How to show that two quadratic forms are equivalent? To show to quadratic forms are not equivalent, we can find rank, or discriminant or some element which is represented by either one only etc. But Is there a general criterion to show that two binary(right now I am only concerned for binary) quadratic forms are equivalent.
Like here is an example which uses variable transformation, but every time find what transformation will apply is tough.
$\textbf{Example-}$ $X^2-Y^2 $ and $4XY$. replacing $X,Y$ by $X+Y,X-Y$ do the job here. Nnow there are criterion which gives an equivalent condition to check whether two binary forms are equivalent or not , like 
$\textbf{Criterion 1-}$ On algebraically closed fields, same dimension and same rank suffice.
$\textbf{Criterion 2-}$ On Reals, same rank and same signature suffices.
$\textbf{Criterion 3-}$ On finite fields,same discriminant is enough.
But in general if we are given field as some arbitrary $K$, then what to do? I am very new at quadratic forms. Any help will be appreciated. 
I know that we can check similarity of their corresponding matrices, but is there any other way, beside checking matrices are similar or not. Like how would you proceed on these couple of question below. If there is no other method, can somebody please explain me the method to check whether two matrices are similar or not, I am really weak with matrices.
$\textbf{Question 1-}$ Prove $X^2-4XY+3Y^2, X^2-Y^2,aX^2-aY^2$ are equivalent over $K$.
$\textbf{Question 2-}$ Show 
i) $X^2-Y^2 \sim XY$ over $K$,   ii) $3X^2-2Y^2 \sim X^2-6Y^2$ (over $Q$)
 A: Q1) 
a) $X^2-Y^2$ is already in diagonal from. It's matrix is  $\bigl( \begin{smallmatrix} 1& 0 \\ 0 & -1 \end{smallmatrix} \bigr)$.
b) $aX^2-aY^2$ is already in diagonal from. It's matrix is  $\bigl( \begin{smallmatrix} a& 0 \\ 0 & -a \end{smallmatrix} \bigr)$. We have  $\bigl( \begin{smallmatrix} a& 0 \\ 0 & -a \end{smallmatrix} \bigr) \sim \bigl( \begin{smallmatrix} 1& 0 \\ 0 & -1 \end{smallmatrix} \bigr)$
c) $X^2-4XY+3Y^2$ can also be written as $(X-2Y)^2-Y^2$
Result a) and b): if $K=\Bbb{R}$ or $\Bbb{C}$ they are always equivalent. If $\sqrt(a)$ is rational and $K=\Bbb{Q}$ then also, and when $K$ is a finite field then if the characteristic is $2$ then always otherwise only when $a=\pm 1$.
Result (b) and (c) always equivalent (same discriminant, rank and signature).
Q2) 
i) if in the second we replace $X=U+V$ and $Y=U-V$ then we get $XY=(U+V)(U-V)=U^2-V^2$, so equivalence.
ii) $3X^2-2Y^2 \sim 3(3X^2-2Y^2)=(3X)^2-6Y^2=U^2-6V^2$  
A: This is an old thread, but the answer to Question 2(b) in the accepted answer is incorrect and should be corrected.
The forms $f(x,y)=x^2-6y^2$ and $g(x,y)= 2x^2-3y^2$ are not equivalent over $Q$. This can be seen as $g(x,y)$ does not represent the value $1$ over the rationals.  If $g(x,y)$ did represent 1, then the form
$h(x,y,z)= 2x^2-3y^2-z^2$
would have an non-zero integer solution.  However, this does not happen, for when $(x,y,z)=1$, the equation does not have a solution $\mod 8$.
