MHB Are $V_1$ and $V_2$ Vector Spaces According to Defined Properties?

[mathmari]

Hey!

I want to check if the following are true.

1. $V_1=\{a\in \mathbb{R}\mid a>0\}$ with the common multiplication as the vector addition and the scalar multiplication $\lambda \odot v=v^{\lambda}$ is a $\mathbb{R}$-vector space.
2. $V_2=\{(x,y)\in \mathbb{Q}^2 \mid x^2=-y^2\}$ with the addition and scalar multiplication of $\mathbb{Q}^2$ is a $\mathbb{Q}$-vector space.

We have the following:
Let $K$ be a field. A vector space over $K$ (or $K$-vector space) is a set $V$ with an addition $V \times V \rightarrow V : (x, y) \mapsto x + y$ and a scalar multiplication $K \times V \rightarrow V : (\lambda , x) \mapsto \lambda \cdot x$, so that the following holds:

• (V1) : $(V,+)$ is an abelian group, with the neutral element $0$.
• (V2) : $\forall a, b \in K, \forall x \in V : (a + b) \cdot x = a \cdot x + b \cdot x$
• (V3) : $\forall a \in K, \forall x, y \in V : a \cdot (x + y) = a \cdot x + a \cdot y$
• (V4) : $\forall a, b \in K, \forall x \in V : (ab) \cdot x = a \cdot (b \cdot x)$
• (V5) : $\forall x \in V : 1 \cdot x = x$ ( $1 = 1_K$ is the identity in $K$).

So, we have to check at both cases these $5$ properties, right?

1. We have that $x+y=x\cdot y$ and $x\cdot y=x\odot y$, or not?
   For (V2) we have that for $a,b\in \mathbb{R}$ and $x\in V_1$
   $(a\cdot b)\odot x=x^{a\cdot b}$ but $a\odot x \cdot b\odot x=x^a\cdot x^b=x^{a+b}$. So the property (V2) is not satisfied. Therefore, $V_1$ is not a vector space.
   Is this correct? (Wondering)
   
   We could also prove it as follows, or not? (Wondering)
   So that $V_1$ is a $\mathbb{R}$-vector space it must be closed under scalar multiplication. We have that $a\in V_1$, so $a>0$. Then $-1\in \mathbb{R}$ bur $(-1)\cdot a=-a<0$ and so $(-1)\cdot a\notin V_1$.
   
2. The properties (V2)-(V5) are satisfied, or not? How can we check the property (V1) ? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

So, we have to check at both cases these $5$ properties, right?

Additionally, we have to check if vector addition and scalar multiplication are well-defined.
That is, whether the vector space is closed for vector addition, and also closed for scalar multiplication. (Thinking)

1.
We have that $x+y=x\cdot y$ and $x\cdot y=x\odot y$, or not?

Let's phrase it: $x+y=xy$ and $\lambda\cdot x=\lambda\odot x$.
That is, we have a good chance to mix up additions and multiplications.
So we need to be mindful that we consistently apply:
\begin{array}{|c|c|c|}
\hline
Notation & Operation & Result
\\
\hline
x + y & \text{vector addition} & \text{vector} \\
xy & \text{multiplication of reals} & \text{vector} \\
\lambda \cdot x & \text{multiplication of scalar with vector} & \text{vector} \\
\lambda\odot x & \text{multiplication of scalar with vector} & \text{vector} \\
x^\lambda & \text{power operation of reals} & \text{vector}\\
\hline
a+b & \text{addition of reals} & \text{scalar} \\
a\cdot b & \text{multiplication of reals} & \text{scalar} \\
\hline
\end{array}

In case of doubt, we could also make it $\overrightarrow x + \overrightarrow y = \overrightarrow {xy}$. (Thinking)

For (V2) we have that for $a,b\in \mathbb{R}$ and $x\in V_1$
$(a\cdot b)\odot x=x^{a\cdot b}$ but $a\odot x \cdot b\odot x=x^a\cdot x^b=x^{a+b}$. So the property (V2) is not satisfied. Therefore, $V_1$ is not a vector space.
Is this correct? (Wondering)

Shouldn't (V2) be: $(a + b)\odot x \overset ? = a\odot x + b\odot x$? (Wondering)
Then we have:
$$(a + b)\odot x = x^{a+b} = (x^a)(x^b) = (a\odot x)(b\odot x) = a\odot x + b\odot x$$
(Thinking)

 

[mathmari]

