# So V is not vector over field \Bbb{R}

• MHB
• Another1
In summary, the conversation discusses the properties of vector spaces and the scalar multiplication operation in the vector space V over the field $\mathbb{R}$. It is questioned whether the condition of $1_F\cdot v = v$ is satisfied for the scalar multiplication operation $\odot$ in this problem. It is shown that this condition is not satisfied for all $u \in \mathbb{R}^2$.
Another1

$$\displaystyle Let$$ $$\displaystyle V=\Bbb{R}^2$$ and $$\displaystyle {u=(u_1,u_2), v=(v_1.v_2)}\in\Bbb{R}^2$$ , $$\displaystyle {k}\in \Bbb{R}$$ define of operation $$\displaystyle u\oplus v = (u_1+v_1,u_2+v_2)$$ and $$\displaystyle k \odot u =(2ku_1,2ku_2)$$ check V is vector over field $$\displaystyle \Bbb{R}$$ ?
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I think
in a property of the additive inverse $$\displaystyle (-u)+u=0=u+(-u)$$

from define $$\displaystyle k \odot u =(2ku_1,2ku_2)$$
So $$\displaystyle (-1)u = (-1) \odot u =(2(-1)u_1,2(-1)u_2) = (-2u_1,-2u_2)$$
$$\displaystyle (-u)+u = (-2u_1,-2u_2) + (u_1,u_2)\ne 0$$

Hi Another.

In a vector space $V$ over a field $F$, we need to have for every $v\in V$ that
$$1_F\cdot v\ =\ v$$
where $1_F$ is the multiplicative identity in $F$. Is this condition satisfied for the scalar multiplication $\odot$ in this problem? In other words, is
$$1\odot u\ =\ u$$
true for all $u\in\mathbb R^2$?

## 1. What does it mean for V to not be a vector over the field of real numbers?

This means that V does not satisfy the properties of a vector space over the field of real numbers. In other words, the operations of addition and scalar multiplication do not behave as they should according to the rules of vector spaces.

## 2. Can V still be a vector space over a different field?

Yes, V can still be a vector space over a different field, such as the field of complex numbers. This means that the operations of addition and scalar multiplication follow the rules of vector spaces when using complex numbers instead of real numbers.

## 3. What are the consequences of V not being a vector over the field of real numbers?

The consequences can vary depending on the specific properties that V does not satisfy. However, some consequences may include the inability to use certain mathematical techniques or the need to find alternative methods to solve problems involving V.

## 4. How can I determine if V is a vector over the field of real numbers?

To determine if V is a vector over the field of real numbers, you can check if the properties of a vector space hold true. These properties include closure under addition and scalar multiplication, existence of a zero vector and additive inverses, and distributivity and associativity of operations.

## 5. Are there any real-world examples of V not being a vector over the field of real numbers?

Yes, there are real-world examples of V not being a vector over the field of real numbers. For instance, the set of all continuous functions on a closed interval does not form a vector space over the field of real numbers because the sum of two continuous functions may not be continuous. However, it can be a vector space over the field of complex numbers.

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