Definition of equality amongst elements of vector space Suppose we are given an arbitrary vector space $V(\mathbb{F})$. We define by a set of axioms what are the properties that a given vector space has to in the form of binary operations of vector addition and scalar multiplication for it to be considered as a vector space.
However, we do not strictly define what is equality in the context of vectors.
Different vector spaces have different definitions of equality. For eg. the vector space $\mathbb{R}^n(\mathbb R)$ considering elements as n-tuples we call two vectors equal iff they are component-wise equal. Similarly, we define equality on the vector space $P_n(\mathbb R)$ i.e. the set of polynomials with degree $\le n$, as equality of coefficients. While for $M_{m\times n}(\mathbb R)$ it is defined as element-wise equality.
So I wanted to ask that is it not necessary to define equality of vectors beforehand and bind it as a set of axioms just like scalar multiplication or vector addition as the definition of equality is unique to a vector space based on the structure it defines. In a more fundamental sense, what do I mean when I say that $u+v=v+u$ when I have not even defined what equality to that specific vector space means.
 A: This is a really good question! So, what you're talking about is an example of a first-order theory with equality. We define the axioms for equality beforehand (I'll state them in a minute), and then add a set of proper symbols $\mathcal P$ and proper axioms. Other examples of first-order theories with equality include that of groups, abelian groups, lattices, posets, etc.

Axioms for Equality:

*

*$\forall x (x=x)$

*$\forall x_1...\forall x_n\forall y_1...\forall y_n ((x_1=y_1 \land ...\land x_n=y_n)\to (p^nx_1...x_n \leftrightarrow p^ny_1...y_n))$ for each n-ary predicate symbol $p^n$ in the first order language.

*$\forall x_1...\forall x_n\forall y_1...\forall y_n ((x_1=y_1 \land ...\land x_n=y_n)\to (f^nx_1...x_n = f^ny_1...y_n))$ for each n-ary function symbol $f^n$ in the first order language.


Now, we add to the above axioms a set of proper axioms and proper symbols to describe what is known as a vector space:

Proper symbols, functions, relations:

*

*There is an element $0\in V$, called zero.

*A function symbol $\alpha: V\times V\to V$, written as $\alpha(v,w) = v+w$.

*Another function symbol $\mu: \mathbb F\times V\to V$, written as $\mu(\lambda,v) = \lambda\cdot v$.


You may be wondering why I'm calling this a first-order theory. The following proper axioms will make that evident:

Vector Space (Proper) Axioms:

*

*$\forall x\forall y (x+y = y+x)$

*$\forall x\forall y\forall z (x+(y+z) = (x+y)+z)$

*$\forall x(0+x = x)$

*$\forall x\exists y (x+y = 0)$

*$\forall x(1\cdot x = x)$

*$\forall a\forall b \forall x((ab)\cdot x = a\cdot (b\cdot x))$

*$\forall a\forall x\forall y (a\cdot(x+y) = a\cdot x + a\cdot y)$

*$\forall a\forall b\forall x ((a+b)\cdot x = a\cdot x + b\cdot x)$

Whenever I say $\forall x, \forall y, \exists y$ I really mean $\forall x\in V, \forall y\in V, \exists y\in V$ respectively. Whenever I say $\forall a,\forall b$ I really mean $\forall a\in\mathbb F,\forall b\in\mathbb F$, respectively.
Clearly, all the axioms of vector spaces are expressible in terms of the first-order language (note the quantifiers, $\forall,\exists$) endowed with the special symbol of equality (and its axioms). Hence, this is a first-order theory with equality. I hope this helps. Please let me know if you've any questions!

To summarize: We don't worry about defining $=$ among the vector space axioms, because the intention is to complement the vector space axioms with the equality axioms - and their amalgamation is the first-order theory (with equality) of vector spaces.
A: The easy variant is to take equality as identity and as integral parto of logic (First Order Logic with Equality). In effect $a=b$ is true whenever $\phi(a)\leftrightarrow\phi(b)$ for every logical predicate $\phi$.
