HW 1 - Vector Spaces

1.A

# 1

Inverse Complex Number
Suppose $a, b \in R$ and one $a, b$ is not equal to 0. Find $c, d \in R$ such that:

\frac{1}{a + b i} = c + d i

Proof

I claim that the values:

c = \frac{a}{a^{2} + b^{2}}, d = \frac{- b}{a^{2} + b^{2}}

Works. First, let $a + b i$ be arbitrary, such that one of $a, b \neq 0$ . Then notice that:

\begin{aligned} \frac{1}{a + b i} & = \frac{a - b i}{(a + b i) (a - b i)} \\ = \frac{a - b i}{a^{2} + b^{2}} \\ = \frac{a}{a^{2} + b^{2}} + \frac{- b}{a^{2} + b^{2}} i \end{aligned}

Thus, choose $c = \frac{a}{a^{2} + b^{2}} \in R$ and $d = \frac{- b}{a^{2} + b^{2}} \in R$ .

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# 3

Find two distinct square roots of $i$ .

Proof
We are asked to find $\sqrt{i}$ , but more specifically, what value $x$ has it such that $x^{2} = i$ . Notice first that if we find some $x$ value that satisfies this, then $- x$ also satisfies this since:

(- x)^{2} = x^{2} = i

So therefore, let's just consider the $+ x$ case. Suppose, for instance, that $x$ can be represented as a complex number, so $x \in C$ . Then $x = a + b i$ where $a, b \in R$ . Therefore:

x^{2} = (a + b i)^{2} = a^{2} - b^{2} + 2 a b i = i

So clearly, then $2 a b = 1$ and we also require that $a^{2} - b^{2} = 0$ . Looking at the latter, then:

a^{2} - b^{2} = 0 \Rightarrow a^{2} = b^{2} \Rightarrow | a | = | b |

Now notice the first equation:

2 a b = 1 \Rightarrow a b = \frac{1}{2}

Now since $\frac{1}{2} > 0$ then either both $a, b$ are negative, or $a, b$ are positive. This is okay, as if we replace $a, b$ with $- a, b$ in the definition for $x$ , we expect that our solution is still correct:

x^{2} = (- a - b i)^{2} = [(- 1) (a + b i)]^{2} = (a + b i)^{2} = \dots

Thus, we can just consider the $+ a, + b$ case. Thus, we know $| a | = a$ and $| b | = b$ , so since $| a | = | b |$ then $a = b$ . Thus, let's just consider the $a$ variable:

\begin{aligned} a b & = \frac{1}{2} \\ a \cdot a & = \frac{1}{2} \\ a^{2} & = \frac{1}{2} \\ a & = \pm \frac{\sqrt{2}}{2} \end{aligned}

Therefore, since $a = b$ and we need to consider the $- a, - b$ case, then our two solutions are:

x_{1} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i, x_{2} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i

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# 8

Existence of the multiplicative inverse.

For all $α \in C$ with $α \neq 0$ , there exists a unique $β \in C$ such that $α β = 1$ .

Proof
Let $α \in C$ be arbitrary, where $α \neq 0$ . Then, by definition, $α = a + b i$ for some $a, b \in R$ . Notice that for $a + b i$ that both cannot be 0, since if that's the case then $a + b i = 0 = α$ but that contradicts $α \neq 0$ .

But from HW 1 - Vector Spaces#1 we know that there exists the number $c + d i \in C$ such that $\frac{1}{a + b i} = c + d i$ . Choose $β = c + d i$ . Then, replace $a + b i$ with $α$ and $c + d i$ with $β$ and we get that $\frac{1}{α} = β$ . Multiply both sides by $α$ and we get that $1 = α β$ .

Thus, we can always choose $β$ such that $α β = 1$ .
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# 11

Theorem

There doesn't exist some $λ \in C$ such that:

λ (2 - 3 i, 5 + 4 i, - 6 + 7 i) = (12 - 5 i, 7 + 22 i, - 32 - 9 i)

Proof
Assume for contradiction that there is such a $λ \in C$ . Then, we can use the operation of scalar multiplication to show that:

λ (2 - 3 i, 5 + 4 i, - 6 + 7 i) = (2 λ - 3 λ i, 5 λ + 4 λ i, - 6 λ + 7 λ i) = (12 - 5 i, 7 + 22 i, - 32 - 9 i)

By definition, these vectors are equal if their components equal, so then:

2 λ - 3 λ i = 12 - 5 i, . . .

