The following proposition uses the definition of the vector spaces and demonstrates that in most cases, you can calculate with vectors as if they were usual numbers. This is ensured by some basic rules of calculation and the existence and uniqueness of zero and inverse vectors in vector spaces.

# Proposition: Simple Consequences from the Definition of a Vector Space

Let $V$ be a vector space over a field $F$. Then the following rules apply:

1. There is only one unique vector $o\in V$, called the zero vector, such that $v+o=o+v=v$ for all $v\in V.$
2. Each vector $x\in V$ has a unique inverse vector $-x$ such that $x+(-x)=o.$
3. The inverse vector $o$ if $o$ itself, i.e. $-o=o.$
4. The inverse of the inverse vector is the vector itself, i.e. $-(-x)=x$ for all $x\in V.$
5. The equation $a + x = b$ is uniquely solvable for any given two vectors $a,b\in V$, and its solution is $x:=b-a.$
6. The equation $-(x+y)=-x-y$ holds for all vectors $x,y\in V.$
7. When adding a finite number of vectors, we can ignore the any paranthesis general associative law): $x_1+x_2+\ldots + x_n=(\ldots((x_1+x_2)+x_2)+\ldots+x_n.$
8. When adding a finite number of vectors, we can arbitrarily change the order general commutative law): $x_1+x_2+\ldots + x_n=x_{k_1}+x_{k_2}+\ldots+x_{k_n}$ for any permutation $k_1,\ldots,k_n.$
9. $-x=(-1)x$ for all $x\in V$.
10. $0x=o$ holds for all $x\in V$, where $0\in F$ is the zero element of the field, and $o\in V$ is the zero vector of $V$.
11. $\lambda o=o$ for all $\lambda \in F$.
12. $\lambda o=o$ for all $\lambda \in F$.

Proofs: 1

Github: ### References

#### Bibliography

1. Knauer Ulrich: "Diskrete Strukturen - kurz gefasst", Spektrum Akademischer Verlag, 2001
2. Knabner, P; Barth, W.: "Lineare Algebra - Grundlagen und Anwendungen", Springer Spektrum, 2013