Like "zero vector" (all elements $0$), do we have a "one vector" (all elements $1$), "two vector" (all elements $2$), etc.?

Question

The zero vector is defined as simply the vector consisting of pure zeros as elements. It is labeled as the number zero with an arrow above $\vec 0$ or as a boldface zero $\boldsymbol 0$ like any other vector, and sometimes I see the vector symbol omitted so it is just a zero $0$.

$$\vec v=\begin{pmatrix}0 \\ 0 \\ 0 \\ \vdots\end{pmatrix}=\vec 0=\boldsymbol 0=0$$

All these are simply conventional notation styles for the zero vector.

As far as I know we do not and have never defined any other vectors that purely contain the same number as all elements in this same way. But would there be any hinder from doing so and defining that? Couldn't we simply define a one vector, a three vector, a seventeen vector etc. in the same way:

$$\vec v=\begin{pmatrix}1 \\ 1 \\ 1 \\ \vdots\end{pmatrix}=\vec 1=\boldsymbol 1=1\qquad\vec v=\begin{pmatrix}3 \\ 3 \\ 3 \\ \vdots\end{pmatrix}=\vec 3=\boldsymbol 3=3\qquad\vec v=\begin{pmatrix}17 \\ 17 \\ 17 \\ \vdots\end{pmatrix}=\vec{17}=\boldsymbol{17}=17$$

Etc. Why do we never define such vectors like this and use such notation when we do so for the zero vector? Is there a reason for not doing so that I am not aware of, or has it just never been found relevant to do?

Answer 1

The case full of $1$s is actually occasionally useful, e.g. to write sums of entries in a vector as a dot product in constrained optimization, or we may write means in a similar way if we divide by the vectors' dimension. The question of how to denote such a vector has been discussed before here (as well as here). As for e.g. using $17$s, you might as well put a coefficient of $17$ in front of $\vec{1}$ (or other preferred notation).

Incidentally, there's an infinite-dimensional case where an all-ones quantity is labelled $1$, namely the Dirichlet inverse of the Möbius function.

Answer 2

In the case of fields (for example the field of real numbers), $0$ and $1$ are special numbers, because for all $a$ in the field, we have that $$a+0=a$$ and $$a\cdot 1 = a.$$ So they are the identity elements of the operations. On the other hand, when you are dealing with vector spaces, you don't have multiplication, only addition, so you can't define the multiplicative identity.

However, you can equip $\mathbb{R}^2$ with a multiplication: $$(a,b)\cdot (c,d)=(ac-bd, bc+ad)$$ and get the field of complex numbers, but there $1=(1,0)$ (and $i=(0,1)$), and not $(1,1)$.

Answer 3

Every vector space is an abelian group with respect to vector addition and every abelian group requires a (unique) identity element. Therefore one always finds the zero vector, whether you operate in an abstract vector space, $\mathbb{R}^n$ or a vector space of functions, the zero vector is always present.

Answer 4

The reason is that there is nothing particularly "special" "vector-wise" about the one vector that couldn't also be said of the seventeen vector or any other non-zero vector. The zero-vector is, of course, special, because it is the solution to $\alpha v = 0$ for all scalars $\alpha$.

Actually, as J.G. notes, this is not 100% true in $\mathbb R^n$, which brings along all sorts of extraneous baggage that you wouldn't see more "vanilla" vector spaces. And, obviously, the multiplicative identity in the scalar space is very important, because it is the solution to $\alpha v=v$ for all vectors $v$. But as vectors themselves, not so much.