Chapter 1 Vector Spaces

1A $𝐑^𝑛$ and $𝐂^𝑛$

1A.1

Show that $𝛼 + 𝛽 = 𝛽 + 𝛼$ for all $𝛼, 𝛽 ∈ 𝐂$.

Proof:

Let

$$𝛼 = a + bi, 𝛽 = c + di$$

Then

$$𝛼 + 𝛽 = (a + bi) + (c + di) = (a + c) + (b + d)i \\ = (c+a) + (d+b)i \\ = (c + di) + (a + bi) \\ = 𝛽 + 𝛼$$

$\square$

1A.2

Show that $(𝛼 + 𝛽) + 𝜆 = 𝛼 + (𝛽 + 𝜆)$ for all $𝛼, 𝛽, 𝜆 ∈ 𝐂$.

Proof:

$$𝛼 = a + bi, 𝛽 = c + di, 𝜆 = e + fi$$

Then

$$(𝛼 + 𝛽) + 𝜆 = ((a+bi) + (c+di)) + (e + fi)\\ = ((a+c) + (b+d)i) + (e + fi)\\ = (a+c+e) + (b+d+f)i$$

On the other hand

$$𝛼 + (𝛽 + 𝜆) = (a+bi) + ((c+di) + (e+fi)) \\ = (a+bi) + ((c+e)+(d+f)i)\\ = (a+c+e)+(b+d+f)i$$

$\square$

1A.3

Show that $(𝛼𝛽)𝜆 = 𝛼(𝛽𝜆)$ for all $𝛼, 𝛽, 𝜆 ∈ 𝐂$.

Proof:

$$(𝛼𝛽)𝜆 = ((a+bi)(c+di))(e+fi) \\ = ((ac-bd) + (ad+bc)i)(e+fi) \\ = ((ac-bd)e -(ad+bc)f) + ((ac-bd)f + (ad+bc)e)i \\ = (ace - bde - adf - bcf) + (acf - bdf + ade + bce)i$$

On the other hand

$$𝛼(𝛽𝜆) = (a+bi)((c+di)(e+fi)) \\ = (a+bi)((ce-df)+(cf+de)i)\\ = (a(ce-df) - b(cf+de)) + (a(cf+de)+b(ce-df))i \\ = (ace - adf - bcf - bde) + (acf + ade + bce - bdf)i$$

$\square$

1A.4

Show that $𝜆(𝛼 + 𝛽) = 𝜆𝛼 + 𝜆𝛽$ for all $𝜆, 𝛼, 𝛽 ∈ 𝐂$.

Proof:

$$𝜆(𝛼 + 𝛽) = (e+fi)((a+bi)+(c+di)) \\ = (e+fi)((a+c)+(b+d)i)\\ = ((a+c)e - (b+d)f) + ((a+c)f + (b+d)e)i \\ $$

On the other hand

$$𝜆𝛼 + 𝜆𝛽 = (e+fi)(a+bi) + (e+fi)(c+di) \\ = (ea-bf) + (eb+fa)i + (ec-fd) + (ed+fc)i \\ = (ea + ec - bf - df) + (af + cf + be + de)i$$

$\square$

1A.5

Show that for every $𝛼 ∈ 𝐂$, there exists a unique $𝛽 ∈ 𝐂$ such that $𝛼 + 𝛽 = 0$.

Proof:

If $𝛼 = a + bi$, then let $𝛽 = (-a) + (-b)i$, we can see $𝛼 + 𝛽 = 0$.

On the other hand, if $𝛼 + 𝛽 = 0$, then $𝛽 = (-a) + (-b)i$.

$\square$

1A.6

Show that for every $𝛼 ∈ 𝐂$ with $𝛼 ≠ 0$, there exists a unique $𝛽 ∈ 𝐂$ such that $𝛼𝛽 = 1$.

Proof:

Let $𝛼 = a + bi$, $𝛽 = c + di$.

$$𝛼𝛽 = (ac - bd) + (ad + bc)i$$

So

$$\begin{cases} ac - bd &= 1 \\ ad + bc &= 0 \\ \end{cases}$$

Assume $a \neq 0$, then $d = -\frac{b}{a}c$

So

$$ac + \frac{b^2}{a}c = 1 \\ \Rightarrow \\ c = \frac{a}{a^2 + b^2}, d = - \frac{b}{a^2 + b^2}$$

$\square$

1A.7

Show that

$$\frac{-1 + \sqrt[]{3} i}{2}$$

is a cube root of 1 (meaning that its cube equals 1).

