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Q13E

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Found in: Page 176

Linear Algebra With Applications

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 ${\mathbf{\left(}}{\mathbf{a}}{\mathbf{,}}{\mathbf{ar}}{\mathbf{,}}{{\mathbf{ar}}}^{{\mathbf{2}}}{\mathbf{,}}{{\mathbf{ar}}}^{{\mathbf{3}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\right)}}$, for some constants and K.

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

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: Verification whether the subset is closed under addition and scalar multiplication.

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

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

Consider two arbitrary elements from X namely x and y.

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

Find $x+y.$

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

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.