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.

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


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