Proof:

let

$$r = \frac{-1 + \sqrt[]{3} i}{2}$$

Then

$$r^2 = \frac{1}{4} ((1-3) + 2 \sqrt[]{3}i) \\ = \frac{-1 - \sqrt[]{3}i}{2}$$

Then

$$r^3 = \frac{1 + 3}{4} = 1$$

$\square$

1A.8

Find two distinct square roots of $𝑖$.

Solution:

Let $(a+bi)^2 = i$, then

$$a^2 - b^2 + 2abi = i$$

So we can have $$r_1 = \frac{\sqrt[]{2} + \sqrt[]{2}i}{2}, r_2 = \frac{-\sqrt[]{2} + -\sqrt[]{2}i}{2},$$

$\square$

1A.9

Find $𝑥 ∈ 𝐑^4$ such that

$$(4, −3, 1, 7) + 2𝑥 = (5, 9, −6, 8).$$

Solution:

$$x = (\frac{1}{2}, 6, -\frac{7}{2}, \frac{1}{2})$$

$\square$

1A.10

Explain why there does not exist $𝜆 ∈ 𝐂$ such that $𝜆(2 − 3𝑖, 5 + 4𝑖, −6 + 7𝑖) = (12 − 5𝑖, 7 + 22𝑖, −32 − 9𝑖)$

Proof:

There is no $𝜆$, such that

$$2𝜆 = 12 \text{ and } -3𝜆 = -5$$

$\square$

1A.11

Show that $(𝑥 + 𝑦) + 𝑧 = 𝑥 + (𝑦 + 𝑧)$ for all $𝑥, 𝑦, 𝑧 ∈ 𝐅^𝑛$.

Proof:

$$(𝑥 + 𝑦) + 𝑧 = (x_1 + y_1, \cdots, x_n + y_n) + z \\ = ((x_1 + y_1) + z_1, \cdots, (x_n + y_n) + z_n) \\ = (x_1 + y_1 + z_1, \cdots, x_n + y_n + z_n) \\ $$

It's same for $𝑥 + (𝑦 + 𝑧)$.

$\square$

1A.12

Show that $(𝑎𝑏)𝑥 = 𝑎(𝑏𝑥)$ for all $𝑥 ∈ 𝐅^𝑛$ and all $𝑎, 𝑏 ∈ 𝐅$.

Proof:

$$(𝑎𝑏)𝑥 = (ab)(x_1, \cdots, x_n) \\ =((ab)x_1, \cdots, (ab)x_n)$$

$𝑎(𝑏𝑥)$ is the same.

$\square$

1A.13

Show that $1𝑥 = 𝑥$ for all $𝑥 ∈ 𝐅^𝑛$

Proof:

$$1𝑥 = 1 (x_1, \cdots, x_n) \\ = (1 \cdot x_1, \cdots, 1 \cdot x_n) \\ = (x_1, \cdots, x_n) = x$$

$\square$

1A.14

Show that $𝜆(𝑥 + 𝑦) = 𝜆𝑥 + 𝜆𝑦$ for all $𝜆 ∈ 𝐅$ and all $𝑥, 𝑦 ∈ 𝐅^𝑛$.

Proof:

$$𝜆(𝑥 + 𝑦) = 𝜆 ((x_1, \cdots, x_n) + (y_1, \cdots, y_n)) \\ = 𝜆 (x_1+y_1, \cdots, x_n+y_n) \\ = (𝜆(x_1+y_1), \cdots, 𝜆(x_n+y_n)) \\ = (𝜆 x_1 + 𝜆 y_1, \cdots , 𝜆 x_n + 𝜆 y_n) \\ = 𝜆 (x_1, \cdots, x_n) + 𝜆 (y_1, \cdots, y_n) \\ = 𝜆 x + 𝜆 y$$

$\square$

1A.15

Show that $(𝑎 + 𝑏)𝑥 = 𝑎𝑥 + 𝑏𝑥$ for all $𝑎, 𝑏 ∈ 𝐅$ and all $𝑥 ∈ 𝐅^𝑛$

Proof:

$$\begin{split} (𝑎 + 𝑏)𝑥 &= (a+b)(x_1, \cdots, x_n) \\ &= ((a+b)x_1, \cdots, (a+b)x_n) \\ &= (ax_1 + bx_1, \cdots, ax_n + bx_n) \\ &= (ax_1, \cdots, ax_n) + (bx_1, \cdots, bx_n) \\ &= a (x_1, \cdots, x_n) + b (x_1, \cdots, x_n) \\ &= ax + bx \end{split}$$

$\square$

1B Definition of Vector Space

1B.1

Prove that $−(−𝑣) = 𝑣$ for every $𝑣 ∈ 𝑉$.

