for example.

• $T^2=0\implies \operatorname R(T)\subset \operatorname N(T)$.
– user403337
Feb 15 '20 at 18:15
• I know that, but how can that help?
– user739612
Feb 15 '20 at 18:20

My standard example of a linear map with $$N(T) = R(T)$$ is $$\begin{bmatrix}0&1\\0&0\end{bmatrix}$$. Of course this operates on $$\mathbb{R}^2$$, and you want to operate on $$F[x]$$. You could think of $$F[x]$$ as $$\langle 1,x\rangle \oplus\langle x^2,x^3\rangle \oplus\langle x^4,x^5\rangle \oplus \dots$$ and have it act like $$\begin{bmatrix}0&1\\0&0\end{bmatrix}$$ on each two-dimensional subspace. It would be kind of ugly to describe but it would work.

You could also try to find a "nicer" way to write the same transformation by observing its action on odd and even exponents and using the operators that pick out the odd and even parts of a function:

$$E(p)(x) = (p(x) + p(-x))/2$$ $$O(p)(x) = (p(x) - p(-x))/2$$

(I'll leave this to you. Ask if you want more ideas on how to do this.) Honestly, it'll probably be harder to prove that $$T$$ behaves the way you want with the "nicer" definition, but it will look cleaner.

Hint One way to approach this problem is to generalize your example, $$V = \Bbb R^2 \qquad T = \pmatrix{0&1\\0&0}$$ by viewing $$V$$ as a direct sum of two subspaces, $$V = V_0 \oplus V_1 ,$$ and finding a transformation $$T$$ whose decomposition with respect to that decomposition has the block form $$T = \pmatrix{0&A\\0&0}$$ for which $$A$$ is invertible. In practice, it might be easier to construct an $$A$$, e.g., by finding an invertible transformation $$S$$ that interchanges $$V_0, V_1$$, so that it has block form $$S = \pmatrix{0&A\\B&0} ,$$ satisfying the condition that $$A$$ is invertible, and declaring $$T = \pi_0 \circ S ,$$ where $$\pi_0 : V \to V$$ is the projection onto $$V_1$$, i.e., the map with decomposition $$\pi_0 = \pmatrix{\operatorname{id_{V_0}}&0\\0&0} .$$

A natural decomposition of $$\Bbb F[x]$$ into two infinite dimensional subspaces is into even and odd polynomials.

If we take $$V_0$$ to be the subspace of even polynomials and $$V_1$$ to be the subspace of odd polynomials, then the derivative operator $$D := p \mapsto p'$$ exchanges $$V_0$$ and $$V_1$$, and its restriction $$V_1 \to V_0$$ is invertible (since $$\ker D$$ consists of the constant polynomials, but the only constant, odd polynoimal is $$0$$), so we make $$S = D$$. For $$\operatorname{char} \mathbb F = 2$$, we can write $$\pi_0$$ as $$p(x) \mapsto \frac{1}{2}[p(x) + p(-x)]$$). Thus, one such operator is $$T := \pi_0 \circ D : p(x) \mapsto \frac{1}{2}[p'(x) + p'(-x)] .$$ We can also check directly from this definition that $$T^2 = 0$$.)