[Math] Determine if the whether a set equipped with some given operations is a vector space.

linear algebra

I should determine whether a set equipped with some given operations is a vector space and I do not know how I could prove that. Here is the first statement:

"The set of all real numbers with the standard operations of addition and multiplication."

So is this a vector space? Now I know that I should prove the axioms 1 – 10 to determine whether this actually is a vector space, but how?

First axiom is "If u and v are objects in V then u+v is also in V". I know that if you add two real numbers you should get a new real number, therefore the result is in V, but how should I write to prove it? I wonder the same thing for the rest of the axioms…

The ten axioms that a set (V) of objects should meet to be a vector space:

  1. if u and v are objects in V, then u+v is in V
  2. u+v = v+u
  3. u+(v+w) = (u+v)+w
  4. There is an object 0 in V called a zero vector for V,
    Such that 0+u = u+0 = u
  5. For each u in V, there is an object -u in V, called a negative of u, such that u+(-u) = (-u)+u = 0
  6. If k is any scalar and u is any object in V, then ku is in V.
  7. k(u+v) = ku + kv
  8. (k+m)u = ku+mu
  9. k(mu) = (km)(u)
  10. 1u = u

Best Answer

The set of real numbers with standard addition and scalar multiplication can be expressed as $(\Bbb{R},+,\times). $

This can also be expressed as a subspace of $\Bbb{R}^n$, namely $\Bbb{R}^1$. Thus, if we prove the ten axioms for $\Bbb{R}^n$ then we need to only prove thereon that $\Bbb{R}^1$ is closed under addition and scalar multiplication, as is the virtue of subspaces.

Let $\mathbf{u,v,w} \in \Bbb{R}^n$ given that $\mathbf{u}=(u_1,...,u_2), \mathbf{v}=(v_1,...,v_2)$, and $\mathbf{w}=(w_1,...,w_2)$. Now,

(1) $\mathbf{u+v}=(u_1+v_1,...,u_n+v_n)$.

(2) $k\cdot \mathbf{u} =(ku_1,...,ku_n)$.

These two new vectors are clearly $n$-tuples of vectors in the vector space. Thus $\Bbb{R}^n$ is closed under addition and scalar multiplication. Now, to prove $\mathbf{u+(v+w)=(u+v)+w}$ we perform the following:

$$\mathbf{u+(v+w)}=(u_1,...,u_n)+((v_1,...,v_n)+(w_1,...,w_n))$$ $$=(u_1+(v_1+w_1))+...+(u_n+(v_n+w_n))$$ $$=((u_1+v_1)+w_1)+...+((u_n+v_n)+w_n)$$ $$=\mathbf{(u+v)+w}.$$

The rest can be proven in a similar matter. Once the ten axioms for $\Bbb{R}^n$ are proven, then to prove that $\Bbb{R}^1$ is a subspace of $\Bbb{R}^n$, and hence a vector space, then you just need to prove closure under addition and scalar multiplication, which I did already for $\Bbb{R}^n$.

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