On $\mathbb R^n$, define two operations
$$\alpha\oplus \beta=\alpha-\beta,$$
$$c.\alpha=-c\alpha$$
The operations on the right are the usual ones. Which of the axioms for a vector space are satisfied by $(\mathbb R^n,\oplus, . )$ ?
I am new to the concept of verctor spaces and I found this question from the book Linear Algebra (2nd Edition) by Kenneth Hoffman and Ray Kunze in Chapter-2, 1st Excercise (Page no.-$34$) as Problem number $5.$
First of all, what is $\mathbb R^n$ ?
From my experience, I know that $\mathbb R^n$ generally denotes the set of all tuples of length $n$ in the set of real numbers, i.e $(x_1,x_2,\cdots,x_n)\in \mathbb R^n$ where $x_i\in \Bbb R.$
If this is the case, then the problem arises in the way they define the operations $\oplus$ and $\cdot $
It is mentioned that,
$$\alpha\oplus \beta=\alpha-\beta.$$
But what is meant by the thing $\alpha-\beta$. The problem is, it might be possible that $\alpha-\beta=\alpha+(-\beta)$, and this $\alpha+\beta$ may be defined as,( if $\alpha=(x_1,x_2,\cdots,x_n)$ and $\beta=(y_1,y_2,\cdots,y_n)$ ) $\alpha+\beta=(x_1+y_1,…,x_n+y_n)$ and $-\beta$ may be defined as $\beta=(-y_1,-y_2,…,y_n)$ (,if $\beta=(y_1,y_2,\cdots,y_n)$ ). But again, this was nowhere mentioned , these are just some of my thoughts based on previous exercises.
Same goes for $$c.\alpha=-c\alpha$$. The term, $-c\alpha$ again creates ambiguity. It seems it's possible that $-c\alpha=(-cx_1,-cx_2,\cdots,-cx_n)$ if $\alpha=(x_1,x_2,\cdots,x_n).$ But again, it's nowhere given. It's just a thought of mine.
With these assumptions/thoughts/considerations, I solved the problem as follows:
First we check, that whether closure property is satisfied or not. If $\alpha,\beta\in \mathbb R^n,$ then, $\alpha\oplus\beta=\alpha-\beta.$ If $\alpha=(x_1,…,x_n), \beta=(y_1,…,y_n)$ then, $\alpha\oplus\beta=\alpha-\beta=(x_1-y_1,\cdots,x_n-y_n).$ Again, $x_i-y_i\in \mathbb F$ implies that $\alpha\oplus \beta\in \mathbb R^n.$
- So, closure property is satisfied.
Next, we check for commutative property. If $\alpha,\beta\in \mathbb R^n$ such that $\alpha=(x_1,x_2,\cdots,x_n)$ and $\beta=(y_1,y_2,\cdots,y_n)$ then, $\alpha\oplus\beta=\alpha-\beta=(x_1-y_1,\cdots,x_n-y_n)$ and $\beta\oplus\alpha=\beta-\alpha=(y_1-x_1,\cdots,y_n-x_n).$
So, $(x_1-y_1,\cdots,x_n-y_n)\neq (y_1-x_1,\cdots,y_n-x_n)$.
- Thus, the commutative property is not satisfied.
Now, we check for the associative property i.e if $\alpha=(x_1,x_2,\cdots,x_n),$$\beta=(y_1,y_2,\cdots,y_n)$$\gamma=(z_1,z_2,\cdots,z_n)$, then, $\alpha-(\beta-\gamma)=(x_1-y_1+z_1,\cdots, x_n-y_n+z_n)$ and $(\alpha-\beta)-\gamma=(x_1-y_1-z_1,\cdots, x_n-y_n-z_n)$ and so, $\alpha-(\beta-\gamma)\neq (\alpha-\beta)-\gamma.$
- Thus, the associative property is not satisfied.
We note that $0\in \mathbb R^n$ and $0=\underbrace{(0,0,\cdots,0)}_{\text{n times}}$ and $\forall\alpha\in \mathbb R^n$ we have, $\alpha\oplus\alpha=0.$
Hence, we conclude that the identity and inverse property are satisfied.
Now, we come to the operation of multiplication.
If $1\in \mathbb F$, i.e $1$ is the unit element of $\mathbb F$ then $1.\alpha=-(1\alpha)=-\alpha$ and we have, $\alpha\neq -\alpha$ as, $-\alpha=(-x_1,-x_2,-x_3,…,-x_n)$ if $\alpha=(x_1,\cdots,x_n).$
- So, $1.\alpha\neq \alpha$
Now, we check $(c_1c_2).\alpha$ where $c_1,c_2$ are the scalars in $\mathbb F.$ We have, $(c_1c_2).\alpha=-(c_1c_2)\alpha=(-c_1c_2x_1,…,-c_1c_2x_n).$ Now, $c_1.(c_2.\alpha)=c_1(-c_2\alpha)=-(c_1(-c_2\alpha))=c_1c_2\alpha.$
- So, $(c_1c_2).\alpha\neq c_1.(c_2.\alpha)$
Now, we try to calculate the value of $c.(\alpha+\beta)$ (, where $c$ is a scalar in $\mathbb F$ and $\alpha,\beta\in \mathbb R^n$ where, $\alpha=(x_1,x_2,\cdots,x_n)$ and $\beta=(y_1,y_2,\cdots,y_n)$). Now, $c.(\alpha+\beta) =-c(\alpha+\beta)=-c\beta-c\beta.$
Also, $c.\alpha+c.\beta=-c\alpha-c\beta.$
- Thus, $c.(\alpha+\beta)=c.\alpha+c.\beta$
We now try to compute $(c_1+c_2).\alpha.$ We have, $(c_1+c_2).\alpha=-(c_1+c_2)\alpha=-c_1\alpha-c_2\alpha.$
Also, $c_1.\alpha+c_2.\alpha=-c_1\alpha-c_2\alpha.$
- Finally, we have, $(c_1+c_2).\alpha=c_1.\alpha+c_2.\alpha$
Is my above solution based on the assumptions, correct/valid?
Best Answer
The first half of your answer looks fine. There are some issues about the second half.
Your disproof of $(c_1c_2)\cdot\alpha=c_1\cdot(c_2\cdot\alpha)$ is essentially correct, but there is a missing dot in the step $c_1\cdot(c_2\cdot\alpha)=c_1\color{red}{\cdot}(-c_2\alpha)$.
Your inspections of the two distributive axioms are incorrect. What you should verify are $c\cdot(\alpha\oplus\beta)=(c\cdot\alpha)\oplus (c\cdot\beta)$ and $(c_1+c_2)\alpha=(c_1\alpha)\oplus(c_2\alpha)$, but you have mistaken them as $c\cdot(\alpha\color{red}{+}\beta)=(c\cdot\alpha)\color{red}{+}(c\cdot\beta)$ and $(c_1+c_2)\alpha=(c_1\alpha)\color{red}{+}(c_2\alpha)$.