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Q25E

Expert-verifiedFound in: Page 176

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**Find the set of all polynomial ${\mathit{f}}{\left(t\right)}$** **in ${{\mathit{P}}}_{{\mathbf{2}}}$** **such that ${\mathit{f}}{\left(1\right)}{\mathbf{=}}{\mathbf{0}}$**, **and determine its dimension.**

The dimension of ${P}_{2}$ such that $f\left(1\right)=0$ is which is spanned by $Span\left\{\right(t-1),({t}^{2}-1\left)\right\}$.

Consider the set of all polynomial such that .

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

Any polynomial $f\left(t\right)\in {P}_{2}$ is defined as follows.

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

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

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

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

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

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

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

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

Compare the equations ${b}_{1}(t-1)+{b}_{2}({t}^{2}-1)=0$ both side as follows.

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

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

Therefore, the set of all polynomial ${P}^{2}$ such that $f\left(1\right)=0$ has dimension 2 and spanned by $Span\left\{(t-1),({t}^{2}-1)\right\}$.

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