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?

  • $\begingroup$ 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. $\endgroup$ Jul 31, 2023 at 5:27
  • 6
    $\begingroup$ 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 $\endgroup$ Jul 31, 2023 at 5:27
  • 3
    $\begingroup$ 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. $\endgroup$
    – copper.hat
    Jul 31, 2023 at 5:32
  • $\begingroup$ @copper.hat Ah, but I am asking about the meaning not the solution? $\endgroup$ Jul 31, 2023 at 5:34
  • 6
    $\begingroup$ You are overthinking this. $\endgroup$
    – copper.hat
    Jul 31, 2023 at 5:36

2 Answers 2


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

  • $\begingroup$ 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? $\endgroup$ Jul 31, 2023 at 10:01
  • $\begingroup$ 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, $\alpha=(x_1,...,x_n)$. $\endgroup$ Jul 31, 2023 at 10:13
  • $\begingroup$ @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$\alpha=(a_1,...,a_n)$ ? $\endgroup$ Jul 31, 2023 at 10:17
  • $\begingroup$ @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 $\ominus(c\alpha)$. $\endgroup$
    – user1551
    Jul 31, 2023 at 10:20
  • $\begingroup$ 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? $\endgroup$ 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.)

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

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