Fdtxjr

New brakes for 90s road bike

What if a revenant (monster) gains fire resistance?

Is it better practice to read straight from sheet music rather than memorize it?

How can "mimic phobia" be cured or prevented?

Pre-mixing cryogenic fuels and using only one fuel tank

Can Legal Documents Be Siged In Non-Standard Pen Colors?

Terse Method to Swap Lowest for Highest?

Why Shazam when there is already Superman?

How could a planet have erratic days?

Open a doc from terminal, but not by its name

How to follow the Halacha?

"Spoil" vs "Ruin"

How much character growth crosses the line into breaking the character

is this legal and f i dont come up with extra money is the deal off

How to explain what's wrong with this application of the chain rule?

Is the U.S. Code copyrighted by the Government?

Is it safe to use olive oil to clean the ear wax?

How do you respond to a colleague from another team when they're wrongly expecting that you'll help them?

Multiplicative persistence

Travelling outside the UK without a passport

Closed-form expression for certain product

Freedom of speech and where it applies

Is there any references on the tensor product of presentable (1-)categories?

How did Rebekah know that Esau was planning to kill his brother in Genesis 27:42?

Determinant of the Adjoint of a matrix

Prove the determinant of this matrixPositive definite matrix and self-adjoint invertible matrixMaximum determinant of Latin squaresMinimum absolute determinant of a regular latin square matrixDeterminant of a standard magic squareWhat is the determinant of the sum of a diagonal matrix and a matrix of ones?Determinant of skew- symmetric matrixGreatest natural number which divides the determinant of a matrixWhat is the determinant of cofactor matrix of a matrix?Coincidence of the determinant and eigenvalue of Hermitian matrix or unitary matrix

0

$begingroup$

The determinant of the adjoint of a matrix $A$ is given by $|A|^n-1$ where $n$ is the order of the square matrix.

So, for odd $n$, the determinant is always positive.

Is there an intuitive explanation for this? It is surprising that every odd ordered matrix irrespective of its elements would have an adjoint with a positive determinant.

linear-algebra matrices determinant

share|cite|improve this question

asked Mar 15 at 16:23

Anubhab DasAnubhab Das

175

$endgroup$

add a comment | 

0

$begingroup$

The determinant of the adjoint of a matrix $A$ is given by $|A|^n-1$ where $n$ is the order of the square matrix.

So, for odd $n$, the determinant is always positive.

Is there an intuitive explanation for this? It is surprising that every odd ordered matrix irrespective of its elements would have an adjoint with a positive determinant.

linear-algebra matrices determinant

share|cite|improve this question

asked Mar 15 at 16:23

Anubhab DasAnubhab Das

175

$endgroup$

add a comment | 

$begingroup$

The determinant of the adjoint of a matrix $A$ is given by $|A|^n-1$ where $n$ is the order of the square matrix.

So, for odd $n$, the determinant is always positive.

Is there an intuitive explanation for this? It is surprising that every odd ordered matrix irrespective of its elements would have an adjoint with a positive determinant.

linear-algebra matrices determinant

share|cite|improve this question

asked Mar 15 at 16:23

Anubhab DasAnubhab Das

175

$endgroup$

share|cite|improve this question

asked Mar 15 at 16:23

Anubhab DasAnubhab Das

175

share|cite|improve this question

asked Mar 15 at 16:23

Anubhab DasAnubhab Das

175

asked Mar 15 at 16:23

Anubhab DasAnubhab Das

175

asked Mar 15 at 16:23

Anubhab DasAnubhab Das

175

1 Answer
1

active

oldest

votes

1

$begingroup$

Here is a way to understand this in $mathbbR^3$. Suppose the rows of your original matrix $A$ are coordinates of vectors $a,b,c$ in some orthonormal basis. The rows of the adjoint matrix are the coordinates of the vectors $btimes c$, $ctimes a$ and $atimes b$ (or even better, the coordinates of the corresponding bivectors). The determinant of the adjoint matrix is thus the oriented volume of the parallelepiped defined by those cross-products.

We can assume that $a,b,c$ are linearly independent, otherwise at least two of the cross-products will be parallel an the adjoint determinant zero. Now observe that if $a,b,c$ is a positively oriented basis, these cross products also form a positively oriented basis (think of $i$, $j$, $k$) and hence the determinant is positive. If, however,$a,b,c$ is a negatively oriented basis, then the cross products belong to the cone, polar to the cone generated by $a,b,c$ (instead of the dual cone they belong in the former case). They are thus opposite to vectors with negative orientation (preserving their order) hence they form a positively oriented basis again. In the general multidimensional case I think you can use geometric algebra to prove it. A proof using geometric algebra will necessarily convey some geometric intuition.

share|cite|improve this answer

answered Mar 15 at 17:42

GReyesGReyes

2,25815

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for answering! I am not familiar with dual cone and cannot follow after that. Can you make it a bit simpler from there? Thanks
  $endgroup$
  – Anubhab Das
  Mar 15 at 18:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure. Linear combinations of your vectors $a,b,c$ with nonnegative coefficients form a cone. For ex. $i,j,k$ generate the first octant. The dual cone is the set of vectors which form an acute angle with all the vectors in your cone. In the $i,j,k$ example, the dual is the first octant again (self-dual). "Acute" cones (pointy) have obtuse duals and viceversa. The polar cone is the opposite one (reverse each vector). For the first octant, it is the opposite octant. $k=itimes j$, $i=jtimes k$ and $j=ktimes i$ are in the first octant, while $-k=jtimes i$ etc. are in the opposite.
  $endgroup$
  – GReyes
  Mar 15 at 20:28
  
  

  
  
  
  

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149494%2fdeterminant-of-the-adjoint-of-a-matrix%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Here is a way to understand this in $mathbbR^3$. Suppose the rows of your original matrix $A$ are coordinates of vectors $a,b,c$ in some orthonormal basis. The rows of the adjoint matrix are the coordinates of the vectors $btimes c$, $ctimes a$ and $atimes b$ (or even better, the coordinates of the corresponding bivectors). The determinant of the adjoint matrix is thus the oriented volume of the parallelepiped defined by those cross-products.

