# A Confusion Regarding the Definitions of Operations Defined in a Vector Space.

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?

• It's clear we still get all the same scalar multiples (since $c \in \mathbb K$ iff $-c$ is for whatever field by definition) and thus $-\beta$ is defined. The exercise wants you to go through the list of axioms for a vector space and see if these operations define one. Jul 31, 2023 at 5:27
• When you say it was "nowhere" mentioned, if they wrote "The operations on the right are the usual ones," that's where it was mentioned. If you wrote it, it is a natural inference for somebody to make, but without the entire textbook in front of me, I can't say more about why it isn't in the book, or where it would be found there Jul 31, 2023 at 5:27
• What would $0 \oplus x$ be and what axiom would this contradict? There is no shortcut to just grinding through the axioms until one fails. Jul 31, 2023 at 5:32
• @copper.hat Ah, but I am asking about the meaning not the solution? Jul 31, 2023 at 5:34
• You are overthinking this. Jul 31, 2023 at 5:36

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 $$(c_1+c_2)\alpha=(c_1\alpha)\color{red}{+}(c_2\alpha)$$.

• Ok, but the thing is, we might not think of $-\alpha$ to be an inverse of $\alpha$ for, the inverse of $\alpha$ is $\alpha$, isn't it? So, we should NOT interpret it like, "$\alpha$ is an element of $R^n$ and inverse of inverse of an element is the element itself i.e $-(-\alpha)=\alpha$." Isnt it? Jul 31, 2023 at 10:01
• Does, $-\alpha$ denote, the inverse of $\alpha$ for in the book, at the definition of a vector space it's written that if $0$ is an identity element of $V$ then, $\forall \alpha$ $\exists -\alpha\in V$ such that $\alpha+(-\alpha)=0$ . The other meaning of $-\alpha$ maybe $(-x_1,...,-x_n)$ if, $\alpha=(x_1,...,x_n)$. Jul 31, 2023 at 10:13
• @user1511 Ok, so there are two $-\alpha$ 's in the story? One $-\alpha$ is the inverse of $\alpha$ and the other one is $-\alpha=(-a_1,-a_2,...,-a_n)$ where$\alpha=(a_1,...,a_n)$ ? Jul 31, 2023 at 10:17
• @ThomasFinley Yes. You may denote the additive inverse of $\alpha$ with respect to $\oplus$ as $\ominus\alpha$. In defining $c\cdot\alpha=-c\alpha$, the authors have explicitly stated that the operations on the right are the usual ones in $R^n$. So, they mean the usual inverse of $c\alpha$, not $\ominus(c\alpha)$. Jul 31, 2023 at 10:20
• But in $c.(\alpha)\oplus c.\beta=-c\alpha-(-c\beta)$ here, $-(-c\beta)$ to be interpreted as follows: In $-\color{red}{(-c\beta)}$ the $\color{red}{(-c\beta)}$ is $(-cb_1,...,-cb_n)$ and then $-(-c\beta)$ as again, $(cb_1,cb_2,...,cb_n)$ ,right? Jul 31, 2023 at 10:25

My feeling is that you're overthinking matters.

Assuming $$R$$ denotes the real numbers, then we answer your questions in turn:

• $$R^n$$ indeed refers to the collection of $$n$$-tuples of real numbers, i.e. $$R^n := \{ (x_1,\cdots,x_n) \mid x_1,\cdots,x_n \in R \}$$ (I like to adopt the shorthand $$(x_i)_{i=1}^n$$ for $$(x_1,\cdots,x_n)$$, and will continue using it from here on out.)

• We define several standard operations on it: given that $$c \in R$$ and $$x,y \in R^n$$, with $$x := (x_i)_{i=1}^n, y := (y_i)_{i=1}^n$$, we have $$x+y := (x_i+y_i)_{i=1}^n \qquad cx := (cx_i)_{i=1}^n$$ One may deduce in turn that $$x-y = x+(-y) = (x_i - y_i)_{i=1}^n$$ and $$-cx = (-cx_i)_{i=1}^n$$

I also feel the need to note that these are discussed in Example $$1$$ of Section $$2.1$$ of your textbook (on page $$29$$). (Note that the real numbers form a field. The addition and field multiplication used there are those inherited from the field in question.)

• I have edited my post. Mind taking a look at it? Jul 31, 2023 at 8:16