Chapter 1 Vector Spaces

1B Definition of Vector Space

1.24 notation $F^S$

• If $𝑆$ is a set, then $𝐅^𝑆$ denotes the set of functions from $𝑆$ to $𝐅$.
• For $𝑓, 𝑔 ∈ 𝐅^𝑆$, the sum $𝑓 + 𝑔 ∈ 𝐅^𝑆$ is the function defined by

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

for all $𝑥 ∈ 𝑆$.

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

$$(𝜆𝑓)(𝑥) = 𝜆 𝑓 (𝑥)$$

for all $𝑥 ∈ 𝑆$.

1.25 example: $𝐅^𝑆$ is a vector space

• If $𝑆$ is a nonempty set, then $𝐅^𝑆$ (with the operations of addition and scalar multiplication as defined above) is a vector space over $𝐅$.

commutativity:

Proof:

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

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 $0(x) = 0, \forall x \in S$.

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

$\square$

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:

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

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

$\square$

• The additive identity of $𝐅^𝑆$ is the function $0 ∶ 𝑆 → 𝐅$ defined by

$$0(x) = 0$$

for all $𝑥 ∈ 𝑆$.

Already proved above.

• For $𝑓 ∈ 𝐅^𝑆$, the additive inverse of $𝑓$ is the function $− 𝑓 ∶ 𝑆 → 𝐅$ defined by

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

for all $𝑥 ∈ 𝑆$.

Already proved above.

The vector space $𝐅^𝑛$ is a special case of the vector space 𝐅𝑆 because each $(𝑥_1, …, 𝑥_𝑛) ∈ 𝐅^𝑛$ can be thought of as a function $𝑥$ from the set $\{1, 2, …, 𝑛\}$ to $𝐅$.

1.26 unique additive identity

1.27 unique additive inverse

1.30 the number 0 times a vector

$$0𝑣 = 0, \forall v \in V$$

1.31 a number times the vector 0

$$𝑎0 = 0, \forall a \in F$$

1.32 the number −1 times a vector

$$(−1)𝑣 = −𝑣, \forall v \in V$$

1C Subspaces

1.33 definition: subspace

A subset $𝑈$ of $𝑉$ is called a subspace of $𝑉$ if $𝑈$ is also a vector space with the same additive identity, addition, and scalar multiplication as on $𝑉$ .

1.34 conditions for a subspace

A subset $𝑈$ of $𝑉$ is a subspace of $𝑉$ if and only if $𝑈$ satisfies the following three conditions.

additive identity:

$$0 ∈ 𝑈.$$

closed under addition:

𝑢, 𝑤 ∈ 𝑈 implies 𝑢 + 𝑤 ∈ 𝑈.

closed under scalar multiplication:

𝑎 ∈ 𝐅 and 𝑢 ∈ 𝑈 implies 𝑎𝑢 ∈ 𝑈.

1.35 example: subspaces

(a) If $𝑏 ∈ 𝐅$, then

$$\{(𝑥_1, 𝑥_2, 𝑥_3, 𝑥_4) ∈ 𝐅^4 ∶ 𝑥_3 = 5𝑥_4 + 𝑏\}$$

is a subspace of $𝐅^4$ if and only if $𝑏 = 0$.

Proof:

Let $U = \{(𝑥_1, 𝑥_2, 𝑥_3, 𝑥_4) ∈ 𝐅^4 ∶ 𝑥_3 = 5𝑥_4 + 𝑏\}$. If $U$ is a subspace, then $0 \in U$, then

$$0 = 5 \cdot 0 + b \Rightarrow b = 0$$

Conversely, assume $b = 0$. Then $0 \in U$.

If $x = (𝑥_1, 𝑥_2, 𝑥_3, 𝑥_4), y = (y_1, y_2, y_3, y_4) \in U$, then

$$𝑥_3 = 5𝑥_4, y_3 = 5y_4 \\ \Rightarrow \\ 𝑥_3 + y_3 = 5 (𝑥_4 + y_4)$$

So $x+y \in U$.

Also we have

$$𝑥_3 = 5𝑥_4 \\ \Rightarrow \\ a𝑥_3 = a5𝑥_4 = 5 (a𝑥_4) \\ $$

So $ax \in U$.

$\square$

(b) The set of continuous real-valued functions on the interval $[0, 1]$ is a subspace of $𝐑^{[0, 1]}$.

additive identity:

Since $0$ is a continuous function, so

$$0 ∈ 𝑈.$$

closed under addition:

If $f, g$ are continuous function, then $f+g$ is also continuous function on $[0,1]$.

closed under scalar multiplication:

If $f$ is continuous and $a \in \mathbb{R}$, then $af$ is also continuous.

