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Expert-verified Found in: Page 176 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Find the set of all polynomial ${\mathbit{f}}\left(t\right)$ in ${{\mathbit{P}}}_{{\mathbf{3}}}$ such that ${\mathbit{f}}\left(1\right){\mathbf{=}}{\mathbf{0}}$ and${{\mathbf{\int }}}_{\mathbf{-}\mathbf{1}}^{{\mathbf{1}}}{\mathbit{f}}\left(t\right){\mathbit{d}}{\mathbit{t}}{\mathbf{=}}{\mathbf{0}}$, and determine its dimension.

The dimension of ${P}_{3}$ such that $f\left(1\right)=0$ and ${{\int }}_{-1}^{{1}}{f}\left(\mathrm{t}\right){d}{t}{=}{0}$ is 3 which is spanned by $span\left\{1,\left({t}^{2}-\frac{2}{3}t\right),\left({t}^{3}-t\right)\right\}$.

See the step by step solution

## Step 1: Determine the span set.

Consider the set of all polynomial such that and .

The set $\left\{1,t,{t}_{2},\dots ,t\right\}$is linear independent set of V if there exist constant ${{\mathbit{a}}}_{{\mathbf{i}}}{\mathbf{\in }}{\mathbit{R}}$ such that ${{\mathbit{a}}}_{{\mathbf{0}}}{\mathbf{+}}{{\mathbit{a}}}_{{\mathbf{1}}}{{\mathbit{t}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbit{a}}}_{{\mathbf{2}}}{{\mathbit{t}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbit{a}}}_{{\mathbf{n}}}{{\mathbit{t}}}_{{\mathbf{n}}}{\mathbf{=}}{\mathbf{0}}$ where ${{\mathbf{a}}}_{{\mathbf{0}}}{\mathbf{=}}{{\mathbf{a}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{a}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{=}}{{\mathbf{a}}}_{{\mathbf{n}}}{\mathbf{=}}{\mathbf{0}}$.

Any polynomial $f\left(t\right)\in {P}_{3}$ is defined as follows.

$f\left(t\right)={b}_{0}+{b}_{1}t+{b}_{2}{t}^{2}+{b}_{3}{t}^{3}$

Substitute the value 1 for t in the equation $f\left(t\right)={b}_{0}+{b}_{1}t+{b}_{2}{t}^{2}+{b}_{3}{t}^{3}$ as follows.

$f\left(t\right)={b}_{0}+{b}_{1}t+{b}_{2}{t}^{2}+{b}_{3}{t}^{3}\phantom{\rule{0ex}{0ex}}f\left(1\right)={b}_{0}+{b}_{1}+{b}_{2}+{b}_{3}$

As $f\left(1\right)=0$, substitute the value 0 for $f\left(1\right)$ in the equation $f\left(1\right)={b}_{0}+{b}_{1}+{b}_{2}+{b}_{3}$ as follows.

$\begin{array}{l}f\left(1\right)={b}_{0}+{b}_{1}+{b}_{2}+{b}_{3}\\ {b}_{1}=-\left({b}_{0}+{b}_{2}+{b}_{3}\right)\end{array}$

Take integration both side in the equation $f\left(t\right)={b}_{0}+{b}_{1}t+{b}_{2}{t}^{2}+{b}_{3}{t}^{3}$as follows.

role="math" localid="1659407877530" $f\left(t\right)={b}_{0}+{b}_{i}t+{b}_{2}{t}^{2}+{b}_{3}{t}^{3}\phantom{\rule{0ex}{0ex}}{\int }_{-1}^{1}f\left(t\right)dt={\int }_{-1}^{1}{b}_{0}+{b}_{1}t+{b}_{2}{t}^{2}+{b}_{2}{t}^{3}dt\phantom{\rule{0ex}{0ex}}={\left\{{b}_{0}t+{b}_{1}\frac{{t}^{2}}{2}+{b}_{2}\frac{{t}^{3}}{3}+{b}_{3}\frac{{t}^{4}}{4}\right\}}_{-1}^{1}\phantom{\rule{0ex}{0ex}}{\int }_{-1}^{1}f\left(t\right)dt={b}_{0}+{b}_{1}\frac{1}{2}+{b}_{2}\frac{1}{3}+{b}_{3}\frac{1}{4}+{b}_{0}-{b}_{1}\frac{1}{2}+{b}_{2}\frac{1}{3}-{b}_{3}\frac{1}{4}$

Further, simplify the equation as follows.