Proof:

From $1.32$

$$−(−𝑣) = (-1)(-v) \\ \Rightarrow \\ −(−𝑣) + (−𝑣) = (-1)(-v) + (1)(-v) = (-1 + 1)(-v) = 0(-v) = 0$$

On the other hand, $v + (−𝑣) = 0$.

So both of them are the additive inverse of the $(−𝑣)$.

From $1.27$, they are the same.

$\square$

1B.2

Suppose $𝑎 ∈ 𝐅, 𝑣 ∈ 𝑉$ , and $𝑎𝑣 = 0$. Prove that $𝑎 = 0$ or $𝑣 = 0$.

Proof:

If $a =0$, then it's valid. If $a \neq 0$, then $a$ has a multiplicative inverse $a^{-1}$.

$a^{-1}(av) = (a^{-1}a)v = v$.

On the other hand, from 1.31, $a^{-1}0 = 0$.

So we have $v = 0$.

$\square$

1B.3

Suppose $𝑣, 𝑤 ∈ 𝑉$. Explain why there exists a unique $𝑥 ∈ 𝑉$ such that $𝑣 + 3𝑥 = 𝑤$.

Proof:

We add $v$'s additive inverse on both side and get

$$(-v) + (𝑣 + 3𝑥) = ((-v) + v) + 3x = 0 + 3x = 3x = w - v$$

Then we multiply the multiplicative inverse of $3$, we got

$$x = 3^{-1} (w-v)$$

So $x$ is unique and satisfy this equation.

$\square$

1B.4

The empty set is not a vector space. The empty set fails to satisfy only one of the requirements listed in the definition of a vector space (1.20). Which one?

Solution:

It must be "additive identity": There exists an element $0 ∈ 𝑉$ such that $𝑣 + 0 = 𝑣$ for all $𝑣 ∈ 𝑉$.

$\square$

1B.5

Show that in the definition of a vector space (1.20), the additive inverse condition can be replaced with the condition that

$$0𝑣 = 0 \text{ for all } 𝑣 ∈ 𝑉.$$

Here the $0$ on the left side is the number $0$, and the $0$ on the right side is the additive identity of $𝑉$.

Proof:

We just need to use this condition to prove the existence of additive inverse.

Let $𝑣 ∈ 𝑉$, and let $w = (-1)v$, then

$$w + v = (-1)v + 1v = ((-1) + 1)v = 0v = 0$$

So $w$ is the additive inverse of $v$.

$\square$

1B.6

With these operations of addition and scalar multiplication, is $𝐑 ∪ {∞, −∞}$ a vector space over $𝐑$? Explain.

Solution:

commutativity: yes.

associativity: no.

Let

$$u = 1, v = ∞, w = -∞ \\ (u + v) + w = ∞ + (-∞) = 0 \\ u + (v+w) = 1 + (∞ + (-∞)) = 1 + 0\\ $$

So it's not a vector space over $\mathbb{R}$.

$\square$

1B.7

Suppose $𝑆$ is a nonempty set. Let $𝑉^𝑆$ denote the set of functions from $𝑆$ to $𝑉$. Define a natural addition and scalar multiplication on $𝑉^𝑆$ , and show that $𝑉^𝑆$ is a vector space with these definitions.

Solution:

Similar as 1.24, we define

• For $𝑓, 𝑔 ∈ V^𝑆$, the sum $𝑓 + 𝑔 ∈ V^𝑆$ is the function defined by

$$(f+g)(x) = f(x) + g(x)$$

for all $𝑥 ∈ 𝑆$.

• For $𝜆 ∈ 𝐅$ and $𝑓 ∈ V^𝑆$, the product $𝜆 𝑓 ∈ V^𝑆$ is the function defined by

$$(𝜆𝑓)(𝑥) = 𝜆 𝑓 (𝑥)$$

for all $𝑥 ∈ 𝑆$.

commutativity:

Proof:

$$(f+g)(x) = f(x) + g(x) = g(x) + f(x) = (g+f)(x)$$

$\square$

associativity:

Proof:

$$((f+g)+h)(x) = (f+g)(x) + h(x) = (f(x) + g(x)) + h(x) \\ =f(x) + (g(x) + h(x)) = f(x) + (g+h)(x) \\ = (f + (g+h))(x)$$

$$((ab)f)(x) = (ab) f(x) = a (bf(x)) = a ((bf)(x)) = (a(bf))(x)$$

$\square$

additive identity:

Proof:

Define $0(x) = 0, \forall x \in V$.

$$(f+0)(x) = f(x) + 0(x) = f(x) + 0 = f(x)$$

$\square$

additive inverse:

Proof:

Define $g(x) = -f(x)$, then

$$(f+g)(x) = f(x) + g(x) = f(x) + (-f(x)) = 0$$

multiplicative identity:

Proof:

$$(1 \cdot f)(x) = 1 \cdot f(x) = f(x)$$

$\square$

distributive properties:

$$(𝜆(𝑓+g))(𝑥) = 𝜆 ((𝑓+g)(x)) = 𝜆 (f(x) + g(x)) \\ = 𝜆f(x) + 𝜆g(x) = (𝜆f)(x) + (𝜆g)(x) = (𝜆f + 𝜆f)(x)$$

$$((𝜆 + μ)f)(x) = (𝜆 + μ) f(x) = 𝜆f(x) + μf(x) \\ = (𝜆f)(x) + (μf)(x) = (𝜆f + μf)(x)$$

$\square$

1B.8

Suppose $𝑉$ is a real vector space.

• The complexification of $𝑉$, denoted by $𝑉^𝐂$, equals $𝑉×𝑉$. An element of $𝑉^𝐂$ is an ordered pair $(𝑢, 𝑣)$, where $𝑢, 𝑣 ∈ 𝑉$, but we write this as $𝑢 + 𝑖𝑣$.
• Addition on $𝑉^𝐂$ is defined by

$$(𝑢_1 + 𝑖𝑣_1) + (𝑢_2 + 𝑖𝑣_2) = (𝑢_1 + 𝑢_2) + 𝑖(𝑣_1 + 𝑣_2)$$

for all $𝑢_1, 𝑣_1, 𝑢_2, 𝑣_2 ∈ 𝑉$.

• Complex scalar multiplication on $𝑉^𝐂$ is defined by

$$(𝑎 + 𝑏𝑖)(𝑢 + 𝑖𝑣) = (𝑎𝑢 − 𝑏𝑣) + 𝑖(𝑎𝑣 + 𝑏𝑢)$$

for all $𝑎, 𝑏 ∈ 𝐑$ and all $𝑢, 𝑣 ∈ 𝑉$.

Proof:

commutativity:

$$\begin{split} (𝑢_1 + 𝑖𝑣_1) + (𝑢_2 + 𝑖𝑣_2) &= (𝑢_1 + 𝑢_2) + 𝑖(𝑣_1 + 𝑣_2) \\ &= (𝑢_2 + 𝑢_1) + 𝑖(𝑣_2 + 𝑣_1) \\ &= (𝑢_2 + 𝑖𝑣_2) + (𝑢_1 + 𝑖𝑣_1) \end{split}$$

$\square$

associativity:

Proof:

$$\begin{split} ((𝑢_1 + 𝑖𝑣_1) + (𝑢_2 + 𝑖𝑣_2)) + (𝑢_3 + 𝑖𝑣_3) \\ &= ((𝑢_1 + 𝑢_2) + 𝑖(𝑣_1 + 𝑣_2)) + (𝑢_3 + 𝑖𝑣_3) \\ &= ((𝑢_1 + 𝑢_2) + 𝑢_3) + 𝑖((𝑣_1 + 𝑣_2) + 𝑣_3) \\ &= (𝑢_1 + (𝑢_2 + 𝑢_3)) + 𝑖(𝑣_1 + (𝑣_2 + 𝑣_3)) \\ &= (𝑢_1 + 𝑖𝑣_1) + ((𝑢_2 + 𝑢_3) + 𝑖(𝑣_2 + 𝑣_3)) \\ &= (𝑢_1 + 𝑖𝑣_1) + ((𝑢_2 + 𝑖𝑣_2) + (𝑢_3 + 𝑖𝑣_3)) \end{split}$$

$\square$

additive identity:

$$(𝑢_1 + 𝑖𝑣_1) + (0 + 𝑖0) = (𝑢_1 + 0) + 𝑖(𝑣_1 + 0) = 𝑢_1 + 𝑖𝑣_1$$

$\square$

additive inverse:

$$\begin{split} (𝑢_1 + 𝑖 𝑣_1) + ((-𝑢_1) + i(-𝑣_1)) \\ &= (u_1 + (-u_1)) + i (v_1 + (-v_1)) \\ &= 0 + i 0 \\ &= 0 \end{split}$$

