Kristen Stokes

2022-07-01

$T\left(p\left(x\right)\right)={\int }_{0}^{1}p\left(x\right)dx.$
(a) Show $T$ is a linear transformation.
(b) Compute $\mathcal{N}\left(T\right).$ Is $T$ one-to-one?
(c) Show that $T$ is onto.
(d) Let $B$ be the standard basis for ${\mathcal{P}}_{2}$ and let ${B}^{\mathrm{\prime }}=\left\{1\right\}$ be a basis for $\mathbb{R}$. Find $\left[T{\right]}_{B}^{{B}^{\mathrm{\prime }}}$.
(e) Use the matrix found in part (d) to compute $T\left(-{x}^{2}-3x+2\right)$

Nicolas Calhoun

Expert

(b) As explained in the comment above,
$\mathrm{ker}T=\left\{{a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}\in {\mathbb{R}}_{2}\left[x\right]:{a}_{0}+\frac{{a}_{1}}{2}+\frac{{a}_{2}}{3}=0\right\}$
We see, for instance, that $1+x-\frac{9}{2}{x}^{2}\in \mathrm{ker}T$, so $T$ is not injective.
(c) To see that $T$ is onto, let $r\in \mathbb{R}$ be arbitrary and consider the constant polynomial $p\left(x\right)=r$. Then $Tp=r$.
(d) To find $\left[T{\right]}_{B}^{{B}^{\prime }}$, we take the basis $1,x,{x}^{2}$ and evaluate $T$ at each of these polynomials. We then get $1,\frac{1}{2},\frac{1}{3}$, so
$\left[T{\right]}_{B}^{{B}^{\prime }}=\left(1\phantom{\rule{0.5cm}{0ex}}\frac{1}{2}\phantom{\rule{0.5cm}{0ex}}\frac{1}{3}\right).$
(e) Finally, $T\left(-{x}^{2}-3x+2\right)=-T\left({x}^{2}\right)-3T\left(x\right)+2T\left(1\right)=-\frac{1}{3}-\frac{3}{2}+2=\frac{1}{6}.$

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