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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 a basis of a linear space and thus determine its dimension. Examine whether a subset of a linear space is a subspace. Which of the subsets of ${{\mathbf{P}}}_{{\mathbf{2}}}$ given in Exercises 1 through 5 are subspaces of ${{\mathbf{P}}}_{{\mathbf{2}}}$(see Example 16)? Find a basis for those that are subspace,${\mathbf{\left\{}}{\mathbf{p}}\left(t\right){\mathbf{:}}{{\mathbf{\int }}}_{{\mathbf{0}}}^{{\mathbf{1}}}{\mathbf{p}}\left(t\right){\mathbf{dt}}{\mathbf{=}}{\mathbf{0}}{\mathbf{\right\}}}$.

$\left\{\mathrm{p}\left(\mathrm{t}\right):{\int }_{0}^{1}\mathrm{p}\left(\mathrm{t}\right)\mathrm{dt}=0\right\}$ is a subset and a subspace of ${\mathrm{P}}_{2}$.

See the step by step solution

Step 1: Definition of subspace

A subset W of a linear space V is called a subspace of V if

(a) W contains the neutral element 0 of V.

(b) W is closed under addition (if f and g are in W then so is f+g )

(c) W is closed under scalar multiplication (if f is in W and k is scalar, then kf is in W ).

we can summarize parts b and c by saying that W is closed under linear combinations.

Step 2: Application of given conditions on vector space V.

General expression for a quadratic polynomial with one variable is ${\mathbf{p}}\left(t\right){\mathbf{=}}{{\mathbf{at}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{bt}}{\mathbf{+}}{\mathbf{c}}$

Given ${\int }_{0}^{1}\mathrm{p}\left(\mathrm{t}\right)\mathrm{dt}=0$.

From given,

${\int }_{0}^{1}\mathrm{p}\left(\mathrm{t}\right)\mathrm{dt}={\int }_{0}^{1}{\mathrm{at}}^{2}+\mathrm{bt}+\mathrm{c}$.

Then,

${\int }_{0}^{1}{\mathrm{at}}^{2}+\mathrm{bt}+\mathrm{c}={\left[\frac{{\mathrm{at}}^{3}}{3}+\frac{{\mathrm{bt}}^{2}}{2}+\mathrm{ct}\right]}_{0}^{1}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{a}}{3}+\frac{\mathrm{b}}{2}+\mathrm{c}\phantom{\rule{0ex}{0ex}}\mathrm{c}=-\frac{\mathrm{a}}{3}+\frac{\mathrm{b}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{p}\left(\mathrm{t}\right)={\mathrm{at}}^{2}+\mathrm{bt}-\frac{\mathrm{a}}{3}+\frac{\mathrm{b}}{2}.$

Assume, $\mathrm{q}\left(\mathrm{t}\right)={\mathrm{et}}^{2}+\mathrm{dt}-\frac{\mathrm{e}}{3}-\frac{\mathrm{d}}{2}.$.

Also, let us assume two scalars $\alpha ,\beta$,

Then, we can write as:

$\alpha p\left(t\right)+\beta q\left(t\right)=\alpha \left(a{t}^{2}+bt-\frac{a}{3}-\frac{b}{2}\right)+\beta \left(e{t}^{2}+dt-\frac{e}{3}-\frac{d}{2}\right)\phantom{\rule{0ex}{0ex}}=\alpha a{t}^{2}+\alpha bt-\frac{a}{3}\alpha -\frac{b}{2}\alpha +\beta e{t}^{2}+\beta dt-\frac{e}{3}\beta -\frac{d}{2}\beta \phantom{\rule{0ex}{0ex}}={t}^{2}\left(\alpha a+\beta e\right)+t\left(\alpha b+\beta d\right)-\frac{1}{3}\left(\alpha a+\beta e\right)-\frac{1}{2}\left(\alpha b+\beta d\right)$

Therefore, the given set is closed under linear combinations. Hence, it is a subspace of ${\mathrm{P}}_{2}$.