Symmetric matrices property
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Reading "Mathematical Physics: Classical Mechanics" by A. Knauf, I found the following statement:
The positive symmetric matrices with determinant 1 can be written as
$$
begin{vmatrix}
A & B \
B & C \
end{vmatrix}
$$ with $A>0$, $B in mathbb R$, $C=frac{1+B^2}{A}$.
I'm on self-teaching, so even if it's maybe a well-known property, I've never encountered it.
From what "process" stems this equality, and is it valid for any "positive symmetric matrix with determinant 1", or am I missing some context?
linear-algebra matrices self-learning positive-definite symmetric-matrices
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add a comment |
$begingroup$
Reading "Mathematical Physics: Classical Mechanics" by A. Knauf, I found the following statement:
The positive symmetric matrices with determinant 1 can be written as
$$
begin{vmatrix}
A & B \
B & C \
end{vmatrix}
$$ with $A>0$, $B in mathbb R$, $C=frac{1+B^2}{A}$.
I'm on self-teaching, so even if it's maybe a well-known property, I've never encountered it.
From what "process" stems this equality, and is it valid for any "positive symmetric matrix with determinant 1", or am I missing some context?
linear-algebra matrices self-learning positive-definite symmetric-matrices
$endgroup$
1
$begingroup$
Write out the expression for the determinant and solve for $C$.
$endgroup$
– amd
Dec 7 '18 at 8:37
add a comment |
$begingroup$
Reading "Mathematical Physics: Classical Mechanics" by A. Knauf, I found the following statement:
The positive symmetric matrices with determinant 1 can be written as
$$
begin{vmatrix}
A & B \
B & C \
end{vmatrix}
$$ with $A>0$, $B in mathbb R$, $C=frac{1+B^2}{A}$.
I'm on self-teaching, so even if it's maybe a well-known property, I've never encountered it.
From what "process" stems this equality, and is it valid for any "positive symmetric matrix with determinant 1", or am I missing some context?
linear-algebra matrices self-learning positive-definite symmetric-matrices
$endgroup$
Reading "Mathematical Physics: Classical Mechanics" by A. Knauf, I found the following statement:
The positive symmetric matrices with determinant 1 can be written as
$$
begin{vmatrix}
A & B \
B & C \
end{vmatrix}
$$ with $A>0$, $B in mathbb R$, $C=frac{1+B^2}{A}$.
I'm on self-teaching, so even if it's maybe a well-known property, I've never encountered it.
From what "process" stems this equality, and is it valid for any "positive symmetric matrix with determinant 1", or am I missing some context?
linear-algebra matrices self-learning positive-definite symmetric-matrices
linear-algebra matrices self-learning positive-definite symmetric-matrices
asked Dec 7 '18 at 6:02
Lo ScrondoLo Scrondo
18210
18210
1
$begingroup$
Write out the expression for the determinant and solve for $C$.
$endgroup$
– amd
Dec 7 '18 at 8:37
add a comment |
1
$begingroup$
Write out the expression for the determinant and solve for $C$.
$endgroup$
– amd
Dec 7 '18 at 8:37
1
1
$begingroup$
Write out the expression for the determinant and solve for $C$.
$endgroup$
– amd
Dec 7 '18 at 8:37
$begingroup$
Write out the expression for the determinant and solve for $C$.
$endgroup$
– amd
Dec 7 '18 at 8:37
add a comment |
1 Answer
1
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oldest
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$begingroup$
By Sylvester's criterion, a real symmetric $2times2$ matrix $M=pmatrix{A&B\ B&C}$ is positive definite if and only if $A>0$ and $det M=AC-B^2>0$. So, if you want $det M=1$, you need $C=frac{1+B^2}A$.
$endgroup$
add a comment |
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$begingroup$
By Sylvester's criterion, a real symmetric $2times2$ matrix $M=pmatrix{A&B\ B&C}$ is positive definite if and only if $A>0$ and $det M=AC-B^2>0$. So, if you want $det M=1$, you need $C=frac{1+B^2}A$.
$endgroup$
add a comment |
$begingroup$
By Sylvester's criterion, a real symmetric $2times2$ matrix $M=pmatrix{A&B\ B&C}$ is positive definite if and only if $A>0$ and $det M=AC-B^2>0$. So, if you want $det M=1$, you need $C=frac{1+B^2}A$.
$endgroup$
add a comment |
$begingroup$
By Sylvester's criterion, a real symmetric $2times2$ matrix $M=pmatrix{A&B\ B&C}$ is positive definite if and only if $A>0$ and $det M=AC-B^2>0$. So, if you want $det M=1$, you need $C=frac{1+B^2}A$.
$endgroup$
By Sylvester's criterion, a real symmetric $2times2$ matrix $M=pmatrix{A&B\ B&C}$ is positive definite if and only if $A>0$ and $det M=AC-B^2>0$. So, if you want $det M=1$, you need $C=frac{1+B^2}A$.
answered Dec 7 '18 at 8:52
user1551user1551
72.4k566127
72.4k566127
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$begingroup$
Write out the expression for the determinant and solve for $C$.
$endgroup$
– amd
Dec 7 '18 at 8:37