$\square$

multiplicative identity:

$$\begin{split} (𝑢_1 + 𝑖 𝑣_1)(1 + i0) \\ &= (u_1 \cdot 1 - v_1 \cdot 0) + i (u_1 \cdot 0 + v_1 \cdot 1) \\ &= u_1 + i v_1 \end{split}$$

distributive properties:

$$\begin{split} 𝜆 ((𝑢_1 + 𝑖𝑣_1) + (𝑢_2 + 𝑖𝑣_2)) \\ &= 𝜆 ((𝑢_1 + 𝑢_2) + i(𝑣_1 + 𝑣_2)) \\ &= 𝜆(𝑢_1 + 𝑢_2) + i(𝜆(𝑣_1 + 𝑣_2)) \\ &= (𝜆𝑢_1 + 𝜆𝑢_2) + i((𝜆𝑣_1 + 𝜆𝑣_2)) \\ &= (𝜆𝑢_1 + 𝑖𝜆𝑣_1) + (𝜆𝑢_2 + 𝑖𝜆𝑣_2) \\ &= 𝜆 (𝑢_1 + 𝑖𝑣_1) + 𝜆 (𝑢_2 + 𝑖𝑣_2) \end{split}$$

$$\begin{split} (𝜆 + μ)(𝑢_1 + 𝑖𝑣_1) \\ &= (𝜆 + μ) u_1 + i ((𝜆 + μ)v_1) \\ &= (𝜆u_1 + μu_1) + i ((𝜆 v_1+ μ v_1)) \\ &= (𝜆u_1 + i 𝜆 v_1) + (μu_1 + iμ v_1) \\ &= 𝜆 (𝑢_1 + 𝑖𝑣_1) + μ (𝑢_1 + 𝑖𝑣_1) \end{split}$$

$\square$

1C Subspaces

1C.1

For each of the following subsets of $𝐅^3$, determine whether it is a subspace of $𝐅^3$.

(a) $\{(𝑥_1, 𝑥_2, 𝑥_3) ∈ 𝐅^3 ∶ 𝑥_1 + 2𝑥_2 + 3𝑥_3 = 0\}$

Solution:

Yes. Let it be $U$.

$$0 + 2 \cdot 0 + 3 \cdot 0 = 0$$

So $(0,0,0) \in U$.

$$(x_1 + y_1) + 2(x_2 + y_2) + 3(x_3 + y_3) = \\ (𝑥_1 + 2𝑥_2 + 3𝑥_3) + (y_1 + 2y_2 + 3y_3) = 0 + 0 = 0$$

So $U$ is closed under addition.

$$a x_1 + 2 (a x_2) + 3 (a x_3) = \\ a (𝑥_1 + 2𝑥_2 + 3𝑥_3) = a 0 = 0$$

So $U$ is closed under scalar multiplication.

$\square$

(b) $\{(𝑥_1, 𝑥_2, 𝑥_3) ∈ 𝐅^3 ∶ 𝑥_1 + 2𝑥_2 + 3𝑥_3 = 4\}$

Solution:

$(0,0,0) \not\in U$.

(c) $\{(𝑥_1, 𝑥_2, 𝑥_3) ∈ 𝐅^3 ∶ 𝑥_1𝑥_2𝑥_3 = 0\}$

Solution:

No. Consider

$$u = (1, 0, 1), v = (1, 1, 0)$$

$u + v = (1, 1, 1)$, then $u + v \not\in U$.

$\square$

(d) $\{(𝑥_1, 𝑥_2, 𝑥_3) ∈ 𝐅^3 ∶ 𝑥_1 = 5 𝑥_3\}$

Solution

Yes. $0 \in U$.

$x_1 + y_1 = 5 x_3 + 5 y_3 = 5 (x_3 + y_3)$.

$$ax_1 = a (5 x_3) = (a5) x_3 = (5a) x_3 = 5 (a x_3)$$

$\square$

1C.2

Verify all assertions about subspaces in Example 1.35.

Proof: See my notes.

1C.3

Show that the set of differentiable real-valued functions $𝑓$ on the interval $(−4, 4)$ such that $𝑓'(−1) = 3𝑓(2)$ is a subspace of $𝐑^{(−4, 4)}$.

Proof:

First, the $f(x) = 0 \in U$. Because $f'(x) = 0$, so $f'(-1) = 3 f(2) = 0$.

Secondly,

$$(f+g)'(-1) = f'(-1) + g'(-1) = 3 f(2) + 3 g(2) = 3(f+g) (2)$$

Thirdly,

$$(af)'(-1) = a(f'(-1)) = a(3f(2)) = 3 ((af)(2))$$

$\square$

1C.4

Suppose $𝑏 ∈ 𝐑$. Show that the set of continuous real-valued functions $𝑓$ on the interval $[0, 1]$ such that $\int_{0}^{1} 𝑓 = 𝑏$ is a subspace of $𝐑^{[0, 1]}$ if and only if $𝑏 = 0$.

Proof:

$\Rightarrow$

Since $U$ is a subspace, then $0 \in U$. $\int_{0}^{1} 0 = 0$, so $b = 0$.

