Nullity and Rank of a Linear Transformation

Question

Let $T: P_2(\mathbb{R})\to\mathbb{R}$ be a linear transformation defined by $$T[f(x)]= \int_0^1 f(t) \, dt $$

Hi everyone, how do I find the rank and nullity of T and a basis for the nullspace? My book doesn't have a step by step solution/explanation for this problem and I'm 90% sure this problem will appear on my exam next week. Thanks in advance.

Answer 1

Let $a+bX+cX^2\in P_2(\mathbb{R})$. Then $T(a+bX+cX^2)=a+\frac{b}{2}+\frac{c}{3}$. So if $a+bX+cX^2\in \ker(T)$, then $a+\frac{b}{2}+\frac{c}{3}=0$. Thus the kernel is determined by only one equation in 3 variables. It follows that $\dim(\ker(T))=2$ and you can easily find a basis. What can you say about $\dim(\text{im}(T))$?

Answer 2

Hint:

If $P_2(\mathbf R)$ denotes the set of polynomials with real coefficients and degree $\le 2$, it is a vector space of dimension $3$ , isomorphic to $\mathbf R^3$ by the isomorphism \begin{align} \mathbf R^3&\longrightarrow P_2(\mathbf R)\\ (a,b,c)&\longmapsto ax^2+bx+c \end{align} Now compute the integral ; you'll obtain a linear form in $a, b, c$. The kernel corresponds to those $(a,b,c)$ which satisfy this (unique within a non-zero factor) linear relation, hence it has codimension $1$, i.e. dimension $3-1=2$.

Further, the subspace of those $(a,b,c)$ which satisfy this linear relation is isomorphic to $\mathbf R^2$ in the following way: express one of $a, b$ or $c$ as a linear function of the other two. You may take as a basis of the kernel in $\mathbf R^3$ the image of a basis of $\mathbf R^2$ by this isomorphism. Then take the polynomials in $P_2(\mathbf R)$ which correspond to this basis of the kernel in $\mathbf R^3$.