# Vector spaces - Non-uniqueness of element with property of scalar-multiplicative identity element?

I am dabbling in vector spaces, thinking about the axioms on Wikipedia. Notably, $$1 \mathbf{v} = \mathbf{v},$$ i.e. identity element of scalar multiplication (IEOSM), attracted my attention. I am wondering if it might happen that there is another element $1'$ with $1' \ne 1$ such that $1'$ also has this property (i.e. $1' \mathbf{v} = \mathbf{v}$ for arbitrary $\mathbf{v}$). I came up with the following example. However, I am not sure whether my thoughts are valid.

Let $V = \{\mathbf{0}\}$ be a vector space over $\mathbb{R}$. It looks like $V$ is indeed a vector space. So the only element of $V$ is the zero vector. Now, not only $1$ has the property; instead, every real number has the property.

Incidentally, there is a related question: For an arbitrary vector space, we have $0 \mathbf{v} = \mathbf{0}$. May it be true that there exists an element $0'$ with the same property? Again, I think the answer is yes.

P.S.: At the moment, further interesting conjectures arise. As an example, we could investigate the above questions assuming that $V \ne \{\mathbf{0}\}$. Maybe I should just look up suchlike results. (Where?)

If $\mathbf{v}\ne\mathbf{0}$ and $\alpha\mathbf{v}=\mathbf{v}$, then $\alpha\mathbf{v}-1\mathbf{v}=\mathbf{0}$ or $$(\alpha-1)\mathbf{v}=\mathbf{0}$$ which shows that $\alpha-1=0$. This follows from the vector space axioms: if $\gamma\ne0$ and $\gamma\mathbf{v}=\mathbf{0}$, then $$\gamma^{-1}(\gamma\mathbf{v})=\gamma^{-1}\mathbf{0}$$ that is $$(\gamma^{-1}\gamma)\mathbf{v}=\mathbf{0}$$ or $$\mathbf{0}=1\mathbf{v}=\mathbf{v}.$$

Note that this also shows that, if $\alpha\mathbf{v}=\mathbf{0}$, for $\mathbf{v}\ne\mathbf{0}$, then $\alpha=0$.

$V$ being the zero vector space is kind of a special case. You are correct that for all $\lambda \in \mathbb{R}$, we have that $\lambda \cdot \vec{v} = \vec{v}$ if $\vec{v} \in V$ in this case. It should be noted that the axioms don't say otherwise, they simply prescribe that this must be true for the element $1 \in \mathbb{R}$.

However, if $\vec{0} \neq \vec{v} \in V$, then the only element $\lambda \in \mathbb{R}$ that satisfies $\lambda \cdot \vec{v} = \vec{v}$ is $\lambda = 1$. Suppose otherwise, that is, suppose that $1 \neq \lambda$ is such that $\lambda \cdot \vec{v} = \vec{v}$. Then: $$(\lambda - 1) \cdot \vec{v} = \vec{0}$$ Since $\lambda \neq 1$, it follows that $\lambda - 1$ is invertible. Hence $$\vec{v} = \frac{1}{\lambda - 1}(\lambda - 1) \cdot \vec{v} = \frac{1}{\lambda - 1} \cdot \vec{0} = \vec{0}$$ and hence $\vec{v} = \vec{0}$.

By definition a vector space $V$ always comes built in with a set of scalars $\mathbb{F}$; here $\mathbb{F}$ is a field, which means we have notions of the four basic operations $(+, -, \cdot, /)$ that satisfy certain satisfy natural properties; we say that $V$ is a vector space over $\mathbb{F}$. $\mathbb{R}$ is typically the field one uses when first learning linear algebra, but it is very far from the only option.

One of the features of a field is that it have a multiplicative identity $1_{\mathbb{F}}$, and this is the IEOSM of any vector field $V$ over $\mathbb{F}$, so there's no choice in the matter, even if, like you say, the identity $a \mathbf{v} = \mathbf{v}$ holds for other $a$ when $\dim V = 0$. Anyway this phenomenon can only occur in this case: If the identity holds (for all $\mathbf{v}$ for two distinct elements of $\mathbb{F}$, say, $1, 1'$, then

$(1' - 1) \mathbf{v} = 1' \mathbf{v} - 1 \mathbf{v} = \mathbf{v} - \mathbf{v} = 0$.

Since $1, 1'$ are distinct, $1' - 1$ is nonzero and hence it has an inverse; multiplying both sides of $(1' - 1) \mathbf{v} = 0$ by that inverse gives, $\mathbf{v} = 0$. Since the identity holds for all $\mathbf{v}$, $0$ is thus the only element of $V$.

Similarly, one can show that if there are distinct elements $0$ and $0'$ of $\mathbb{F}$ that both satisfy $a \mathbf{v} = \mathbf{0}$ for all $\mathbf{v} \in V$, then $V = \{ 0 \}$.

Remark One can ask if we can formulate objects that satisfy axioms like a vector space, but where we allow something other than a field to be our scalars---indeed we can: if we take our set of scalars to be a ring (which has addition, subtraction, and multiplication by not, in general, division), then instead of a vector space we get the notion of module (over a ring).

• I just noticed the "textit" and "mathbf" in your answer. Should this be edited? Commented Aug 26, 2014 at 12:24