$\Leftarrow$

First, $0 \in U$.

Secondly,

$$\int_{0}^{1} (f+g) = \int_{0}^{1} f + \int_{0}^{1} g = 0 + 0 = 0$$

Thirdly,

$$\int_{0}^{1} (af) = a \int_{0}^{1} f = a 0 = 0$$

$\square$

1C.5

Is $𝐑^2$ a subspace of the complex vector space $𝐂^2$?

Solution:

No. Consider $a = i, u = (1, 0)$, then $au = (i,0) \not\in 𝐑^2$.

So addition is not closed.

$\square$

1C.6

(a) Is $\{(𝑎, 𝑏, 𝑐) ∈ 𝐑^3 ∶ 𝑎^3 = 𝑏^3\}$ a subspace of $𝐑^3$?

Solution:

If $a, b \in \mathbb{R}^3$, and $a^3 = b^3$, then $a = b$.

So if

$$u = (a_1, b_1, c_1) \in U, v = (a_2, b_2, c_2) \in U,$$

then $a_1 = b_1, a_2 = b_2$, then $a_1+b_1 = a_2+b_2$. So $u+v \in U$.

So it's a subspace of $𝐑^3$.

(b) Is $\{(𝑎, 𝑏, 𝑐) ∈ 𝐂^3 ∶ 𝑎^3 = 𝑏^3\}$ a subspace of $𝐂^3$.

Consider

$$u = (1, 1, 0), v = (\frac{-1 + i\sqrt[]{3}}{2}, 1)$$

Note

$$\left( \frac{-1 + i\sqrt[]{3}}{2} \right)^2 \\ = \frac{1 - 2 i \sqrt[]{3} - 3}{4} \\ = \frac{-1 - i \sqrt[]{3}}{2}$$

So

$$\left( \frac{-1 + i\sqrt[]{3}}{2} \right)^3 = 1$$

Now

$$\left( \frac{1 + i\sqrt[]{3}}{2} \right)^2 \\ = \frac{1 + 2 i \sqrt[]{3} - 3}{4} \\ = \frac{-1 + i \sqrt[]{3}}{2}$$

Then

$$\left( \frac{1 + i\sqrt[]{3}}{2} \right)^3 = -1$$

So $u+v \not\in U$.

$\square$

1C.7

Prove or give a counterexample: If $𝑈$ is a nonempty subset of $𝐑^2$ such that $𝑈$ is closed under addition and under taking additive inverses (meaning $−𝑢 ∈ 𝑈$ whenever $𝑢 ∈ 𝑈$), then $𝑈$ is a subspace of $𝐑^2$.

Solution:

Counterexample: Consider

$$U = \{ (n, 0) | n \in \mathbb{Z} \}$$

$\square$

1C.8

Give an example of a nonempty subset $𝑈$ of $𝐑^2$ such that $𝑈$ is closed under scalar multiplication, but $𝑈$ is not a subspace of $𝐑^2$.

Solution

Consider

$$U = \{(a, 0) | a \in \mathbb{R}\} \cup \{(0, b) | b \in \mathbb{R}\}$$

$\square$

1C.9

A function $𝑓 ∶ 𝐑 → 𝐑$ is called periodic if there exists a positive number $𝑝$ such that $𝑓(𝑥) = 𝑓(𝑥 + 𝑝)$ for all $𝑥 ∈ 𝐑$. Is the set of periodic functions from $𝐑$ to $𝐑$ a subspace of $𝐑^𝐑$ ? Explain.

Solution:

This example is inspired from "Lectures on the Fourier Transform and Its Applications" exercise 1.2.

Let

$$f(t) = \begin{cases} 1 &\text{if } t \in \mathbb{Z} \\ 0 &\text{otherwise} \\ \end{cases}$$

$$g(t) = \begin{cases} 1 &\text{if } t = n \sqrt[]{2}, n \in \mathbb{Z}\\ 0 &\text{otherwise} \\ \end{cases}$$

$f+g$ cannot be a periodic function, since $(f+g)(t) \neq 2, t \neq 0$.

If $f(t) = g(t) = 1$, then $t = n = m \sqrt[]{2}$. That means $n/m = \sqrt[]{2}$.

$\square$

1C.10

Suppose $𝑉_1$ and $𝑉_2$ are subspaces of $𝑉$. Prove that the intersection $𝑉_1 ∩ 𝑉_2$ is a subspace of $𝑉$.

Proof:

$0 \in V_1, 0 \in V_2$, so $0 \in 𝑉_1 ∩ 𝑉_2$.

