- #1
brunotolentin.4
- 6
- 0
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?
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?