Understanding question 77 in Golan's "Linear Algebra".

Question

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?

Answer 1

1. Yes, $V$ with the standard addition, has no 0 element.
2. 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.
3. 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.

Answer 2

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, $+^{'}$.

Answer 3

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.