For my homework, the official question reads:
Find bases for the null space and range of $T:P_2({\Bbb R}) \rightarrow P_3({\Bbb R})$ given by $$(Tf)(x) = xf(x) – \int_0^x f(t) dt [sic]$$
Firstly, I suspect the $f(t)dt$ is a typo and am going to assume that it's $f(x) dx$. If this is incorrect and there's something I'm clueless about, this whole question is now pointless…
So assuming I'm working with $$(Tf)(x) = xf(x) – \int_0^x f(x) dx.$$
I figure $f(x) = a_1x^2 + a_2x+a_3$, so then $T(f(x))=(2/3)a_1x^3 + (1/2)a_2x^2$. Arbitrarily setting the $P_2$ vector to $\{1, x, x^2\}$, I calculate $$T(1) = (2/3)a_1+(1/2)a_2$$ $$T(x)=(2/3)a_1x^3+(1/2)a_2x$$ $$T(x^2)=(2/3)a_1x^6+(1/2)a_2x^2$$
So I should have this:
$$\begin{bmatrix}
(2/3)a_1+(1/2)a_2&0\\
0&(1/2)a_2&0\\
0&0&(1/2)a_2\\
0&(2/3)a_1&0\\
0&0&0\\
0&0&(2/3)a_1\\
\end{bmatrix}$$
Putting that into a rref, I get all 3 columns (of $M_{5×3})$) as pivotal, and since $dim(P_3) = 4$ and $dim(R(T)) = 3$:$$\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
0&0&0\\
\end{bmatrix}$$
So $x_1\begin{bmatrix}1\\0\\0\\0\end{bmatrix}+x_2\begin{bmatrix}0\\1\\0\\0\end{bmatrix}+x_3\begin{bmatrix}0\\0\\1\\0\end{bmatrix}$ right?
Now for my question: Did I do this all correctly? This should technically mean that my $dim(R(T))=3$, and since I am transposing into $P_3$, I should have $dim(N(T)) =1$… which would mean it'd just be $x_4\begin{bmatrix}\\0\\0\\0\\1\end{bmatrix}$?
Best Answer
What you did seems correct (except the integral part):
$$f(x)=a_0+a_1x+a_2x^2\implies Tf=a_0x+a_1x^2+a_2x^3-\int_0^x(a_0+a_1t+a_2t^2)dt=$$
$$=a_0x+a_1x^2+a_2x^3-a_0x-\frac{a_1}2x^2-\frac{a_2}3x^3=\frac{a_1}2x^2+\frac{2a_2}3x^3$$
Thus
$$Tf=0\iff a_1=a_2=0\,,\,\,\text{so for example}\ker T=\text{Span}\,\{1\}$$
I'll leave it to you to find out the image of $\;T\;$ , but pay attention: you don't need at all to calculate the matrix of $\;T\;$ wrt some basis in the domain and codomain: that'd only be too much timeand work wasted and it is irrelevant for the question itself.