$\square$

(c) The set of differentiable real-valued functions on 𝐑 is a subspace of $𝐑^𝐑$.

additive identity:

Since $0$ is a differentiable function, so

$$0 ∈ 𝑈.$$

closed under addition:

If $f, g$ are differentiable function, then $f+g$ is also differentiable function on $\mathbb{R}$.

closed under scalar multiplication:

If $f$ is differentiable and $a \in \mathbb{R}$, then $af$ is also differentiable.

$\square$

(d) The set of differentiable real-valued functions $𝑓$ on the interval $(0, 3)$ such that $𝑓' (2) = 𝑏$ is a subspace of $𝐑^(0, 3)$ if and only if $𝑏 = 0$.

$\Rightarrow$

Since $0 \in U$, and $0'(2) = 0$ then $b = 0$.

$\Leftarrow$

$0$ is differentiable and $0'(2) = 0 = b$. So $0 \in U$.

If $f,g$ are differentiable and $f'(2) = g'(2) = 0$, then

$$(f+g)'(2) = f'(2) + g'(2) = 0 + 0 = 0$$

If $f$ is differentiable and $f'(2) = 0$, then

$$(af)'(2) = a (f'(2)) = a 0 = 0$$

$\square$

(e) The set of all sequences of complex numbers with limit 0 is a subspace of $𝐂^∞$ .

Proof:

First, $(0, 0, 0, \dots) \rightarrow 0$. So $0 \in U$.

Then if $(a_n), (b_n) \rightarrow 0$, then

$$\lim_{n \to \infty} (a_n+b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = 0$$

Lastly,

$$\lim_{n \to \infty} a b_n = a \lim_{n \to \infty} b_n = 0$$

$\square$

• The subspaces of $𝐑^2$ are precisely $\{0\}$, all lines in $𝐑^2$ containing the origin, and $𝐑^2$ .
• The subspaces of $𝐑^3$ are precisely $\{0\}$, all lines in $𝐑^3$ containing the origin, all planes in $𝐑^3$ containing the origin, and $𝐑^3$ .

Sums of Subspaces

1.36 definition: sum of subspaces

Suppose $𝑉_1, \cdots, 𝑉_𝑚$ are subspaces of 𝑉 . The sum of $𝑉_1, \cdots, 𝑉_𝑚$, denoted by $𝑉_1 + ⋯ + 𝑉_𝑚$, is the set of all possible sums of elements of $𝑉_1, \cdots, 𝑉_𝑚$. More precisely,

$$V_1 + \cdots + V_m = \{ v_1 + \cdots + v_m : v_1 \in V_1, \cdots, v_m \in V_m \}.$$

1.37 example: a sum of subspaces of $𝐅^3$

Suppose $𝑈$ is the set of all elements of $𝐅^3$ whose second and third coordinates equal $0$, and 𝑊 is the set of all elements of $𝐅^3$ whose first and third coordinates equal 0:

$$𝑈 = \{(𝑥, 0, 0) ∈ 𝐅^3 ∶ 𝑥 ∈ 𝐅\} \quad \text{and} \quad W = \{(0, y, 0) ∈ 𝐅^3 ∶ 𝑥 ∈ 𝐅\}$$

Show that $U+W = \{(𝑥, 𝑦, 0) ∈ 𝐅^3 ∶ 𝑥, 𝑦 ∈ 𝐅\}$.

Proof:

Let $V = \{(𝑥, 𝑦, 0) ∈ 𝐅^3 ∶ 𝑥, 𝑦 ∈ 𝐅\}$.

If $u = (x,0,0), w = (0,y,0)$, then $u+w = (x,y,0) \in V$. So $U+W \subseteq V$.

Let $v = (x,y,0) \in V$, then let $u = (x,0,0) \in U, w = (0,y,0) \in W$, so $v = u+w \in U+W$, i.e. $V \subseteq U+W$.

So $U+W = V$.

$\square$

1.40 sum of subspaces is the smallest containing subspace

Suppose $V_1, \cdots, V_m$ are subspaces of 𝑉. Then $V_1 + \cdots + V_n$ is the smallest subspace of 𝑉 containing $V_1, \cdots, V_m$.

Proof:

Let $U = V_1 + \cdots + V_n$, we first prove $U$ is a subspace.

First $0 \in V_i, i = 1, \cdots, m$, so $0 \in U$.

Secondly, given $a, b \in U$, then

$$a = a_1 + \cdots + a_m \\ b = b_1 + \cdots + b_m \\ \Rightarrow \\ a+b = (a_1 + b_1) + \cdots + (a_n + b_n) \in U$$

Finally, if $a \in F, u \in U$, then

$$u = v_1 + \cdots + v_m \\ \Rightarrow \\ au = a(v_1 + \cdots + v_m) = av_1 + \cdots + av_m \in U$$

So $U$ is a subspace of $V$.

