# Find a basis for $\operatorname{null}(T)$ for $T\in\mathscr{L}(\mathscr{P}_2(\Bbb{R}))$ s.t. $T(f)=xf'(x)-f(1)x$

Find a basis for $\operatorname{null}(T)$ for $T\in\mathscr{L}(\mathscr{P}_2(\Bbb{R}))$ s.t. $T(f)=xf'(x)-f(1)x$

First, I find the basis for the range(T). If $f=a_0+a_1x+a_2x^2$, then $T(f)=x(a_1+2a_2x)-(a_0+a_1+a_2)x=-a_0x+a_2(2x^2-x)$. Thus a basis of the range(T) is the span of $-x,2x^2-x$.

To find the basis for null(T), I set $-a_0x+a_2(2x^2-x)=0$ because I need to know what functions will cause this to be zero. I know that any basis of null(T) must have 1 vector because $\dim\mathscr{P}_2=\dim(\operatorname{range}T)+\dim( \operatorname{null}T)$ and $\dim( \operatorname{range}T)=2$ as shown above, and clearly the dimension of the vector space is 3. However, I'm not sure how to actually find the basis for the null space.

Well, note that for $T(f)=0,$ we need $a_0=a_2=0.$ Can you see why? This means that every element of the null space of $T$ has the form $a_1x$ for some real $a_1.$ Can you take it from there?

• I understand how to finish it out from the last part, but how do you know for sure that $a_0=a_2=0$?
– mmm
Commented Jul 4, 2018 at 21:15
• Well, $\{1,x,x^2\}$ is a basis for $\mathscr{P}_2(\Bbb{R}),$ yes? Rewrite $T(f)$ as a linear combination of those vectors, set it equal to $0,$ and use linear independence. If that doesn't get you there, or if you just want to check your thinking, let me know. Commented Jul 4, 2018 at 21:18

Compute the matrix of $T$ with respect to the basis $\{p_1=1,p_2=x,p_3=x^2\}$.

We have $$T(p_1)=-x,\qquad T(p_2)=0,\qquad T(p_3)=2x^2-x$$ so the matrix is $$\begin{bmatrix} 0 & 0 & 0 \\ -1 & 0 & -1 \\ 0 & 0 & 2 \end{bmatrix}$$ A standard Gaussian elimination reveals the RREF $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$ A basis for the null space of the matrix is clearly given by the vector $$\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$ which corresponds to the polynomial $x=p_2$. A basis for the range consists of the vectors $T(p_1)$ and $T(p_3)$, because the pivot columns are the first and the third.

Of course the particular problem could be directly attacked by showing that $T(p_1)$ and $T(p_3)$ are linearly independent. Since we also have $T(p_2)=0$, we can conclude that the dimension of the range is $\ge2$ and the dimension of the null space is $\ge1$; the rank-nullity theorem forces the range to have dimension $2$ and the null space to have dimension $1$.

However, the method outlined before is fairly general and provides the answer for every linear map between finite dimensional vector spaces.