$a,b \in V_1, a,b \in V_2$ so $a+b \in 𝑉_1 ∩ 𝑉_2$.

$k \in F, a \in 𝑉_1 ∩ 𝑉_2$, so $k a \in 𝑉_1 ∩ 𝑉_2$.

$\square$

1C.11

Prove that the intersection of every collection of subspaces of 𝑉 is a subspace of 𝑉 .

Proof: Very similar to the previous one.

$\square$

1C.12

Prove that the union of two subspaces of $𝑉$ is a subspace of $𝑉$ if and only if one of the subspaces is contained in the other.

Proof:

$\Rightarrow$

Assume $U, W, U \cup W$ are subspaces. If $a \in U, a \not\in W$. Now assume $b \in W$, since $a+b \in U \cup W$. If $a+b \in W$, then $(a+b) + (-b) = a \in W$, we have contradition.

So $a+b \in U$, then $(a+b) + (-a) = b \in U$, so $W \subseteq U$.

$\Leftarrow$ is obvious.

$\square$

1C.13

Prove that the union of three subspaces of $𝑉$ is a subspace of $𝑉$ if and only if one of the subspaces contains the other two.

This exercise is surprisingly harder than Exercise 12, possibly because this exercise is not true if we replace $𝐅$ with a field containing only two elements.

Proof:

Assume $a, b, c$ are unique elements in $A, B, C$.

Now, consider $a+b$.

If $a+b \in A$, then $b \in A$, we have contradition.

If $a+b \in B$, then $a \in B$, we have contradition.

So $a+b \in C$. For the same reason, $a-b \in C$.

That means $a, b \in C$. We have contradition.

This means, one the following must be true:

$$A \subseteq B \cup C \text{ or } B \subseteq C \cup A \text{ or } C \subseteq A \cup B$$

In either case, this problem is reduced to 1C.12.

$\square$

1C.14

Suppose

$$𝑈 = \{(𝑥, −𝑥, 2𝑥) ∈ 𝐅^3 ∶ 𝑥 ∈ 𝐅\} \quad \text{and} \quad 𝑊 = \{(𝑥, 𝑥, 2𝑥) ∈ 𝐅^3 ∶ 𝑥 ∈ 𝐅\}.$$

Describe $𝑈 + 𝑊$ using symbols, and also give a description of $𝑈 + 𝑊$ that uses no symbols.

Solution:

$$U+W = \{ (x, y, 2x) \in 𝐅^3 ∶ 𝑥 ∈ 𝐅\}$$

$\square$

1C.15

Suppose $𝑈$ is a subspace of $𝑉$ . What is $𝑈 + 𝑈$?

Solution:

It's still $U$.

$\square$

1C.16

Is the operation of addition on the subspaces of $𝑉$ commutative? In other words, if $𝑈$ and $𝑊$ are subspaces of $𝑉$ , is $𝑈 + 𝑊 = 𝑊 + 𝑈$?

Solution:

Yes, this is because the vector space itself is commutative.

$\square$

1C.17

Is the operation of addition on the subspaces of $𝑉$ associative? In other words, if $𝑉_1, 𝑉_2, 𝑉_3$ are subspaces of $𝑉$ , is

$$(V_1 + V_2) + V_3 = V_1 + (V_2 + V_3)?$$

Solution:

Yes, this is because the vector space itself is associative.

$\square$

1C.18

Does the operation of addition on the subspaces of $𝑉$ have an additive identity? Which subspaces have additive inverses?

Solution:

Yes. The subspace contains only $0$. Only itself has additive inverse, which is itself.

$\square$

1C.19

Prove or give a counterexample: If $𝑉_1, 𝑉_2, 𝑈$ are subspaces of $𝑉$ such that

$$V_1 + U = V_2 + U$$

then $𝑉_1 = 𝑉_2$.

Solution:

Counterexample, consider

$$V_1 = \{(x, 0, 0) \in F^3\} \\ V_2 = \{(0, y, 0) \in F^3\} \\ U = \{(x, y, 0) \in F^3\} \\ $$

Then $V_1 + U = U = V_2 + U$, but $V_1 \neq V_2$.

$\square$

1C.20

Suppose

$$𝑈 = \{(𝑥, 𝑥, 𝑦, 𝑦) ∈ 𝐅^4 ∶ 𝑥, 𝑦 ∈ 𝐅\}.$$

Find a subspace $𝑊$ of $𝐅^4$ such that $𝐅^4 = 𝑈 ⊕ 𝑊$.

Solution:

Consider

$$W = \{(𝑥, -𝑥, 𝑦, -𝑦) ∈ 𝐅^4 ∶ 𝑥, 𝑦 ∈ 𝐅\}.$$

Then $U \cap W = 0$.