Then assume $V_i \subseteq U', i = 1, \cdots, m$. Let $u \in V_1 + \cdots + V_n$, so

$$u = v_1 + \cdots + v_m \\ \Rightarrow \\ v_i \in V_i \subseteq U' \\ \Rightarrow \\ u \in U'$$

So $V_1 + \cdots + V_n \subseteq U$.

$\square$

Direct Sums

1.41 definition: direct sum, $⊕$

Suppose $V_1, \cdots, V_m$ are subspaces of $𝑉$.

• The sum $V_1 + \cdots + V_m$ is called a direct sum if each element of $V_1 + \cdots + V_m$ can be written in only one way as a sum $v_1 + \cdots + v_m$, where each $𝑣_𝑘 ∈ 𝑉_𝑘$.

• If $V_1 + \cdots + V_m$ is a direct sum, then $V_1 ⊕ \cdots ⊕ V_m$ denotes $V_1 + \cdots + V_m$, with the $⊕$ notation serving as an indication that this is a direct sum.

1.42 example: a direct sum of two subspaces

$$𝑈 = \{(𝑥, 𝑦, 0) ∈ 𝐅^3 ∶ 𝑥, 𝑦 ∈ 𝐅\} \quad \text{and} \quad 𝑊 = \{(0, 0, 𝑧) ∈ 𝐅^3 ∶ 𝑧 ∈ 𝐅\}.$$

Then $𝐅^3 = 𝑈 ⊕ 𝑊$.

The symbol $⊕$, which is a plus sign inside a circle, reminds us that we are dealing with a special type of sum of subspaces—each element in the direct sum can be represented in only one way as a sum of elements from the specified subspaces.

1.44 example: a sum that is not a direct sum

Suppose

$$𝑉_1 = \{(𝑥, 𝑦, 0) ∈ 𝐅^3 ∶ 𝑥, 𝑦 ∈ 𝐅\}, \\ 𝑉_2 = \{(0, 0, 𝑧) ∈ 𝐅^3 ∶ 𝑧 ∈ 𝐅\}, \\ 𝑉_3 = \{(0, 𝑦, 𝑦) ∈ 𝐅^3 ∶ 𝑦 ∈ 𝐅\}. \\ $$

Note we have

$$(0, 0, 0) = (0, 1, 0) + (0, 0, 1) + (0, −1, −1) \\ (0, 0, 0) = (0, 0, 0) + (0, 0, 0) + (0, 0, 0)$$

$\square$

1.45 condition for a direct sum

Suppose $V_1, \cdots, V_m$ are subspaces of $𝑉$. Then $V_1 + \cdots + V_m$ is a direct sum if and only if the only way to write $0$ as a sum $𝑣_1 +⋯ + 𝑣_𝑚$, where each $𝑣_𝑘 ∈ 𝑉_𝑘$, is by taking each $𝑣_𝑘$ equal to $0$.

Proof:

$\Rightarrow$

By definition, if $V_1 + \cdots + V_m$ is a direct sum, then $0$ can be written in only one way, since $0 + \cdots + 0 = 0$, then it is the only way.

$\Leftarrow$

Assume

$$v = v_1 + \cdots + v_m \\ v = v_1' + \cdots + v_m' \\ $$

Then

$$0 = v - v = (v_1 - v_1') + \cdots + (v_m - v_m')$$

Then

$$v_1 = v_1', \cdots, v_m = v_m'$$

So $V_1 + \cdots + V_m$ is a direct sum.

$\square$

1.46 direct sum of two subspaces

Suppose $𝑈$ and $𝑊$ are subspaces of $𝑉$. Then

$$U+W \text{ is a direct sum} \Longleftrightarrow 𝑈 ∩ 𝑊 = \{0\}.$$

Proof:

$\Rightarrow$

If $U+W$ is a direct sum, and assume $u \in 𝑈 ∩ 𝑊$, then $-u \in 𝑈 ∩ 𝑊$. So

$$u + (-u) = 0$$

Then $u = 0$. So $𝑈 ∩ 𝑊 = \{0\}$.

$\Leftarrow$

If $u + w = 0, u \in U, w \in W$, then $w = -u \in U$. So $w \in 𝑈 ∩ 𝑊 \Rightarrow w = 0$. So $u = 0$. Then from 1.45, $U+W$ is a direct sum.

$\square$

Sums of subspaces are analogous to unions of subsets. Similarly, direct sums of subspaces are analogous to disjoint unions of subsets. No two sub- spaces of a vector space can be disjoint, because both contain 0. So disjoint- ness is replaced, at least in the case of two subspaces, with the requirement that the intersection equal {0}.