Kristen Stokes

Answered

2022-07-01

$T(p(x))={\int}_{0}^{1}p(x)dx.$

(a) Show $T$ is a linear transformation.

(b) Compute $\mathcal{N}(T).$ Is $T$ one-to-one?

(c) Show that $T$ is onto.

(d) Let $B$ be the standard basis for ${\mathcal{P}}_{2}$ and let ${B}^{\mathrm{\prime}}=\{1\}$ be a basis for $\mathbb{R}$. Find $[T{]}_{B}^{{B}^{\mathrm{\prime}}}$.

(e) Use the matrix found in part (d) to compute $T(-{x}^{2}-3x+2)$

(a) Show $T$ is a linear transformation.

(b) Compute $\mathcal{N}(T).$ Is $T$ one-to-one?

(c) Show that $T$ is onto.

(d) Let $B$ be the standard basis for ${\mathcal{P}}_{2}$ and let ${B}^{\mathrm{\prime}}=\{1\}$ be a basis for $\mathbb{R}$. Find $[T{]}_{B}^{{B}^{\mathrm{\prime}}}$.

(e) Use the matrix found in part (d) to compute $T(-{x}^{2}-3x+2)$

Answer & Explanation

Nicolas Calhoun

Expert

2022-07-02Added 15 answers

(b) As explained in the comment above,

$\mathrm{ker}T=\{{a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}\in {\mathbb{R}}_{2}[x]:{a}_{0}+\frac{{a}_{1}}{2}+\frac{{a}_{2}}{3}=0\}$

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 $p(x)=r$. Then $Tp=r$.

(d) To find $[T{]}_{B}^{{B}^{\prime}}$, 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

$[T{]}_{B}^{{B}^{\prime}}=\left(1\phantom{\rule{0.5cm}{0ex}}\frac{1}{2}\phantom{\rule{0.5cm}{0ex}}\frac{1}{3}\right).$

(e) Finally, $T(-{x}^{2}-3x+2)=-T({x}^{2})-3T(x)+2T(1)=-\frac{1}{3}-\frac{3}{2}+2=\frac{1}{6}.$

$\mathrm{ker}T=\{{a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}\in {\mathbb{R}}_{2}[x]:{a}_{0}+\frac{{a}_{1}}{2}+\frac{{a}_{2}}{3}=0\}$

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 $p(x)=r$. Then $Tp=r$.

(d) To find $[T{]}_{B}^{{B}^{\prime}}$, 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

$[T{]}_{B}^{{B}^{\prime}}=\left(1\phantom{\rule{0.5cm}{0ex}}\frac{1}{2}\phantom{\rule{0.5cm}{0ex}}\frac{1}{3}\right).$

(e) Finally, $T(-{x}^{2}-3x+2)=-T({x}^{2})-3T(x)+2T(1)=-\frac{1}{3}-\frac{3}{2}+2=\frac{1}{6}.$

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