Notice then that we require that real components equal with real components, and likewise for imaginary (derived from the definition of complex numbers). Then:

2 λ = 12, - 3 λ = - 5

So the first equation would indicate $λ = 6$ , but plugging into the second equation we get that $(- 3) λ = (- 3) (6) = 18 \neq - 5$ . This is a contradiction, so via proof by contradiction, then the theorem holds.
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# 12

Theorem

$(x + y) + z = x + (y + z)$ for all $x, y, z \in F^{n}$ .

Proof
We use the fact that $F$ is a field, so it contains the property of associativity. Thus, let $x, y, z \in F^{n}$ be arbitrary. We may say that:

x = (x_{1}, . . ., x_{n}), y = (y_{1}, . . ., y_{n}), z = (z_{1}, . . ., z_{n})

Now, consider the following:

\begin{aligned} (x + y) + z & = [(x_{1}, . . ., x_{n}) + (y_{1}, . . ., y_{n})] + (z_{1}, . . ., z_{n}) \\ = (x_{1} + y_{1}, . . ., x_{n} + y_{n}) + (z_{1}, . . ., z_{n}) & (Vector Addition Operator) \\ = ((x_{1} + y_{1}) + z_{1}, . . ., (x_{n} + y_{n}) + z_{n}) & (Vector Addition Operator) \\ = (x_{1} + (y_{1} + z_{1}), . . ., x_{n} + (y_{n} + z_{n})) & (Associativity of F) \\ = (x_{1}, . . ., x_{n}) + (y_{1} + z_{1}, . . ., y_{n} + z_{n}) & (Vector Addition) \\ = (x_{1}, . . ., x_{n}) + [(y_{1}, . . ., y_{n}) + (z_{1}, . . ., z_{n})] & (Vector Addition) \\ = x + (y + z) \end{aligned}

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1.B

# 1

Theorem

$- (- \vec{v}) = \vec{v}$ for all $\vec{v} \in V$ .

Proof
Notice that:

\begin{aligned} (- \vec{v}) + (- (- \vec{v})) & = \vec{0} & (Additive Inverse) \\ \vec{v} + (- \vec{v}) + (- (- \vec{v})) & = \vec{v} & (Add \vec{v} to both sides) \\ \vec{0} + (- (- \vec{v})) & = \vec{v} & (Definition of \vec{0}) \\ - (- \vec{v}) & = \vec{v} & (Definition of \vec{0}) \end{aligned}

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# 2

Theorem

Suppose that $a \in F$ and $\vec{v} \in V$ , and $a \vec{v} = \vec{0}$ . Then either $a = 0$ or $\vec{v} = \vec{0}$ .

Proof
If $a = 0$ already then we are done, so instead suppose that $a \neq 0$ . Notice then that, by the properties of the field $F$ that the multiplicative inverse $\frac{1}{a}$ exists. Thus:

\begin{aligned} \frac{1}{a} (a \vec{v}) & = \vec{v} & (\frac{a}{a} = 1) \\ \frac{1}{a} \vec{0} & = \vec{v} & (Given) \\ \vec{0} & = \vec{v} & (Theorem 1.30 from the Textbook) \end{aligned}

Therefore, we either get $a = 0$ , or we get $\vec{v} = \vec{0}$ .
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# 4

The empty set $\emptyset$ is not a vector space. It doesn't even include $\vec{0}$ , since if it did then $\vec{0} \in \emptyset$ is itself a contradiction. But $\vec{0}$ is needed in the set to consider it a vector space. Thus, $\emptyset$ is not a vector space.

# 5

Theorem

The conditions of a vector space $V$ are unchanged if, instead of confirming the additive inverse for every $\vec{v} \in V$ , we instead say that $0 \vec{v} = \vec{0}$ for all $\vec{v} \in V$ .

Proof

We try to prove that the additive inverse exists for all $\vec{v} \in V$ using only the definition of a vector space (except for the additive inverse's existence itself) and the condition that $0 \vec{v} = \vec{0}$ .

Let $\vec{v} \in V$ be arbitrary. Notice that:

\begin{aligned} 0 \vec{v} & = \vec{0} & (Given) \\ (1 + (- 1)) \vec{v} & = \vec{0} & (0 \equiv 1 + (- 1)) \\ \vec{v} + (- \vec{v}) & = \vec{0} & (Distributive Property) \end{aligned}

Thus, we found the vector $- \vec{v}$ such that the properties of the additive inverse exists
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1.C

# 1

a

The set:

S = {(x_{1}, x_{2}, x_{3}) \in F^{3} : x_{1} + 2 x_{2} + 3 x_{3} = 0}

Is a subspace.

