Finding a basis of the null space and range of a transformation of polynomial spaces $T: P_2(\mathbb{R}) \to P_3(\mathbb{R})$

Question

Let $P_n(\mathbb{R})$ be the collection of polynomials of degree less than or equal to $n$ with real coefficients and $T\colon P_2(\mathbb{R}) \to P_3(\mathbb{R})$ be given by $$ Tp(x) = \int_0^x p(t)\,dt - \frac{1}{2}x^2p'(x) $$

a) Find a basis for $\operatorname{null}(T)$.

b) Find a basis for $\operatorname{range}(T)$.

I got basis for $\operatorname{null}(T)$ as $\{x\}$ and range as $\{x,x^2\}$.

No idea if that's correct though.

The $p(t)\,dt$ part confuses me since $T$ is being applied to $p(x)$ are the coefficients in the polynomial the same?

Any help would be greatly appreciated, cheers.

Answer 1

Let's check if $x$ is in the null space.

\begin{align} T(x) &= \int_0^x t \, dt - \frac12 x^2 \\ &= \frac{x^2}{2} - \frac{x^2}2 \\ &=0 \end{align}

Let's check for the range,

\begin{align} T(1) &= \int_0^x 1 \, dt - \frac12x^2 (0) \\ &= x \end{align}

\begin{align} T(x^2) &= \int_0^x t^2 \, dt - \frac12x^2 (2x) \\ &= \frac{x^3}3-x^3 \end{align}

Hence while your basis for the null space is correct, a basis for the range should be $\{ x, x^3\}$.

Note that $T$ is being applied to $p$, the quadratic polynomial. If $p(x)=x^2$, then $p(t)=t^2$.

Answer 2

The simplest strategy, in my opinion, is working with matrices. The domain and codomain have natural bases, $\{1,x,x^2\}$ and $\{1,x,x^2,x^3\}$ respectively.

Let's compute the matrix of $T$ with respect to these bases.

\begin{align} T(1)&=\int_0^x 1\,dt-\frac{1}{2}x^2\cdot 0=x \\[6px] T(x)&=\int_0^x t\,dt-\frac{1}{2}x^2\cdot 1=\frac{x^2}{2}-\frac{x^2}{2}=0 \\[6px] T(x^2)&=\int_0^x t^2\,dt-\frac{1}{2}x^2\cdot 2x=\frac{x^3}{3}-x^3=-\frac{2}{3}x^3 \end{align} Therefore the matrix we are looking for is $$ A=\begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -2/3 \end{bmatrix} $$ This matrix has rank $2$ and the first and third columns are a basis for the column space. Therefore $\{T(1),T(x^2)\}=\{x,-\frac{2}{3}x^3\}$ is a basis for the range of $T$.

The null space of $A$ has dimension $1$ (by the rank nullity theorem) and we already know a nonzero vector in the null space, so a basis for the null space of $T$ is $\{x\}$.