Degree of liberty of a matrix 2x2

[brunotolentin.4]

How many degree of liberty exist, actually, in a matrix 2x2 ?

I think that is three! Because the conic equation can be wrote like this:

$$\begin{bmatrix}<br /> A & B\\ <br /> C & D<br /> \end{bmatrix}<br /> :\begin{bmatrix}<br /> x^2 & xy\\ <br /> yx & y^2<br /> \end{bmatrix}<br /> +<br /> \begin{bmatrix}<br /> E\\ <br /> F<br /> \end{bmatrix}<br /> \cdot<br /> \begin{bmatrix}<br /> x\\ <br /> y<br /> \end{bmatrix}<br /> +G=0$$

But, xy = yx, thus ... + Bxy + Cyx +... = ... + (B+C)xy + ...

So: $$\begin{bmatrix}<br /> A & (B+C)\\ <br /> 0 & D<br /> \end{bmatrix}<br /> :\begin{bmatrix}<br /> x^2 & xy\\ <br /> yx & y^2<br /> \end{bmatrix}<br /> +<br /> \begin{bmatrix}<br /> E\\ <br /> F<br /> \end{bmatrix}<br /> \cdot<br /> \begin{bmatrix}<br /> x\\ <br /> y<br /> \end{bmatrix}<br /> +G=0$$

Another example: the coefficients of the equation Ay'' + By' + Cy = 0 has three degree of liberty (A, B and C) and it can be converted in a matrix:

y' = a y + b y'
y'' = c y + d y'

So, exist more and more examples that I could give here. But, the felling that I have is the a matrix 2x2 has 3 degree of liberty, although of has four coefficients... My feeling is correct?

 

[JorisL]

You are looking for degrees of freedom.
Are the matrices you are looking at the most general?

Look for example at the general linear group $Gl(n, \mathbb{R})$
This a group containing the $n\times n$ invertible matrices.
You can prove that this group has dimension $n^2$.

If we take n = 2 you can show this in several ways.
The main thing is that the condition that the matrix is invertible reduces to $\text{det}A\neq 0$.
So let's say we have such a matrix $A = \left[a_{ij}\right]$.

The determinant condition is $\text{det}A = a_{11}a_{22} - a_{12}a_{21} \neq 0$.
It is clear that when we know three components ($a_{11},\,a_{22}\text{ and }a_{12}$), the fourth still has a lot of freedom.
$$a_{21} \neq \frac{a_{11}a_{22}}{a_{12}}$$

Clearly there is some symmetry in your examples.
An example is found by considering the Special linear group.
This is the subgroup $Sl(n, \mathbb{R}) \subset Gl(n, \mathbb{R})$ with $\text{det}A = 1$.
You can see how knowledge of 3 elements gives you the fourth in the case of n = 2.

I'm not entirely familiar with your first notation (what is the colon?).
I also don't understand your point.

The second example is trivial, you start with three coefficients so that will be reflected in your matrix.

 

[WWGD]

I agree with Joris L . The number degrees of freedom depend on the context. Would you elaborate on what you are after?