$\vec{0} = (0, 0, 0) \in S$ .
Let $x, y \in S$ be arbitrary. You may say that $x = (x_{1}, x_{2}, x_{3})$ such that $x_{1} + 2 x_{2} + 3 x_{3} = 0$ . Similarly, $y = (y_{1}, y_{2}, y_{3})$ and $y_{1} + 2 y_{2} + 3 y_{3} = 0$ . Notice then that: $$
x + y = (x_1 + y_1, x_2 + y_2, x_3 + y_3)

B u t n o t i c e t h a t :

\begin{align}
(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
\end

S o b y d e f i n i t i o n t h e n $ x + y \in S $ . 3. L e t $ α \in F $ b e a r b i t r a r y . N o t i c e t h a t $ α x = (α x_{1}, α x_{2}, α x_{3}) $, a n d t h a t :

\begin{align}
\alpha x_1 + 2\alpha x_2 + 3\alpha x_3 &= \alpha(x_1 + 2x_2 + 3x_3) \
&= \alpha(0) \
&= 0
\end

So then by definition again $\alpha x \in S$. #### b The set:

S = {(x_1, x_2, x_3) \in \mathbb{F} : x_1 + 2x_2 + 3x_3 = 4}

Is *not* a subspace. The $\vec{0} = (0,0,0) \notin S$ since $0 + 2\cdot 0 + 3\cdot 0 = 0 \neq 4$. #### c The set:

S = {(x_1, x_2, x_3) \in \mathbb{F}^3 : x_1x_2x_3 = 0}

Is *not* a subspace, since if we consider $v = (1,1,0)$ and $w = (0,0,1)$, then $v + w = (1,1,1) \notin S$. #### d The set:

S = {(x_1, x_2, x_3) \in \mathbb{F}^3 : x_1 = 5x_3}

*is* a subspace. Similar to my reasoning for 1.a, $\vec{0} \in S$, the set is closed under vector addition still, and also closed under scalar multiplication. ### \# 3 > [!theorem] > The set of differentiable real-valued functions $f$ on the interval $(-4,4)$ such that $f'(-1) = 3f(2)$ is a subspace of $\mathbb{R}^{(-4,4)}$. *Proof* For convenience, denote our set $S$: 1. $\vec{0}$ in this case is the function $f(x) = 0$ defined on $(-4,4)$. Notice that $f'(x) = 0$ still, so then:

f'(-1) = 0 = 3f(2) = 3\cdot 0

S o t h e n $ \vec{0} \in S $ . 2. L e t $ f, g \in S $ b e a r b i t r a r y, s o t h u s $ f^{'} (- 1) = 3 f (2) $ a n d l i k e w i s e f o r $ g $ a s w e l l . N o t i c e t h e n t h a t $ (f + g)^{'} (x) = f^{'} (x) + g^{'} (x) $, s o t h e n :

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

S o t h e n $ f + g \in S $ . 3. L e t $ α \in R $ b e a r b i t r a r y, a n d $ f \in S $ . T h e n $ f^{'} (- 1) = 3 f (2) $ . T h u s, s i n c e $ (α f)^{'} (x) = α f^{'} (x) $ t h e n :

(\alpha f)'(-1) = \alpha f'(-1) = \alpha 3f(2) = 3\alpha f(2) = 3(\alpha f)(2)

So therefore $\alpha f \in S$. Thus, $S$ is a subspace of $\mathbb{R}^{(-4,4)}$. ☐ ### \# 5 > [!theorem] > $\mathbb{R}^2$ is *not* a subspace of $\mathbb{C}^2$. *Proof* Namely, the scalar multiplication fails to be closed. Notice that we can have $\alpha = i$, but if we had some vector, say $\vec{x} = (1,1)$, then $\alpha \vec{x} = i(1,1) = (i,i) \notin \mathbb{R}^2$. ☐ ### \# 7 We need to find a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition and taking additive inverses ($-u \in U$), but $U$ isn't a subspace of $\mathbb{R}^2$. Notice that as a result we'd require that $\vec{0} \in U$ since $u + (-u) = \vec{0} \in U$ by definition. Thus, we need to find a way to make $U$ not be closed under scalar multiplication. The only way this could happen is that we scale by $\alpha$ such that the values of $U$ that make the vector are of a smaller "resolute" set. An example of this would be $U = \{(x_1, x_2) : x_1, x_2 \in \mathbb{Q}\}$. Here we see that $\vec{0} \in U$ while also $U$ is closed under vector addition for similar reasons it is in $\mathbb{R}$. However, $U$ isn't closed under scalar multiplication since notice that, as an example, the vector $\vec{u} = (1, 1) \in U$, but if $\alpha = \sqrt{2}$, then:

\alpha \vec{u} = (\sqrt{2}, \sqrt{2}) \notin U

Thus $U$ isn't a subspace. ### \# 8 We need to give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ isn't a subspace of $\mathbb{R}$. Notice too that by definition we could have $0\vec{u} = \vec{0} \in U$, so the zero vector is in $U$. By being forced to be closed under scalar multiplication, we can have things like $x_2 = x_1 + 1$, $x_2 = f(x_1)$ where $f$ isn't a linear function, and if we instead had $x_2 = cx_1$ then we just get a vector space which we don't want. Thus, let's try a different approach. We know a $U \cup W$ may not necessarily be a subspace. This happens when we have $0$'s in one variable that affect the other variables be non-zero. Thus, consider if $U = \{(x_1, x_2) : x_1, x_2 \in \mathbb{R} \land (x_1 = 0 \lor x_2 = 0\}$. So for instance $(1,0), (0,1) \in U$ but $(1,1) \notin U$. Notice by definition that $(0,0) \in U$ and better still if we have $\alpha (x_1, 0) = (\alpha x_1, 0) \in U$ and likewise if we did $\alpha (0, x_2) \in U$. But notice that if we have $(0,1) = v \in U$ and $(1,0) = u \in U$, then $u + v = (1,1) \notin U$, so $U$ is not a subspace. ### \# 11 > [!theorem] > The intersection of every collection of subspaces of $V$ is a subspace of $V$. *Proof* We prove this in a similar way we'd prove any subspace of $V$. Denote the collection of all intersected subspaces of $V$ as $U_S$. 1. Notice that $\vec{0} \in U_S$ since all subspaces $U \in U_S$ have $\vec{0} \in U$ by definition of subspace. 2. Let $\vec{v}, \vec{u} \in U_S$ be arbitrary. It follows that $\vec{v}, \vec{u} \in U_S$ are in all subspaces $U \in U_S$. But for each subspace $U \in U_S$ since $U$ is a subspace then since $\vec{v}, \vec{u} \in U$ then clearly $\vec{v} + \vec{u} \in U$ for each subspace. Thus, since it's true for all subspaces $U \in U_S$ then clearly when we intersect them all then $\vec{v} + \vec{u}$ is also in $U_S$ as required. 3. Let $\alpha \in \mathbb{F}$ and $\vec{v} \in U_S$ be arbitrary. It follows that $\vec{v} \in U_S$ is in all subspaces $U \in U_S$. But for each subspace $U \in U_S$ since $U$ is a subspace then since $\vec{v} \in U$ then clearly $\alpha \vec{v} \in U$ for each subspace. Then, since it's true for all subspaces $U \in U_S$ then clearly when we intersect them all then $\alpha \vec{v}$ is also in $U_S$ as required. ☐ ### \# 12 > [!theorem] > Let $U, W$ be subspaces of $V$. Then $U \cup W$ is a subspace of $V$ iff one of the subspaces is contained in the other (ie: $U \subseteq W$ or $W \subseteq U$). *Proof* Let $U, W$ be arbitrary subspaces of $V$. We do the ($\rightarrow$) direction first: 1. Consider that $U \cup W$ is a subspace of $V$. First, notice that if $W \subseteq U$ then we are done, so suppose that $W \not \subseteq U$. If this is the case, then there exists some element, let's call it $y$, such that $y \in U$ and $y \notin W$. We'll try to prove that instead $U \subseteq W$. Notice that, via proof by contradiction, if $U \not \subseteq W$, then there'd be possibly another element $z$ such that $z \in W$ and $z \notin U$. But look! Notice that both $y,z \in U \cup W$. Since $U \cup W$ is a subspace, then clearly $y + z \in U \cup W$. If $y + z \in W$ as case 1, then since $z \in W$ as well then $y + z + (-z) \in W$, so then $y \in W$, which is a contradiction! As a case 2, we'd consider $y + z \in U$ and consider that since $y \in U$ then $y + z + (-y) \in U$ so then $z \in U$ which is also a contradiction. Thus, no matter what, we always get a contradiction, so then $U \not \subseteq W$ is false, so $U \subseteq W$. Therefore, either $W \subseteq U$ or $U \subseteq W$ as required. 2. Consider that $U \subseteq W$ or $W \subseteq U$. Without loss of generality, say $U \subseteq W$. Then if we consider $U \cup W$ then we notice that $U \cup W = W$, which is given as a subspace. So $U \cup W$ is a subspace. ☐ ### \# 15 > [!theorem] > Suppose $U$ is a subspace of $V$. Then $U + U = U$. *Proof* Notice that:

