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In the definition of a vector space, the additive inverse condition requires that for every $v \in V$ (where $V$ is a vector space over $\mathbf{F} = \mathbb{R}$ or $\mathbb{C}$), there exists $w \in V$ such that $$v + w = 0$$ Show that this condition can be replaced with the condition that

$$0v = 0$$

for all $v \in V$. The $0$ on the left side is the number $0$ in $\mathbf{F}$ and the $0$ on the right side is the additive identity of $V$.

(The phrase "a condition can be replaced" in a definition means that the >collection of objects satisfying the definition is unchanged if the >original condition is replaced with a the new condition)

(Taken from "Linear Algebra Done Right (3rd edition), by S. Axler)

I am new to abstract algebra and have close to no clue how to tackle this problem. Here is my futile "attempt" nonetheless:

I tried some substitution and got

$$v + w = 0v$$

but am unable to make any sense out of this, much less determine if I am headed towards the right direction.

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3 Answers 3

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Call the first condition C1 and the second condition C2. Then we have to prove two things:

  • The axioms of a vector space (with C1) imply C2
  • The axioms of a vector space (without C1) together with C2 imply C1

This means that, within the other vector space axioms, C1 and C2 are equivalent. This is already an important first step.

Now let's prove both directions.

  • Suppose that every $v\in V$ has an inverse. Using the axioms of a vector space, we see that: $$0\cdot v = (0+0)\cdot v = 0\cdot v+0\cdot v$$ Now, because we assumed that C1 holds, we know that $0\cdot v$, which is a vector, has an inverse $w\in V$. Now add this inverse to both sides to get that $$ 0\cdot v +w = 0\cdot v+0\cdot v+w \Leftrightarrow 0 = 0\cdot v+0 = 0\cdot v$$ So C2 is proven.
  • Suppose that C2 holds, so $0\cdot v = 0$ for every vector $v\in V$. Again using only vector space axioms (but not using C1!!) we find that $$0 = 0\cdot v = (1+(-1))\cdot v = 1\cdot v+(-1)\cdot v = v+(-1)\cdot v $$ So $(-1)\cdot v$ is an inverse of $v$ for every $v\in V$. So C1 is proven.
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  • $\begingroup$ Forgive me on commenting on an old question, but when supposing that C2 holds, why do we know that (-1)v exists in V? $\endgroup$
    – yhylord
    May 26, 2018 at 18:05
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    $\begingroup$ $-1$ exists in $\mathbf{F}$ and by definition a vector space comes equipped with a scalar multiplication, so $\lambda\cdot v$ is a well-defined element of $V$ for every $\lambda \in \mathbf{F}$ and $v\in V$. $\endgroup$
    – Jef
    May 26, 2018 at 18:15
  • $\begingroup$ How can we be sure that $(-1)\cdot v$ is a unique additive inverse of $v$, through this proof? How do we know there is no other object, $w$, that satisfies $v + w = 0$? $\endgroup$
    – Mailbox
    Mar 15, 2023 at 16:40
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You want to prove that if $0v=0$ for all $v$ then every element has an inverse.

Consider that you can multiply vectors by any scalars. That every field has an element $-1$ such that $-1+1=0$ and that scalar multiplication is distributive in both directions so $av+bv=(a+b)v$.

See if using these hints you can get the proof.

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I believe the question is asking you to show that the statements

  1. For all $v$, there exists a $w$ such that $v + w = 0$, and
  2. For all $v$, we have $0v = 0$

are equivalent, given the other axioms: If you have all the other axioms of a vector space, then it doesn't matter which of these two conditions you choose; they'll both deem exactly the same objects to be (or not to be) vector spaces.

You're definitely heading in the right direction, as you've shown one direction: That 1. $\implies$ 2., given the other axioms. This is because, given the existence of such a $w$, you know

$$0 = w + v = 0v, \text{ hence } 0 = 0v \text{ for all $v$, as desired.}$$

Now all that's left is showing 2. $\implies$ 1., given the other axioms. So, you'll need to show that, if $0v = 0$ for all $v$, then for all $v$, there exists some $w$ such that $w + v = 0$.

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