Symmetric matrices property
$begingroup$
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
asked Dec 7 '18 at 6:02
Lo ScrondoLo Scrondo
18210
$endgroup$
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
asked Dec 7 '18 at 6:02
Lo ScrondoLo Scrondo
18210
$endgroup$
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
asked Dec 7 '18 at 6:02
Lo ScrondoLo Scrondo
18210
$endgroup$
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
asked Dec 7 '18 at 6:02
Lo ScrondoLo Scrondo
18210
asked Dec 7 '18 at 6:02
Lo ScrondoLo Scrondo
18210
asked Dec 7 '18 at 6:02
Lo ScrondoLo Scrondo
18210
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
active
oldest
votes
$begingroup$
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 $C=frac{1+B^2}A$.
answered Dec 7 '18 at 8:52
user1551user1551
72.4k566127
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029537%2fsymmetric-matrices-property%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $C=frac{1+B^2}A$.
answered Dec 7 '18 at 8:52
user1551user1551
72.4k566127
$endgroup$
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 $C=frac{1+B^2}A$.
answered Dec 7 '18 at 8:52
user1551user1551
72.4k566127
$endgroup$
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 $C=frac{1+B^2}A$.
answered Dec 7 '18 at 8:52
user1551user1551
72.4k566127
$endgroup$
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
answered Dec 7 '18 at 8:52
user1551user1551
72.4k566127
answered Dec 7 '18 at 8:52
user1551user1551
72.4k566127
answered Dec 7 '18 at 8:52
user1551user1551
72.4k566127
72.4k566127
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029537%2fsymmetric-matrices-property%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Write out the expression for the determinant and solve for $C$.
$endgroup$
– amd
Dec 7 '18 at 8:37