U + U = {u_1 + u_2 | u_1, u_2 \in U}

B u t s i n c e $ U $ i t s e l f i s a v e c t o r s p a c e (r e a l l y a s u b s p a c e), t h e n s i n c e $ u_{1}, u_{2} \in U $ t h e n $ u_{1} + u_{2} \in U $ a l w a y s, s o r e a l l y :

U + U = {u | u \in U} = U

Thus $U + U = U$. ☐ ### \# 19 It isn't necessarily the case that $U_1 = U_2$, as the following counterexample shows: Consider how $U_1, U_2, W$ are subspaces of $V$. Notice that:

U_1 + W = {u_1 + w | u_1 \in U_1, w \in W}

a n d s i m i l a r f o r $ U_{2} + W $ . I t i s a l s o g i v e n t h a t $ U_{1} + W = U_{2} + W $, s o e q u a t i n g b o t h s i d e s s a y s t h a t :

U_1 + W = {u_1 + w | u_1 \in U_1, w \in W} = {u_2 + w | u_2 \in U_2, w \in W} = U_2 + W

B u t t h i s s u g g e s t s t h a t w e s h o u l d f i n d d i f f e r e n t s u b s p a c e s $ U_{1}, U_{2}, W $ s u c h t h a t $ W $ s e e m s t o " c o n n e c t " t h e t w o t o m a k e t h e l a r g e r v e c t o r s p a c e $ V $, o r s o m e r e l a t i v e l y l a r g e e n o u g h v e c t o r s p a c e . L e t $ W = V = R^{3} $ . F u r t h e r, l e t $ U_{1} = {(x, y, 0) : x, y \in R} $ a n d $ U_{2} = {(0, x, y) : x, y \in R} $ . N o t i c e t h a t h e r e a l l $ U_{1}, U_{2}, W $ a r e s u b s p a c e s o f $ V = R^{3} $ (e v e n $ W $, b e c a u s e $ W \subseteq V $ e v e n i f $ W = V $), b u t s e e t h a t :

U_1 + W = V = U_2 + W

While still $U_1 \neq U_2$. Thus, this is a valid counterexample to the claim that $U_1 = U_2$. > [!note] > I could've had $W = \{(x, 0, y) : x, y \in \mathbb{R}^3\}$, and had $U_1 = \{(x,0,0) : x \in \mathbb{R}\}$ and $U_2 = \{(0,0,x) : x \in \mathbb{R}\}$ and the same argument would apply (except here we don't require that $W = V$). ### \# 20 Suppose

U = {(x,x,y,y) \in \mathbb{F}^4 : x, y \in \mathbb{F}}

I p r o p o s e t h e s u b s p a c e :

W = {(0,x, 0, y) : x ,y \in \mathbb{F}}

w i l l a l l o w $ F^{4} = U \oplus W $ . W e p r o v e t h a t t h i s i s c a s e a s f o l l o w s, u s i n g i f f s t a t e m e n t s o n l y (t o a l l o w f o r a n $ = $ i n s t e a d o f l o o s e $ \subseteq $) . L e t $ \vec{x} \in F^{4} $ b e a r b i t r a r y, s o t h e n :

\vec{x} = (x,y,z,w): x,y,z,w \in \mathbb

N o t i c e t h a t w e c a n s p l i t t h e v e c t o r $ \vec{x} $ i n t o t h e f o l l o w i n g :