${\int }_{-1}^{1}f\left(t\right)dt={b}_{0}+{b}_{2}\frac{1}{3}+{b}_{0}+{b}_{2}\frac{1}{3}\phantom{\rule{0ex}{0ex}}{\int }_{-1}^{1}f\left(t\right)dt=2{b}_{0}+{b}_{2}\frac{2}{3}$

As ${\int }_{-1}^{1}f\left(t\right)dt=0$, substitute the value 0 for${\int }_{-1}^{1}f\left(t\right)t$ in the equation ${\int }_{1}^{1}f\left(t\right)t=2{b}_{b}+{b}_{2}\frac{2}{3}$ as follows.

${\int }_{-1}^{1}f\left(t\right)=2{b}_{0}+{b}_{2}\frac{2}{3}\phantom{\rule{0ex}{0ex}}0=2{b}_{0}+{b}_{2}\frac{2}{3}\phantom{\rule{0ex}{0ex}}{b}_{0}=-{b}_{2}\frac{1}{3}$

Substitute the value $-{b}_{2}\frac{1}{3}$ for ${b}_{0}$ in the equation ${b}_{1}=-\left({b}_{0}+{b}_{2}+{b}_{3}\right)$as follows.

${b}_{-1}=-\left({b}_{0}+{b}_{2}+{b}_{3}\right)\phantom{\rule{0ex}{0ex}}=-\left(-{b}_{2}\frac{2}{3}+{b}_{2}+{b}_{3}\right)\phantom{\rule{0ex}{0ex}}=-\left({b}_{2}\frac{2}{3}+{b}_{3}\right)\phantom{\rule{0ex}{0ex}}{b}_{1}=-{b}_{2}\frac{2}{3}-{b}_{3}$

Finally, substitute the value $-{b}_{2}\frac{2}{3}-{b}_{3}$ for ${b}_{1}$ in the equation $f\left(t\right)={b}_{0}+{b}_{1}t+{b}_{2}{t}^{2}+{b}_{3}{t}^{3}$ as follows.

$f\left(t\right)={b}_{0}+{b}_{1}t+{b}_{2}{t}^{2}+{b}_{3}{t}^{3}\phantom{\rule{0ex}{0ex}}f\left(t\right)={b}_{0}+\left(-{b}_{2}\frac{2}{3}-{b}_{3}\right)t+{b}_{2}{t}^{2}+{b}_{3}{t}^{3}\phantom{\rule{0ex}{0ex}}f\left(t\right)={b}_{0}+{b}_{2}\left({t}^{2}-\frac{2}{3}\right)+{b}_{3}\left({t}^{3}-t\right)$

From the equation, the set $\left\{1\left({t}^{2}-\frac{2}{3}t\right),\left({t}^{3}-t\right)\right\}$span ${P}_{3}$ if $f\left(1\right)=0$ and ${\int }_{1}^{1}f\left(t\right)dt=0$.

## Step 2: Determine the property of independency.

Compare the equations ${b}_{0}+{b}_{2}\left({t}^{2}-\frac{2}{3}t\right)+{b}_{3}\left({t}^{3}-t\right)=0$ both side as follows.

$\begin{array}{l}b+{b}_{2}\left({t}^{2}-\frac{2}{3}t\right)\right)+{b}_{3}\left({t}^{3}-t\right)=0\\ {b}_{0}+{b}_{2}\left({t}^{2}-\frac{2}{3}t\right)+{b}_{3}\left({t}^{3}-t\right)=\left(0\right)+\left(0\right)\left({t}^{2}-\frac{2}{3}t\right)+\left(0\right){t}^{3}-t\right)\end{array}$

By the definition of linear independence, the subset $\left\{1,\left({t}^{2}-\frac{2}{3}t\right),\left({t}^{3}-t\right)\right\}$ is L.I.

Therefore, the set of all polynomial ${P}_{3}$ such that $f\left(1\right)=0$ and ${\int }_{1}^{1}f\left(t\right)dt=0$ has dimension 3 and spanned by $span\left\{1,\left({t}^{2}-\frac{2}{3}t\right),\left({t}^{3}-t\right)\right\}$. ### Want to see more solutions like these? 