Vector space- linear algebra

Question

I don't get how the book solved this problem, because axioms $4$, $5$ and $7$ are about adding not multiplying.

I really need to understand this. Thank you

Answer 1

The key phrase is "vector addition is numerical multiplication"

That is they define the operation $+_V$ as $u+_Vv=u\cdot v$. I explicitly write out the subscript $V$ for the vector addition so as to highlight the addition is not the convention addition that you are familiar with.

We can check that $u+_V1 = u \cdot 1=u$.

I will leave the rest as an exercise.

Answer 2

I certainly understand that it looks confusing when the book writes $u+v=uv$ and $ku=u^k,$ so let us instead introduce two operators $\oplus : V \times V \to V$ and $\odot : \mathbb R \times V \to V$ satisfying some axioms like $(u \oplus v) \oplus w = u \oplus (v \oplus w)$ and $k \otimes (u \oplus v) = (k \odot u) \oplus (k \odot v).$ I won't list all axioms since I don't want to confuse the enumeration with the one in you course literature.

In the example given, they take $V = (0, \infty)$ and define $\oplus$ and $\odot$ by $u \oplus v = uv$ and $k \odot u = u^k.$ Then, for example, they show the second axiom that I listed: $$\begin{align} k \odot (u \oplus v) & = \{\text{ by def. of $\odot$ }\} = (u \oplus v)^k \\ & = \{\text{ by def. of $\oplus$ }\} = (uv)^k \\ & = \{\text{ by one of the power laws }\} = u^k v^k \\ & = \{\text{ by def. of $\oplus$ }\} = u^k \oplus v^k \\ & = \{\text{ by def. of $\odot$ }\} = (k \odot u) \oplus (k \odot v) \end{align}$$

It is very uncommon though with vector spaces where the operations are not derived from ordinary addition and multiplication with a real or complex number. Therefore the operations are normally written as just $+$ and juxtaposition (i.e. $ku$ without an explicit operation).

Answer 3

Here $V$ is the set $\mathbb{R}_{>0}$, so the vector $\mathbf{v}$ is just the real number $v$ (although it is still a vector in $V$).

Now define the operations on $V$: $$u+v=uv \quad\textbf{{Vector addition is numerical multiplication}}\tag{1}$$

$$ku=u^k \quad\textbf{{Scalar multiplication is numerical exponentiation}}\tag{2}$$

Now with operations $(1)$ and $(2)$ set up you have to try each axiom in turn to see if it satisfies them.

For Axiom $4$ we want the definition of the zero vector, i.e., the element $i$ that makes $u+i=u$. It's obviously not $i=0$, as $0\not\in V$ (if it were it would give: $u+0=u\cdot0=0$, and so $u=0$, contradicting the fact $u$ is a positive real). In a sense you can work backwards here, since you know $u\cdot1=u$, where $u$ and $1$ are seen as numbers to be multiplied. Now looking at $(1)$ you see the addition of two vectors gives a numerical product so $u+1=u\cdot1$. But then $u+1=u$, and so the number $1$ acts as the zero-vector in this space.

For Axiom $5$ the negative of a vector $u$ is its reciprocal ($-u=1/u$) $$u+\frac{1}{u}=u\left(\frac{1}{u}\right)=1$$ The point here is we want define negative vectors, so we can have additive inverses. We know when we subtract a vector from itself we get the zero-vector. So since $u+(-u)=1$, we need to solve $u+(-u)=u(-u)=1$, and $-u=1/u$ does this.

For Axiom $7$ we need to show $k(u+v)=ku+kv$, so we can have the property of distributivity when multiplying. We have by $(1)$ that $u+v=uv$, so $k(u+v)=k(uv)$. Now use $(2)$ to give $k(uv)=(uv)^k=u^kv^k$, where some power law is invoked. This gives $u^kv^k=u^k+v^k$ by $(1)$, with $u^k+v^k=ku+kv$ by $(2)$, and so $k(u+v)=ku+kv$ as required.

Now do the rest.