\vec{x} = (x, 0, 0, 0) + (0,y,0,0) + (0,0,z,0) + (0,0,0,w)

v i a t h e o p e r a t i o n s i m p o s e d o n $ F^{4} $ . N o w, n o t i c e t h a t f o r $ y $ s i n c e $ F $ i s a f i e l d t h e n c l e a r l y $ y = x + (y - x) $ . L i k e w i s e, $ w = z + (w - z) $ . W e c a n m a k e t h o s e r e p l a c e m e n t s a s f o l l o w s :

\vec{x} = (x, 0, 0, 0) + (0,x + (y -x),0,0) + (0,0,z,0) + (0,0,0,z + (w - z))

A n d u n d e r t h e o p e r a t i o n s o f v e c t o r a d d i t i o n, $ \vec{x} $ c a n b e o r g a n i z e d a s :

\begin{align}
\vec{x} &= (x, x, 0, 0) + (0,0, z, z) + (0,y-x, 0, 0) + (0, 0, 0, w-z) \
&= (x,x,z,z) + (0,y-x, 0, w-z)
\end

B u t n o t i c e t h a t $ (x, x, z, z) \in U $ a n d $ (0, y - z, 0, w - z) \in W $, s o c l e a r l y a t l e a s t $ U + W = F^{4} $ . N o w w e s h o w i t^{'} s a d i r e c t s u m s p e c i f i c a l l y . N o t i c e t h a t :

U \cap W = {0}

Since if some vector $\vec{y} \neq \vec{0}$ was such that $\vec{y} \in U$ and $\vec{y} \in W$ then clearly, since $\vec{y} = (x,y,z,w)$ then $x = 0$, which would make $y = 0$ and so on for $z,w = 0$, so then $\vec{y} = \vec{0}$ this whole time. Thus, $U, W$ are a direct sum (via 1.45), so then $U \oplus W = \mathbb{F}^4$. ### \# 23 > [!proposition] > If $U_1, U_2, W$ are subspaces of $W$ such that: >

V = U_1 \oplus W = U_2 \oplus W
$T h e n $ U_{1} = U_{2} $ .$

This is a false proposition. Consider $V = {(x, 0, y) : x, y \in R}$ . Then let $U_{1} = {(x, 0, 0) : x \in R}$ , $U_{2} = {(0, 0, y) : y \in R}$ , and $W = {(z, 0, z) : z \in R^{3}}$ . Notice that these follow what the proposition says, but clearly $U_{1} \neq U_{2}$ .

# 24

Theorem

A function $f : R \to R$ is even if $f (- x) = f (x)$ and odd if $f (- x) = - f (x)$ for all $x \in R$ . Let $U_{e}$ denote the set of real-valued even functions on $R$ and likewise $U_{o}$ for all the odd functions on $R$ . Then $R^{R} = U_{e} \oplus U_{o}$ .

Proof
We need to find a way to first see if we can span $R^{R}$ solely by a combination of a single even function, and a single odd function. Then we need to show that these combinations are unique.

First, let $f \in R^{R}$ be an arbitrary function. Then notice that:

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

But notice the function $f_{e}$ on the left is an even function since:

f_{e} (x) = \frac{f (x) + f (- x)}{2}, f_{e} (- x) = \frac{f (- x) + f (x)}{2} = f_{e} (x)

Similarly, the $f_{o}$ function on the right is odd since:

f_{o} (x) = \frac{f (x) - f (- x)}{2}, f_{o} (- x) = \frac{f (- x) - f (x)}{2} = - f_{o} (x)

Thus, we can always write $f$ as a sum of an even and odd function:

f (x) = f_{e} (x) + f_{o} (x)

Now we show that this combination must be unique. We'll try to show that $U_{e} \oplus U_{o}$ via definition 1.45. Notice that the zero function $f_{0} (x) = 0$ is both odd and even (try this yourself), so then $f_{0} \in U_{e} \cap U_{o}$ . We now show that any other non-zero function $f \notin U_{e} \cap U_{o}$ . Assume via contradiction that $f \in U_{e} \cap U_{o}$ . Then $f (x) = f (- x)$ and $f (- x) = - f (x)$ . But notice then that if we combine the equations that:

f (x) = f (- x) = - f (x) \Rightarrow 2 f (x) = 0 \Rightarrow f (x) = 0

But this contradicts $f$ not being the zero-function, so then there mustn't be any non-zero function in the set. Thus $U_{e} \cap U_{o} = {f_{0}}$ , so $U_{e}, U_{o}$ are a direct sum.

Therefore, $U_{e} \oplus U_{o} = R^{R}$ .
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