Proving vector identities in general

Question

There a lot of vector identities which have been proved by writing components in one special coordinate system but how can be sure that it's true in other coordinate systems? Should we prove in each coordinate systems separately? If not, what allows us to believe it's true in all of the coordinate systems?

Answer 1

In fact, when we're talking about $\mathbb R^n$ or $\mathbb C^n$, we can use the "coordinate-free" language. Let me explain what actually it is. First of all, we can define an abstract $\mathbb R$-vector space. We say that $V$ is a $\mathbb R$-vector space if it is:

1) an abelian group (a set with an associative "addition", i.e. with a map $+: V \times V \to V$ which requires some very intuitive conditions. At first, it must be associative and commutative, i.e. for any $a, b, c \in V$ we have $a + (b + c) = (a + b) + c$ and for any $a, b \in V$ we have $a + b = b + a$ (here and further I'll write $a + b$ instead of $+(a, b)$). Secondly, we want $V$ to have a "neutral element" denoted by $0$. "Neutrality" means that $\forall v \in V$ $0 + v = v$. And the third condition: for any $v \in V$ we have an opposite element, i.e. $\forall v \in V \exists w \in W v + w = 0$. An opposite to $v$ is usually denoted by the symbol $-v$. An example of abelian group is $\mathbb R$ itself (with usual notions of addition, zero element and opposite elements);

2) other property of a $\mathbb R$-vector space $V$ is the existence of the map $\circ: R \times V \to V$. Instead of $\circ(\lambda, v)$ here and further in this text I'll write just $\lambda v$;

3) and the last property: we want some "harmony" between $\circ$ and $+$. More explicitly, $\forall a, b \in V$ the following must be true:

• $\forall \lambda, \mu \in \mathbb R$ $\lambda(\mu a) = (\lambda \mu) a$;
• $1a = a$;
• $\forall \lambda, \mu \in \mathbb R$ $(\lambda + \mu)a = \lambda a + \mu a$;
• $\forall \lambda \in \mathbb R$ $\lambda a + \lambda b = \lambda(a + b)$.

A little remark: elements of $\mathbb R$-vector space are often called "vectors", not just "elements".

That's all!

An important exercise: you can easily check that the usual definition of "multiplication" $\circ$ (namely, $\circ(\lambda, v) = \lambda v$) gives $\mathbb R$ a structure of $\mathbb R$-vector space.

Now if $V$ and $W$ are two $\mathbb R$-vector spaces we can define their "direct sum" $V \oplus W$ as an abelian group of all pairs $(v, w)$ $(v \in V, w \in W)$ with the addition defined by the formula $(a, b) + (c, d) = (a + c, b + d)$. Also we can define $\circ$ for $V \oplus W$: $\lambda(v, w) =(\lambda v, \lambda w)$.

An exercise: you can check that these operations give $V \oplus W$ a structure of $\mathbb R$-vector space.

Since, as mentioned above, $\mathbb R$ is the $\mathbb R$-vector space, we can define $\mathbb R^n$ as a vector space $\mathbb R \oplus \mathbb R \oplus \ldots \mathbb R$ ($n$ times).

Using the "coordinate-free language", we can now define "coordinate system" in $\mathbb R^n$ as a way to choose the base (a system of vectors in $\mathbb R^n$ such that any vector in $\mathbb R^n$ can be uniquely written as their linear combination; coefficients in this combination are, in fact, called "coordinates of v") in $\mathbb R^n$. Since a way to find coordinates of v is unique for every $v \in \mathbb R^n$, we can prove various statements by, firstly, choosing the base, and, secondly, by comparing coordinates of some vectors in this base. If they're the same, we can say that the vectors are also the same as elements of $\mathbb R^n$ (since they're linear combinations of the vectors from the base with the same coefficients). Since the definition of $\mathbb R^n$ doesn't depend on the choice of base, in any other base the coordinates of these vectors also'll be the same (since the vectors are the same!!).

I don't know how clear was this explanation, but, anyway, you can find this stuff in any modern Linear algebra or Abstract algebra textbook for undergrad students...