## Linear Algebra and Its Applications, Exercise 2.1.7

Exercise 2.1.7. Of the following, which are subspaces of $\mathbf{R}^{\infty}$?

(a) the set of all sequences that include infinitely many zeros, e.g., $(1,0,1,0,\dotsc)$

(b) the set of all sequences of the form $(x_1,x_2,\dotsc)$ where $x_j = 0$ from some point onward

(c) the set of all decreasing sequences, i.e., $x_{j+1} \le x_{j}$ for all $j$

(d) the set of all sequences such that $x_j$ converges to a limit as $j \rightarrow \infty$

(e) the set of all arithmetic progressions for which $x_{j+1} - x_j$ is constant for all $j$

(f) the set of all geometric progressions of the form $(x_1, kx_1, k^2x_1, \dotsc)$ for any choice of $x_1$ and $k$

Answer: (a) This set is not a subspace because it is not closed under addition: If we have $x = (1,0,1,0,\dotsc)$ and $y = (0,1,0,1,\dotsc)$ then both $x$ and $y$ are in the set but their sum $x+y = (1,1,1,1,\dotsc)$ is not.

(b) This set is closed under scalar multiplication: Consider $x = (x_1, x_2, \dotsc)$ and suppose for some $i$ we have $x_k = 0$ for $k \ge i$. We then have $cx = (cx_1, cx_2, \dotsc)$. Since we have $x_k = 0$ for $k \ge i$ we also have $cx_k = 0$ for $k \ge i$. So $c x$ is also a member of the set.

This set is also closed under vector addition. Consider $x$ from above and $y$ where $y = (y_1, y_2, \dotsc)$ and for some $j$ we have $y_k = 0$ for $k \ge j$, and consider the sum $x+y = (x_1+y_1, x_2+y_2,\dotsc)$. Choose $l$ so that $l \ge i$ and $l \ge j$. Then $x_k = 0$ for $k \ge l$ and $y_k = 0$ for $k \ge l$ so that $x_k+y_k = 0$ for $k \ge l$. The sum $x+y$ 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 $x = (0, -1, -2, \dotsc)$ is a member of the set, but $-x = (0, 1, 2, \dotsc)$ is not.

(d) We first check that the set of converging sequences is closed under scalar multiplication. Let $x$ be a member of this set, so that $l(x) = \lim_{j \rightarrow \infty} x_j$ exists. Then for any $\epsilon > 0$ there exists $n$ such that $\left| l(x) - x_j \right| < \epsilon$ for $j > n$. Now consider $c x$ where $c$ is any scalar. If $c = 0$ then $c x = (0, 0, \dotsc)$ so that it converges to the limit 0.

Suppose that $c \ne 0$ and choose any $\epsilon > 0$. Since $l(x)$ exists we can choose $n$ such that $\left| l(x) - x_j \right| < \epsilon/\left|c\right|$ for $j > n$. Multiplying both sides by $\left|c\right|$ we have $\left|c\right| \left| l(x) - x_j \right| < \epsilon$. But $\left|c\right| \left| l(x) - x_j \right| = \left|c(l(x)-x_j)\right| = \left| c l(x) - c x_j \right|$. We therefore see that for any $\epsilon > 0$ we can choose $n$ such that $\left| c l(x) - c x_j \right| < \epsilon$.

This means that $l(cx) = \lim_{j \rightarrow \infty} c x_j$ exists and is equal to $c l(x)$ so that for any scalar $c \ne 0$ and converging sequence $x$ the sequence $c x$ is also in the set of converging sequences. Since $c x$ converges both for $c = 0$ and $c \ne 0$ the set is therefore closed under scalar multiplication.

We next check that the set of converging sequences is closed under vector addition. Let $y$ also be a member of this set, so that $l(y) = \lim_{j \rightarrow \infty} y_j$ exists. Then for any $\epsilon > 0$ there exists $n$ such that $\left| l(y) - y_j \right| < \epsilon$ for $j > n$.

Now consider $x + y = (x_1+y_1, x_2+y_2, \dotsc)$ and choose any $\epsilon > 0$. Since $l(x)$ exists we can choose $m$ such that $\left| l(x) - x_j \right| < \epsilon/2$ for $j > m$, and since $l(y)$ exists we can choose $n$ such that $\left| l(y) - y_j \right| < \epsilon/2$ for $j > n$. Choose $p$ such that $p \ge m$ and $p \ge n$. Adding both sides of the two inequalities we have $\left| l(x) - x_j \right| + \left| l(y) - y_j \right| < \epsilon$ for $j > p$.

We have $\left| a+b \right| \le \left| a \right| + \left| b \right|$ for any $a$ and $b$, and thus for any $j > p$ we have

$\left| (l(x)+l(y)) - (x_j+y_j) \right| = \left| (l(x)-x_j) + (l(y)-y_j) \right|$

$\le \left| l(x)-x_j \right| + \left| l(y)-y_j \right| < \epsilon$

So for any $\epsilon > 0$ we can choose $p$ such that $\left| (l(x)+l(y)) - (x_j+y_j) \right| < \epsilon$ for all $j > p$. This means that $l(x+y) = \lim_{j \rightarrow \infty} x_j+y_j$ exists and is equal to $l(x)+l(y)$ so that for any two converging sequences $x$ and $y$ the sequence $x+y$ 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 $x$ be a member of this set, so that $x_{j+1} - x_j$ is a constant value $a$ for all $j$. Then for $c x = (c x_1, c x_2, \dotsc)$ we have $c x_{j+1} - c x_j = c (x_{j+1} - x_j) = c a$ for all $j$. The sequence $c x$ 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 $y$ also be a member of this set, so that $y_{j+1} - y_j$ is a constant value $b$ for all $j$. Then for $x+y = (x_1+y_1, x_2+y_2, \dotsc)$ we have

$(x_{j+1}+y_{j+1}) - (x_j+y_j) = (x_{j+1} - x_j) + (y_{j+1} - y_j) = a+b$

for all $j$. The sequence $x+y$ 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 $x = (1, 2, 4, 8, \dotsc)$ (for which $x_1 = 1$ and $k = 2$), and $y = (1, 3, 9, 27, \dotsc)$ (for which $x_1 = 1$ and $k = 3$). We then have $x + y = (2, 5, 13, 35, \dotsc)$. For $x+y$ we have $x_1 = 2$ (from the first element) and $k = 5/2 = 2.5$ (from the second element). If $x + y$ is a geometric progression then the third element should be $k^2x_1 = 2.5^2 \cdot 2 = 6.25 \cdot 2 = 12.5$ instead of the actual value of 13. So $x + y$ is not a geometrical progression for all geometric progressions $x$ and $y$, and the set is not closed under vector addition.

UPDATE: Fixed typos in the question ($\mathbf{R}^3$ should have been $\mathbf{R}^{\infty}$) and in the answers to (b) ($x_k = 0$ should have been $y_k = 0$) and (d) ($\epsilon < 0$ should have been $\epsilon > 0$).

UPDATE 2: Fixed typos in the question and answer for (f) (references to $x_2$ and $x_3$ should have been to $x_1$, and a reference to $x_1$ should have been to $-x_1$).

UPDATE 3: Fixed the answer for (f); the original answer (claiming that $-x$ 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.

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### 7 Responses to Linear Algebra and Its Applications, Exercise 2.1.7

1. O.H. says:

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.

• hecker says:

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).

2. GG says:

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,… )

• hecker says:

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

3. Samuel says:

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

• hecker says:

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!

4. econsamuel says:

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