Isotropic points with a jacobian matrix
$begingroup$
As definition, in order to find the isotropic points of a jacobian matrix, the matrix's columns become orthogonal and equal to the magnitude. I don't quite understand this definition. if i have a matrix looks like this:
a b
c d
Matrix example:
2.x 0
2.y+3 3
How can I find its isotropic points ?
Thank you very much.
matrix-calculus
$endgroup$
add a comment |
$begingroup$
As definition, in order to find the isotropic points of a jacobian matrix, the matrix's columns become orthogonal and equal to the magnitude. I don't quite understand this definition. if i have a matrix looks like this:
a b
c d
Matrix example:
2.x 0
2.y+3 3
How can I find its isotropic points ?
Thank you very much.
matrix-calculus
$endgroup$
add a comment |
$begingroup$
As definition, in order to find the isotropic points of a jacobian matrix, the matrix's columns become orthogonal and equal to the magnitude. I don't quite understand this definition. if i have a matrix looks like this:
a b
c d
Matrix example:
2.x 0
2.y+3 3
How can I find its isotropic points ?
Thank you very much.
matrix-calculus
$endgroup$
As definition, in order to find the isotropic points of a jacobian matrix, the matrix's columns become orthogonal and equal to the magnitude. I don't quite understand this definition. if i have a matrix looks like this:
a b
c d
Matrix example:
2.x 0
2.y+3 3
How can I find its isotropic points ?
Thank you very much.
matrix-calculus
matrix-calculus
asked Jan 21 '14 at 14:31
XitrumXitrum
1012
1012
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add a comment |
1 Answer
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votes
$begingroup$
The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
$$
begin{pmatrix}
a & b \
c & d
end{pmatrix}
$$
these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.
For the concrete example
$$
begin{pmatrix}
2x & 0 \
2y+3 & 3
end{pmatrix}
$$
the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
$$
begin{pmatrix}
a & b \
c & d
end{pmatrix}
$$
these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.
For the concrete example
$$
begin{pmatrix}
2x & 0 \
2y+3 & 3
end{pmatrix}
$$
the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.
$endgroup$
add a comment |
$begingroup$
The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
$$
begin{pmatrix}
a & b \
c & d
end{pmatrix}
$$
these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.
For the concrete example
$$
begin{pmatrix}
2x & 0 \
2y+3 & 3
end{pmatrix}
$$
the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.
$endgroup$
add a comment |
$begingroup$
The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
$$
begin{pmatrix}
a & b \
c & d
end{pmatrix}
$$
these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.
For the concrete example
$$
begin{pmatrix}
2x & 0 \
2y+3 & 3
end{pmatrix}
$$
the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.
$endgroup$
The columns must be orthogonal to each other (i.e., the dot product of any two columns must be equal to $0$), and their vector norms must be equal. For the general $2times 2$ matrix
$$
begin{pmatrix}
a & b \
c & d
end{pmatrix}
$$
these requirements amount to: $ab+cd=0$, $a^2+c^2=b^2+d^2$.
For the concrete example
$$
begin{pmatrix}
2x & 0 \
2y+3 & 3
end{pmatrix}
$$
the above requirements simplify to $2y+3=0$; $(2x)^2+(2y+3)^2 = 9$. Hence $y=-3/2$ and $x=pm 3/2$.
answered Mar 2 '14 at 0:22
user127096user127096
8,17011140
8,17011140
add a comment |
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