Symmetric matrices property












1












$begingroup$


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?










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$endgroup$








  • 1




    $begingroup$
    Write out the expression for the determinant and solve for $C$.
    $endgroup$
    – amd
    Dec 7 '18 at 8:37
















1












$begingroup$


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?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Write out the expression for the determinant and solve for $C$.
    $endgroup$
    – amd
    Dec 7 '18 at 8:37














1












1








1





$begingroup$


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?










share|cite|improve this question









$endgroup$




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






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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














  • 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










1 Answer
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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 $C=frac{1+B^2}A$.






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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 $C=frac{1+B^2}A$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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 $C=frac{1+B^2}A$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $C=frac{1+B^2}A$.






        share|cite|improve this answer









        $endgroup$



        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 $C=frac{1+B^2}A$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 8:52









        user1551user1551

        72.4k566127




        72.4k566127






























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