Exercise 2.1.7. Of the following, which are subspaces of ?

(a) the set of all sequences that include infinitely many zeros, e.g.,

(b) the set of all sequences of the form where from some point onward

(c) the set of all decreasing sequences, i.e., for all

(d) the set of all sequences such that converges to a limit as

(e) the set of all arithmetic progressions for which is constant for all

(f) the set of all geometric progressions of the form for any choice of and

Answer: (a) This set is not a subspace because it is not closed under addition: If we have and then both and are in the set but their sum is not.

(b) This set is closed under scalar multiplication: Consider and suppose for some we have for . We then have . Since we have for we also have for . So is also a member of the set.

This set is also closed under vector addition. Consider from above and where and for some we have for , and consider the sum . Choose so that and . Then for and for so that for . The sum is therefore also a member of the set.

Since the set is closed under both vector addition and scalar multiplication and it is a subset of the vector space of infinite sequences, it is a subspace of that vector space.

(c) This set is not a subspace because it is not closed under scalar multiplication: The sequence is a member of the set, but is not.

(d) We first check that the set of converging sequences is closed under scalar multiplication. Let be a member of this set, so that exists. Then for any there exists such that for . Now consider where is any scalar. If then so that it converges to the limit 0.

Suppose that and choose any . Since exists we can choose such that for . Multiplying both sides by we have . But . We therefore see that for any we can choose such that .

This means that exists and is equal to so that for any scalar and converging sequence the sequence is also in the set of converging sequences. Since converges both for and the set is therefore closed under scalar multiplication.

We next check that the set of converging sequences is closed under vector addition. Let also be a member of this set, so that exists. Then for any there exists such that for .

Now consider and choose any . Since exists we can choose such that for , and since exists we can choose such that for . Choose such that and . Adding both sides of the two inequalities we have for .

We have for any and , and thus for any we have

So for any we can choose such that for all . This means that exists and is equal to so that for any two converging sequences and the sequence is also in the set of converging sequences. The set is therefore closed under vector addition.

Since the set of converging sequences is closed under both vector addition and scalar multiplication and it is a subset of the vector space of infinite sequences, it is a subspace of that vector space.

(e) We first check that the set of arithmetic progressions is closed under scalar multiplication. Let be a member of this set, so that is a constant value for all . Then for we have for all . The sequence is therefore also an arithmetic progression, and the set is closed under scalar multiplication.

We next check that the set of arithmetic progressions is closed under vector addition. Let also be a member of this set, so that is a constant value for all . Then for we have

for all . The sequence is therefore also an arithmetic progression, and the set is closed under vector addition.

Since the set of arithmetic progressions is closed under both vector addition and scalar multiplication and it is a subset of the vector space of infinite sequences, it is a subspace of that vector space.

(f) The set of geometric progressions is not a subspace because it is not closed under vector addition: For example, suppose that (for which and ), and (for which and ). We then have . For we have (from the first element) and (from the second element). If is a geometric progression then the third element should be instead of the actual value of 13. So is not a geometrical progression for all geometric progressions and , and the set is not closed under vector addition.

UPDATE: Fixed typos in the question ( should have been ) and in the answers to (b) ( should have been ) and (d) ( should have been ).

UPDATE 2: Fixed typos in the question and answer for (f) (references to and should have been to , and a reference to should have been to ).

UPDATE 3: Fixed the answer for (f); the original answer (claiming that was not a geometric progression) was incorrect. Thanks go to Samuel for pointing this out.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

Is (a) not a special case of (b)? i.e. if we take one form of (x1,x2…) to be (1,0,1,0…) and (y1,y2,…) to be (0,1,0,1…) as done in (a), we are outside the space of either sequences. But, perhaps, you’re suggesting that since the x’s are arbitrary, the resulting sequence need not be preserved for any particular instantiation, but rather that the x’s in the resulting sequence can take on any value?

Thanks.

Actually (b) is a special case of (a): (a) includes all sequences with an infinite number of zeroes, while (b) includes only such sequences where there are only zeroes (and no other number) past a certain point. So, for example, (a) would include a sequence where every millionth number was a zero, but such a sequence would not be a member of (b).

It’s the fact that sequences in (b) are guaranteed to contain only zeroes above a certain point that makes the set (b) closed under scalar multiplication and addition. This guarantee does not apply to sequences in (a), and as I note you can choose sequences in (a) whose sum is not in (a).

there is something wrong in (f)

it should be -1*x=(-x1,-kx2, -k^2×3,… )

and i found the problem in textbook is

(x1, kx1, k^2×1,… )

You are correct; I have updated the post to fix these errors. Thanks for finding them!

Why is it not ( -x1, k(-x1), k^2(-x1),…) still a geometric progression in (f)?

You are correct. I have fixed the answer. The set of geometric progressions is in fact closed under scalar multiplication, but it is not closed under vector addition. Thanks for finding and reporting this error!

Why is it not (-x1, k(-x1), k^2(-x1),…) still a geometric progression in (f)?