Are there other ways to solve uniqueness theorems?

Question

Everytime I've seen a proof of a uniqueness theorem, I've seen the exact same method. They assume that there is more than one, (usually two) solution of say some equation or some number satisfying some property, etc. and then show that the two are equal. That's well and good, but are there any other methods for proving uniqueness theorems? It may be that this is the general way to tackle uniqueness theorems, and that certain types of uniqueness may be proved in different methods. If there is something like that also, could I know how please? I've looked over the internet but haven't found anything else.

Answer 1

If $G$ is a group, you can show that for given $a,b\in G$, the equation $$\tag1 x\cdot a=b $$ has a unique solution as follows: From the group axioms, we know there exists at least one ight-neutral element of $e\in G$ (i.e., $g\cdot e=g$ for all $g\in G$) and at least one right-inverse $\bar a\in G$ of $a$ (ie., with $a\cdot \bar a=e$). Then, as the group operation is a function $G\times G\to G$, i.e., because we are given that for $u,v\in G$, the product $u\cdot v$ is a uniquely determined element of $G$, we conclude $$(x\cdot a)\cdot \bar a=b\cdot \bar a. $$ By associativity, $$x\cdot (a\cdot \bar a)=b\cdot \bar a, $$ by choice of $\bar a$, $$x\cdot e=b\cdot \bar a, $$ and finally by right-neutrality of $e$, $$ x=b\cdot \bar a.$$ By this construction of $x$, we showed both existence and uniqueness in one go and without the method of shoing that two solutions must be equal. Note that we didn't even assume that the neutral or the inverse are unique (which is no biggy though, given that one can quickly prove their uniqueness). We did need the given uniqueness of the product of two elements, though.

Answer 2

Let $V$ be an $F$-vector space ($F$ a field) with (ordered) basis $\mathscr{B}=\{v_1,v_2,\dots,v_n\}$. Define the map $C_{\mathscr{B}}\colon V\to F^n$ by $$ C_{\mathscr{B}}(v)=\begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} \qquad\text{if and only if}\qquad v=a_1v_1+a_2v_2+\dots+a_nv_n $$ (identify $F^n$ with the space of column vectors). If $\mathscr{D}$ is another basis of $V$, there exists a unique $n\times n$ matrix $M$ such that, for every $v\in V$, $$ C_{\mathscr{D}}(v)=MC_{\mathscr{B}}(v) $$

Proof of uniqueness. Suppose $M$ exists. Then, for $v=v_i$, we easily see that $C_{\mathscr{B}}(v_i)=e_i$ (the $i$-th column of the identity matrix). Therefore $$ C_{\mathscr{D}}(v_i)=MC_{\mathscr{B}}(v_i)=Me_i $$ is the $i$-th column of $M$. Hence, as a block matrix, $$ M=\begin{bmatrix} C_{\mathscr{D}}(v_1) & C_{\mathscr{D}}(v_2) & \dots & C_{\mathscr{D}}(v_n)\end{bmatrix} $$ Uniqueness is proved because the found matrix only depends on $\mathscr{B}$ and $\mathscr{D}$.

Proof of existence. Just show that the matrix found above does the job.