# An isomorphism maps a zero vector to a zero vector?

(Apologies for ascii art math, I don't know latex. Also apologies if this is in the wrong forum.)

## Homework Statement

Why, in this lemma, must there be a vector v in V? That is, why must V be nonempty?

An isomorphism maps a zero vector to a zero vector.
Where f:V->W is an isomorphism, fix any vector v in V. Then f(0 vector represented with respect to V) = f(0 * vector v) = 0 * f(vector v) = 0 vector represented with respect to W.

## Homework Equations

The answer is given as "No vector space has the empty set underlying it. We can take vector v to be the zero vector."

## The Attempt at a Solution

So actually, I'm not trying to solve the problem. I'm just having a hard time understanding the answer.

What does it mean by "no vector space has the empty set underlying it?" Does that mean no vector space consists entirely of the empty set? The way it's phrased makes it sound like the vector space can include the empty set along with other sets. Wouldn't you be able to take vector v to be the zero vector in either case? Or is there no zero vector for the empty set? Even if you couldn't, why would you need to be able to take vector v to be the zero vector? A scalar zero times anything should be the zero vector, right? Or am I misinterpreting that, in that you're not supposed to take v as the zero vector?

I haven't studied in weeks, so these are possibly/probably stupid questions, I feel like I've forgotten all the basic math I learned when I started linear algebra.

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They sure phrased it in a very incomprehensible manner. Let me try to clear things up for you.

A vector space is a certain set with an addition and a scalar multiplication on it which satisfies certain properties. One of these properties is

$$\exists 0\in V:~\forall v\in V:~0+v=v=v+0$$

So in ANY vector space there must exist a zero vector. In particular, every vector space must be nonempty (since there must at least be one vector in it: 0). This is what they mean with "there is no vector space with the empty set underlying it", it's just phrased horribly. A translation is "every vector space is nonempty".

So, now we know that every vector space has a vector. So your lemma works out. In particular, we can always take v=0 in your lemma, and the same proof will hold!

Aha. That clears things up. Thanks!