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Q25E

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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 ${{\mathbit{P}}}_{{\mathbf{2}}}$ such that ${\mathbit{f}}\left(1\right){\mathbf{=}}{\mathbf{0}}$, and determine its dimension.

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

See the step by step solution

Step 1: Determine the span set.

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 ${{\mathbit{a}}}_{{\mathbf{i}}}{\mathbf{\in }}{\mathbit{R}}$ 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 ${{\mathbit{a}}}_{{\mathbf{0}}}{\mathbf{=}}{{\mathbit{a}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbit{a}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{=}}{{\mathbit{a}}}_{{\mathbf{n}}}{\mathbf{=}}{\mathbf{0}}$.

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.

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

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

Step 2: Determine the property of independency.

Compare the equations ${b}_{1}\left(t-1\right)+{b}_{2}\left({t}^{2}-1\right)=0$ 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 $\left\{\left(t-1\right),\left({t}^{2}-1\right)\right\}$is L.I.

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