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Chapter 4: Q56E (page 177)

Show that the space of infinite sequence of real numbers is infinite dimensional.

Short Answer

Expert verified

The solution is the space V of infinite sequence is infinite dimensional.

Step by step solution

01

Explanation of the solution

Assume that the opposite of that the space V of infinite sequences is finite dimensional with as follows.

dim(V)=n.

Consider an example of n+1 linearly independent infinite sequence as follows.

(1,0,0,...),(0,1,00,...),(0,0,1,0,...),...,(0,0,0,...,0,1,0,...,).

02

Draw the conclusion

This contradicts the fact that n is the largest possible dimension for an n-dimensional space that m=n.

Therefore, the assumption is wrong.

Thus, the space V of infinite sequence is infinite dimensional.

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Most popular questions from this chapter

Let Vbe the space of all infinite sequences of real numbers. See Example 5. Which of the subsets of Vgiven in Exercises 12 through 15 are subspaces of V? The square-summable sequences(x0,x1,...) (i.e., those for which鈭�i=0鈭�xi2converges).

If matrix A is similar to B T(M)=AM-MBis an isomorphism from R2脳2toR2脳2

For which constant k is a linear transformation T(M)=[2304]M-M[300k]is an isomorphism formR2脳2 to R2脳2.

Find out which of the transformations in Exercises 1 through 50 are linear. For those that are linear, determine whether they are isomorphism,T(x+iy)=x2+y2 from 鈩�to 鈩�.

Find out which of the transformations in Exercises 1 through 50 are linear. For those that are linear, determine whether they are isomorphism,T(x+iy)=y+ix
from to .
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