$\square$

1C.21

Suppose

$𝑈 = \{(𝑥, 𝑦, 𝑥 + 𝑦, 𝑥 − 𝑦, 2𝑥) ∈ 𝐅^5 ∶ 𝑥, 𝑦 ∈ 𝐅\}.$

Find a subspace $𝑊$ of $𝐅^5$ such that $𝐅^5 = 𝑈 ⊕ 𝑊$.

Solution

Consider

$$W = \{(0, 0, a, b, c) ∈ 𝐅^5 ∶ a, b, c ∈ 𝐅 \}.$$

$\square$

1C.22

Suppose

$𝑈 = \{(𝑥, 𝑦, 𝑥 + 𝑦, 𝑥 − 𝑦, 2𝑥) ∈ 𝐅^5 ∶ 𝑥, 𝑦 ∈ 𝐅\}.$

Find a subspace $𝑊_1, W_2, W_3$ of $𝐅^5$ such that $𝐅^5 = 𝑈 ⊕ 𝑊_1 ⊕ 𝑊_2 ⊕ 𝑊_3$.

Solution:

Consider

$$W_1 = \{(0, 0, x_1, 0, 0) ∈ 𝐅^5 ∶ w_1, x_1, y_1, z_1 ∈ 𝐅 \} \\ W_2 = \{(0, 0, 0, y_2, 0) ∈ 𝐅^5 ∶ y_2, z_2 ∈ 𝐅 \} \\ W_3 = \{(0, 0, 0, 0, z_3) ∈ 𝐅^5 ∶ z_3 ∈ 𝐅 \} \\ $$

So if $u + w_1 + w_2 + w_3 = 0$, then we must have

$$\begin{cases} x + 0 + 0 + 0 = 0 &\text{}\\ y + 0 + 0 + 0 = 0 &\text{}\\ (x + y) + x_1 + 0 + 0 = 0 &\text{}\\ (x - y) + 0 + y_2 + 0 = 0 &\text{}\\ 2x + 0 + 0 + z_3 = 0 &\text{}\\ \end{cases}$$

then we must have

$$x = y = x_1 = y_2 = z_3 = 0 \\ \Rightarrow \\ u = w_1 = w_2 = w_3 = 0$$

So $𝐅^5 = 𝑈 ⊕ 𝑊_1 ⊕ 𝑊_2 ⊕ 𝑊_3$.

$\square$

1C.23

Prove or give a counterexample: If $𝑉_1, 𝑉_2, 𝑈$ are subspaces of $𝑉$ such that

$$𝑉 = 𝑉_1 ⊕ 𝑈 \quad \text{and} \quad 𝑉 = 𝑉_2 ⊕ 𝑈,$$

then $𝑉_1 = 𝑉_2$.

Hint: When trying to discover whether a conjecture in linear algebra is true or false, it is often useful to start by experimenting in $𝐅^2$.

Solution:

Counterexample:

As hinted, let $F = \mathbb{R}$.

$$U = (x, 0) \\ V_1 = (0, y) \\ V_2 = (z, z) \\ $$

So

$$𝑉_1 ⊕ 𝑈 = \mathbb{R}^2 = 𝑉_2 ⊕ 𝑈$$

But $V_1 \neq V_2$.

$\square$

1C.24

A function $𝑓 ∶ 𝐑 → 𝐑$ is called even if

$$f(-x) = f(x)$$

for all $𝑥 ∈ 𝐑$. A function $𝑓 ∶ 𝐑 → 𝐑$ is called odd if

$$f(-x) = -f(x)$$

for all $𝑥 ∈ 𝐑$. Let $𝑉_e$ denote the set of real-valued even functions on $𝐑$ and let $𝑉_o$ denote the set of real-valued odd functions on $𝐑$. Show that $𝐑^𝐑 = 𝑉_e ⊕ 𝑉_o$.

Solution:

We have seen this again in the book of "Lectures on the Fourier Transform and Its Applications" Appendix B.

Let

$$f_e(x) = \frac{f(x) + f(-x)}{2} \\ f_o(x) = \frac{f(x) - f(-x)}{2} \\ $$

Then $f_e$ is even, and $f_o$ is odd. Furthermore, $f = f_e + f_o$. This proves $𝐑^𝐑 = 𝑉_e + 𝑉_o$.

Now if $f \in 𝑉_e \cap 𝑉_o$, then $f(x) = f(-x) = -f(x)$ so $f(x) = 0$, i.e. $𝑉_e \cap 𝑉_o = 0$.

So $𝐑^𝐑 = 𝑉_e ⊕ 𝑉_o$.

$\square$