Vtyjkyuk

Let $P_n(mathbbR)$ be the collection of polynomials of degree less than or equal to $n$ with real coefficients and $Tcolon P_2(mathbbR) to P_3(mathbbR)$ be given by
$$
Tp(x) = int_0^x p(t),dt - frac12x^2p'(x)
$$

a) Find a basis for $operatornamenull(T)$.

b) Find a basis for $operatornamerange(T)$.

I got basis for $operatornamenull(T)$ 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.

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

beginalign
T(x) &= int_0^x t , dt - frac12 x^2 \
&= fracx^22 - fracx^22 \
&=0
endalign

Let's check for the range,

beginalign
T(1) &= int_0^x 1 , dt - frac12x^2 (0) \
&= x
endalign

beginalign
T(x^2) &= int_0^x t^2 , dt - frac12x^2 (2x) \
&= fracx^33-x^3
endalign

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 $p(t)=t^2$.

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.

beginalign
T(1)&=int_0^x 1,dt-frac12x^2cdot 0=x \[6px]
T(x)&=int_0^x t,dt-frac12x^2cdot 1=fracx^22-fracx^22=0 \[6px]
T(x^2)&=int_0^x t^2,dt-frac12x^2cdot 2x=fracx^33-x^3=-frac23x^3
endalign
Therefore the matrix we are looking for is
$$
A=beginbmatrix
0 & 0 & 0 \
1 & 0 & 0 \
0 & 0 & 0 \
0 & 0 & -2/3
endbmatrix
$$
This matrix has rank $2$ and the first and third columns are a basis for the column space. Therefore $T(1),T(x^2)=x,-frac23x^3$ 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 $x$.