Shouldn't (V2) be: $(a + b)\odot x \overset ? = a\odot x + b\odot x$? (Wondering)
Then we have:
$$(a + b)\odot x = x^{a+b} = (x^a)(x^b) = (a\odot x)(b\odot x) = a\odot x + b\odot x$$
(Thinking)

I got confused with "common multiplication as the vector addition". What does this mean? That the addition of that space is the common multiplication? (Wondering)

 

[I like Serena]

I got confused with "common multiplication as the vector addition". What does this mean? That the addition of that space is the common multiplication? (Wondering)

Each vector is represented by a single real number.
Adding two such vectors is executed by multiplying the real numbers they represent.
We can distinguish by writing:
$$\overrightarrow x + \overrightarrow y = \overrightarrow {xy}$$
As an example:
$$\overrightarrow 2 + \overrightarrow 3 = \overrightarrow {2\cdot 3} = \overrightarrow {6}$$
(Thinking)

 

[mathmari]

Each vector is represented by a single real number.
Adding two such vectors is executed by multiplying the real numbers they represent.
We can distinguish by writing:
$$\overrightarrow x + \overrightarrow y = \overrightarrow {xy}$$
(Thinking)

So, how can identity V2 be written? (Wondering)

 

[I like Serena]

So, how can identity V2 be written? (Wondering)

(V2) We want to verify if for all $a,b\in \mathbb R, \overrightarrow x \in V$:
$$(a+b)\cdot \overrightarrow x \overset ?= a\cdot \overrightarrow x + b \cdot \overrightarrow x$$
Working it out we get:
$$(a+b)\cdot \overrightarrow x = \overrightarrow{ (a+b)\odot x} = \overrightarrow{ x^{a+b}} \\
a\cdot \overrightarrow x + b \cdot \overrightarrow x = \overrightarrow{a\odot x} + \overrightarrow{b\odot x}
= \overrightarrow{x^a} + \overrightarrow{x^b}
= \overrightarrow{(x^a)(x^b)} = \overrightarrow{x^{a+b}}
$$
So (V2) checks out. (Thinking)

 

[mathmari]

(V2) We want to verify if for all $a,b\in \mathbb R, \overrightarrow x \in V$:
$$(a+b)\cdot \overrightarrow x \overset ?= a\cdot \overrightarrow x + b \cdot \overrightarrow x$$
Working it out we get:
$$(a+b)\cdot \overrightarrow x = \overrightarrow{ (a+b)\odot x} = \overrightarrow{ x^{a+b}} \\
a\cdot \overrightarrow x + b \cdot \overrightarrow x = \overrightarrow{a\odot x} + \overrightarrow{b\odot x}
= \overrightarrow{x^a} + \overrightarrow{x^b}
= \overrightarrow{(x^a)(x^b)} = \overrightarrow{x^{a+b}}
$$
So (V2) checks out. (Thinking)

Ah ok... (Thinking)

When we symbolize the addition of vectors as $\oplus$ we have the following:

(V2) : Let $a,b\in K$ and $x\in V_1$
$(a+ b)\odot x=x^{a+ b}$
$a\odot x\oplus b\odot x=x^a\cdot x^b=x^{a+b}$

(V3) : Let $a\in K$ and $x,y\in V_1$
$a\odot (x\oplus y)=a\odot (x\cdot y)=(x\cdot y)^a=x^a\cdot y^a$
$a\odot x\oplus a\odot y=x^a\cdot y^a$

(V4) : Let $a,b\in K $ and $x\in V_1$
$(a\cdot b)\odot x=x^{a\cdot b}$
$a\odot (b\odot x)=a\odot x^b=(x^b)^a=x^{a\cdot b}$

(V5) : Let $x\in V_1$
$1\odot x=x^1=x$Is everything correct so far? (Wondering)

It is left to check the first property, right? How could we do that? (Wondering)

 

[I like Serena]

Ah ok... (Thinking)

When we symbolize the addition of vectors as $\oplus$ we have the following:

(V2) : Let $a,b\in K$ and $x\in V_1$
$(a+ b)\odot x=x^{a+ b}$
$a\odot x\oplus b\odot x=x^a\cdot x^b=x^{a+b}$

(V3) : Let $a\in K$ and $x,y\in V_1$
$a\odot (x\oplus y)=a\odot (x\cdot y)=(x\cdot y)^a=x^a\cdot y^a$
$a\odot x\oplus a\odot y=x^a\cdot y^a$