We can assume that $a,b,c$ are linearly independent, otherwise at least two of the cross-products will be parallel an the adjoint determinant zero. Now observe that if $a,b,c$ is a positively oriented basis, these cross products also form a positively oriented basis (think of $i$, $j$, $k$) and hence the determinant is positive. If, however,$a,b,c$ is a negatively oriented basis, then the cross products belong to the cone, polar to the cone generated by $a,b,c$ (instead of the dual cone they belong in the former case). They are thus opposite to vectors with negative orientation (preserving their order) hence they form a positively oriented basis again. In the general multidimensional case I think you can use geometric algebra to prove it. A proof using geometric algebra will necessarily convey some geometric intuition.

share|cite|improve this answer

answered Mar 15 at 17:42

GReyesGReyes

2,25815

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for answering! I am not familiar with dual cone and cannot follow after that. Can you make it a bit simpler from there? Thanks
  $endgroup$
  – Anubhab Das
  Mar 15 at 18:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure. Linear combinations of your vectors $a,b,c$ with nonnegative coefficients form a cone. For ex. $i,j,k$ generate the first octant. The dual cone is the set of vectors which form an acute angle with all the vectors in your cone. In the $i,j,k$ example, the dual is the first octant again (self-dual). "Acute" cones (pointy) have obtuse duals and viceversa. The polar cone is the opposite one (reverse each vector). For the first octant, it is the opposite octant. $k=itimes j$, $i=jtimes k$ and $j=ktimes i$ are in the first octant, while $-k=jtimes i$ etc. are in the opposite.
  $endgroup$
  – GReyes
  Mar 15 at 20:28
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Here is a way to understand this in $mathbbR^3$. Suppose the rows of your original matrix $A$ are coordinates of vectors $a,b,c$ in some orthonormal basis. The rows of the adjoint matrix are the coordinates of the vectors $btimes c$, $ctimes a$ and $atimes b$ (or even better, the coordinates of the corresponding bivectors). The determinant of the adjoint matrix is thus the oriented volume of the parallelepiped defined by those cross-products.

We can assume that $a,b,c$ are linearly independent, otherwise at least two of the cross-products will be parallel an the adjoint determinant zero. Now observe that if $a,b,c$ is a positively oriented basis, these cross products also form a positively oriented basis (think of $i$, $j$, $k$) and hence the determinant is positive. If, however,$a,b,c$ is a negatively oriented basis, then the cross products belong to the cone, polar to the cone generated by $a,b,c$ (instead of the dual cone they belong in the former case). They are thus opposite to vectors with negative orientation (preserving their order) hence they form a positively oriented basis again. In the general multidimensional case I think you can use geometric algebra to prove it. A proof using geometric algebra will necessarily convey some geometric intuition.

share|cite|improve this answer

answered Mar 15 at 17:42

GReyesGReyes

2,25815

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for answering! I am not familiar with dual cone and cannot follow after that. Can you make it a bit simpler from there? Thanks
  $endgroup$
  – Anubhab Das
  Mar 15 at 18:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure. Linear combinations of your vectors $a,b,c$ with nonnegative coefficients form a cone. For ex. $i,j,k$ generate the first octant. The dual cone is the set of vectors which form an acute angle with all the vectors in your cone. In the $i,j,k$ example, the dual is the first octant again (self-dual). "Acute" cones (pointy) have obtuse duals and viceversa. The polar cone is the opposite one (reverse each vector). For the first octant, it is the opposite octant. $k=itimes j$, $i=jtimes k$ and $j=ktimes i$ are in the first octant, while $-k=jtimes i$ etc. are in the opposite.
  $endgroup$
  – GReyes
  Mar 15 at 20:28
  
  

  
  
  
  

add a comment | 

$begingroup$

Here is a way to understand this in $mathbbR^3$. Suppose the rows of your original matrix $A$ are coordinates of vectors $a,b,c$ in some orthonormal basis. The rows of the adjoint matrix are the coordinates of the vectors $btimes c$, $ctimes a$ and $atimes b$ (or even better, the coordinates of the corresponding bivectors). The determinant of the adjoint matrix is thus the oriented volume of the parallelepiped defined by those cross-products.

We can assume that $a,b,c$ are linearly independent, otherwise at least two of the cross-products will be parallel an the adjoint determinant zero. Now observe that if $a,b,c$ is a positively oriented basis, these cross products also form a positively oriented basis (think of $i$, $j$, $k$) and hence the determinant is positive. If, however,$a,b,c$ is a negatively oriented basis, then the cross products belong to the cone, polar to the cone generated by $a,b,c$ (instead of the dual cone they belong in the former case). They are thus opposite to vectors with negative orientation (preserving their order) hence they form a positively oriented basis again. In the general multidimensional case I think you can use geometric algebra to prove it. A proof using geometric algebra will necessarily convey some geometric intuition.

share|cite|improve this answer

answered Mar 15 at 17:42

GReyesGReyes

2,25815

$endgroup$

share|cite|improve this answer

answered Mar 15 at 17:42

GReyesGReyes

2,25815

answered Mar 15 at 17:42

GReyesGReyes

2,25815

answered Mar 15 at 17:42

GReyesGReyes

2,25815