Understanding question 77 in Golan's "Linear Algebra". The question is given below:

Exercise 77
Let $V =
\left\{
    \left.
    \begin{bmatrix}
        a_1    \\
        \vdots \\
        a_5
    \end{bmatrix}\ 
    \right|\ 
    0 < a_i \in \mathbb{R}
\right\}$.
  If $v =
    \begin{bmatrix}
        a_1    \\
        \vdots \\
        a_5
    \end{bmatrix}$
  and $w =
    \begin{bmatrix}
        b_1    \\
        \vdots \\
        b_5
    \end{bmatrix}$
  belong to $V$, and if $c \in \mathbb{R}$, set $v + w =
    \begin{bmatrix}
        a_1b_1 \\
        \vdots \\
        a_5b_5
    \end{bmatrix}$
  and $cv =
    \begin{bmatrix}
        a_1^c  \\
        \vdots \\
        a_5^c
    \end{bmatrix}$.
  Do these operations turn $V$ into a vector space over $\mathbb{R}$?

My questions are:
1-Why was not $V$ a vector space over $\mathbb{R}$? maybe because it did not contain the additive identity element...am I correct?
2-I feel that the answer is yes, but I am unable to totally justify it. I know that the additive identity element under the new defined operations is $\mathbb{1}$ ... am I correct? but how can I find the general form of the inverse of each element?
3-I want to check that the scalar multiplication is distributed over field addition, but what is the definition of the field addition in our case?
 A: Hint: The log function is an isomorphsim $(\Bbb R^{\gt 0}, \times) \to (\Bbb R,+)$. Also, exponentiation distributes over the multiplication of two positive numbers (there are other laws of exponents applicable here).
The OP's $V$ is a commutative group with additive identity the element $(1,1,1,1,1) \in \Bbb R^5$.
The scalar field is the set of real numbers $\Bbb R$ with the familiar operations of addition $+$ and multiplication $\times$.
Of course using the symbol $+$ for $V$ might lead to confusion, and one might want to write, say, $+^{'}$.
A: The inverse of $\begin {pmatrix}b_1\\b_2\\b_3\\b_4\\b_5\end{pmatrix}$ is $\begin {pmatrix}\frac1{b_1}\\\frac1{b_2}\\\frac1{b_3}\\\frac1{b_4}\\\frac1{b_5}\end{pmatrix}$.
The distributivity you seek holds, because $a^{(b+c)}=a^ba^c$.
So far, so good.
The rest of the axioms are easily verified. 
A: *

*Yes, $V$ with the standard addition, has no 0 element.

*Yes, with the new addition, the 0 element is $(1,1\dots,1)^{\mathrm t}$. As to $-(a_1,a_2,\dots,a_n)^{\mathrm t}$, you have to solve $\;a_1x_1=1,\;a_2 x_2=1,\;\dots$. This shouldn't be too hard.

*The usual addition for the exponents. Some details: one has to check that for all $c\in\mathbf R$, $v, w\in V$, $c(v+w)=cv+cw$. This means, using your notations,
$$\begin{bmatrix}
(a_1b_1)^c \\\vdots \\(a_n b_n)^c 
\end{bmatrix}=\begin{bmatrix}
a_1^c\, b_1^c \\\vdots \\a_n^c\,b_n^c
\end{bmatrix}$$
which is, on each component of the vectors, one of the usual rules for exponents.

