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Q13E

Expert-verifiedFound in: Page 176

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**Let V ****be the space of all infinite sequences of real numbers. See Example 5. Which of the subsets of ****given in Exercises 12 through 15 are subspaces of V****? The geometric sequences [i.e., sequences of the form ${\mathbf{(}}{\mathbf{a}}{\mathbf{,}}{\mathbf{ar}}{\mathbf{,}}{{\mathbf{ar}}}^{{\mathbf{2}}}{\mathbf{,}}{{\mathbf{ar}}}^{{\mathbf{3}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{)}}$****, for some constants ****and K****.**

The geometric sequences [i.e., sequences of the form $(\mathrm{a},\mathrm{ar},{\mathrm{ar}}^{2},{\mathrm{ar}}^{3},....)$, for some constants and K are subspaces of V .

**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.**

Let us consider a set X to be the set of all geometric sequences which is a subset of V.

Then, $X=\left\{(a,ar,a{r}^{2},a{r}^{3},...)|a,k\in R\right\}.$

Consider two arbitrary elements from X namely x and y.

Then,$x=(a,ar,a{r}^{2},...),y=(b,bs,b{s}^{2},...).$

Find $x+y.$

$x+y=(a,ar,a{r}^{2},...),(b,bs,b{s}^{2},...).\phantom{\rule{0ex}{0ex}}=(a+b,ar+bs,a{r}^{2}+b{s}^{2},\dots )$

In a geometric sequence, on dividing two consecutive elements we get a common ratio. (i.e.)

$\frac{ar+bs}{a+b}=\frac{a{r}^{2}+b{s}^{2}}{ar+bs}$

Let us assume $\mathrm{a}=1,\mathrm{r}=3,\mathrm{b}=2\mathrm{and}\mathrm{s}=4.$

Then, left side can be expressed as

$\frac{1.3+2.4}{1+2}=\frac{11}{2}$

Right side can be expressed as

$\frac{1.{3}^{2}+2.{4}^{2}}{1.3+2.4}=\frac{9+32}{3+8}\phantom{\rule{0ex}{0ex}}=\frac{41}{11}$

It is obtained that L.H.S $\ne $R.H.S. Thus, it is not closed under addition.

Therefore, it is not a subspace of V.

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