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Degree of liberty of a matrix 2x2

  1. Sep 15, 2015 #1
    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:

    [tex]
    \begin{bmatrix}
    A & B\\
    C & D
    \end{bmatrix}
    :\begin{bmatrix}
    x^2 & xy\\
    yx & y^2
    \end{bmatrix}
    +
    \begin{bmatrix}
    E\\
    F
    \end{bmatrix}
    \cdot
    \begin{bmatrix}
    x\\
    y
    \end{bmatrix}
    +G=0
    [/tex]

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

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

    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?
     
  2. jcsd
  3. Sep 16, 2015 #2
    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 lets 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.
    [tex]a_{21} \neq \frac{a_{11}a_{22}}{a_{12}}[/tex]

    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.
     
  4. Sep 17, 2015 #3

    WWGD

    User Avatar
    Science Advisor
    Gold Member

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