(V4) : Let $a,b\in K $ and $x\in V_1$
$(a\cdot b)\odot x=x^{a\cdot b}$
$a\odot (b\odot x)=a\odot x^b=(x^b)^a=x^{a\cdot b}$

(V5) : Let $x\in V_1$
$1\odot x=x^1=x$Is everything correct so far? (Wondering)

Those power identities are not generally true.
Consider for instance that $-1 = (-1)^{1} = (-1)^{(1/2 \cdot 2)} \ne ((-1)^{1/2})^2 = \text{undefined}$, since we cannot take the square root of a negative number within the reals.
And $-1 = (-1)^{1} = (-1)^{(2 \cdot 1/2)} \ne ((-1)^2)^{1/2} = 1^{1/2} = 1$. (Worried)

It is left to check the first property, right? How could we do that? (Wondering)

Check the axioms of an abelian group?
That is, closure of addition, associativity, existence of neutral element, existence of additive inverse, commutativity. (Thinking)

 

[mathmari]

Those power identities are not generally true.
Consider for instance that $-1 = (-1)^{1} = (-1)^{(1/2 \cdot 2)} \ne ((-1)^{1/2})^2 = \text{undefined}$, since we cannot take the square root of a negative number within the reals.
And $-1 = (-1)^{1} = (-1)^{(2 \cdot 1/2)} \ne ((-1)^2)^{1/2} = 1^{1/2} = 1$. (Worried)

$x$ must be always positive, or not? (Wondering)

Check the axioms of an abelian group?
That is, closure of addition, associativity, existence of neutral element, existence of additive inverse, commutativity. (Thinking)

Let $x,y\in V_1$ then $x\oplus y=xy\in V_1$, since $x>0$ and $y>0$ and so $xy>0$.

The neutral element doesn'tbelong to $V_1$, does it? (Wondering)

 

[I like Serena]

$x$ must be always positive, or not? (Wondering)

Yes.
We require that $x$ is positive and that the scalars are real for those power identities to hold.
They are, but perhaps we should mention that as part of the verification? (Wondering)

Let $x,y\in V_1$ then $x\oplus y=xy\in V_1$, since $x>0$ and $y>0$ and so $xy>0$.

The neutral element doesn'tbelong to $V_1$, does it? (Wondering)

Let's see... we're looking for an element $e$, such that $e\oplus x=x\oplus e = x$.
What does that mean for $e$? (Wondering)

 

[mathmari]

Yes.
We require that $x$ is positive and that the scalars are real for those power identities to hold.
They are, but perhaps we should mention that as part of the verification? (Wondering)

Let's see... we're looking for an element $e$, such that $e\oplus x=x\oplus e = x$.
What does that mean for $e$? (Wondering)

$e$ must be $1$, right? (Wondering)

Doesn't $0$ must also an element of the set? (Wondering)

 

[I like Serena]

$e$ must be $1$, right? (Wondering)

Yep.

Doesn't $-0$ must also an element of the set? (Wondering)

Here's another mixup in additions and multiplications..
The $0$ vector must indeed be an element of the set. That is, the vector that is neutral for addition.
However, that happens to be $\overrightarrow 1$, which is indeed part of the set! (Tauri)

In particular $0$ itself is not part of the set, so neither is its additive inverse.
And the additive inverse of $\overrightarrow 1$ is itself.

 

[mathmari]

Check the axioms of an abelian group?
That is, closure of addition, associativity, existence of neutral element, existence of additive inverse, commutativity. (Thinking)

Closure of addition: $x\oplus y=x\cdot y\in V_1$

Associativity: $(x\oplus y)\oplus z=(x\cdot y)\cdot z=x\cdot (y\cdot z)=x\cdot (y\oplus z)=x\oplus (y\oplus z)$

Existence of neutral element: $e\oplus x=x=x\oplus e \Rightarrow e\cdot x=x=x\cdot e \Rightarrow e=1$

Existence of additive inverse: $x\oplus y=e=y\oplus x \Rightarrow x\cdot y=e=y\cdot x$, so $y$ is $x^{-1}$ ? (Wondering)

Commutativity: $x\oplus y=x\cdot y=y\cdot x=y\oplus x$ Is everything correct? (Wondering)

 

[I like Serena]

Yep. All correct.
Note that all $x^-1 \in V_1$ since we're talking about all positive reals.

 

[mathmari]

Yep. All correct.
Note that all $x^-1 \in V_1$ since we're talking about all positive reals.

Ok! Thank you very much! (Happy)