Determinant is linear as a function of each of the rows of the matrix. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determinant after matrix change issueWhat is the origin of the determinant in linear algebra?The determinant function is the only one satisfying the conditionsIf a NxN matrix has two identical columns will its determinant be zero?Geometric interpretation of determinant when two rows are swappedDeterminant functionDeterminant and determinant function(explanation)How would I answer the following question about the determinant of a matrix?Determinant of a matrix and linear independence (explanation needed)Compute new matrix derterminant when only two rows change
What is the musical term for a note that continously plays through a melody?
Is there a concise way to say "all of the X, one of each"?
Using et al. for a last / senior author rather than for a first author
How to bypass password on Windows XP account?
What's the purpose of writing one's academic bio in 3rd person?
What makes black pepper strong or mild?
Is 1 ppb equal to 1 μg/kg?
Models of set theory where not every set can be linearly ordered
What is this single-engine low-wing propeller plane?
What is the longest distance a 13th-level monk can jump while attacking on the same turn?
Right-skewed distribution with mean equals to mode?
What happens to sewage if there is no river near by?
Bonus calculation: Am I making a mountain out of a molehill?
Do I really need recursive chmod to restrict access to a folder?
Storing hydrofluoric acid before the invention of plastics
Why aren't air breathing engines used as small first stages
What are the motives behind Cersei's orders given to Bronn?
What's the difference between `auto x = vector<int>()` and `vector<int> x`?
Were Kohanim forbidden from serving in King David's army?
Determinant is linear as a function of each of the rows of the matrix.
How widely used is the term Treppenwitz? Is it something that most Germans know?
How much radiation do nuclear physics experiments expose researchers to nowadays?
Is the Standard Deduction better than Itemized when both are the same amount?
What would be the ideal power source for a cybernetic eye?
Determinant is linear as a function of each of the rows of the matrix.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determinant after matrix change issueWhat is the origin of the determinant in linear algebra?The determinant function is the only one satisfying the conditionsIf a NxN matrix has two identical columns will its determinant be zero?Geometric interpretation of determinant when two rows are swappedDeterminant functionDeterminant and determinant function(explanation)How would I answer the following question about the determinant of a matrix?Determinant of a matrix and linear independence (explanation needed)Compute new matrix derterminant when only two rows change
$begingroup$
Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.
I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^n times n$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$
Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?
linear-algebra matrices determinant
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 1 hour ago
StammeringMathematicianStammeringMathematician
2,8121324
$endgroup$
add a comment |
$begingroup$
Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.
I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^n times n$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$
Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?
linear-algebra matrices determinant
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 1 hour ago
StammeringMathematicianStammeringMathematician
2,8121324
$endgroup$
$begingroup$
I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
$endgroup$
– StammeringMathematician
1 hour ago
$begingroup$
Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
$endgroup$
– copper.hat
1 hour ago
add a comment |
$begingroup$
Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.
I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^n times n$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$
Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?
linear-algebra matrices determinant
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 1 hour ago
StammeringMathematicianStammeringMathematician
2,8121324
$endgroup$
Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.
I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $mathbb K^n times n$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^ndet(A)$
Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?
linear-algebra matrices determinant
linear-algebra matrices determinant
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 1 hour ago
StammeringMathematicianStammeringMathematician
2,8121324
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 1 hour ago
StammeringMathematicianStammeringMathematician
2,8121324
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
13.2k41961
asked 1 hour ago
StammeringMathematicianStammeringMathematician
2,8121324
asked 1 hour ago
StammeringMathematicianStammeringMathematician
2,8121324
asked 1 hour ago
StammeringMathematicianStammeringMathematician
2,8121324
2,8121324
$begingroup$
I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
$endgroup$
– StammeringMathematician
1 hour ago
$begingroup$
Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
$endgroup$
– copper.hat
1 hour ago
add a comment |
$begingroup$
I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
$endgroup$
– StammeringMathematician
1 hour ago
$begingroup$
Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
$endgroup$
– copper.hat
1 hour ago
$begingroup$
I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
$endgroup$
– StammeringMathematician
1 hour ago
$begingroup$
I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
$endgroup$
– StammeringMathematician
1 hour ago
$begingroup$
Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
$endgroup$
– copper.hat
1 hour ago
$begingroup$
Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
$endgroup$
– copper.hat
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have
$$detbeginpmatrixr_1 \ vdots \r_i \ vdots \ r_nendpmatrix = detbeginpmatrixr_1 \ vdots \ sa+tb \ vdots \ r_nendpmatrix = sdetbeginpmatrixr_1 \ vdots \ a \ vdots \ r_nendpmatrix + tdetbeginpmatrixr_1 \ vdots \ b \ vdots \ r_nendpmatrix$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
$endgroup$
add a comment |
$begingroup$
Let $M$ be an $ntimes n$ matrix with rows $mathbfr_1,dots,mathbfr_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbfr_1,dots,mathbfr_n).
$$
To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
$$
det(mathbfr_1,dots,mathbfr_i-1,cmathbfr_i,mathbfr_i+1dotsmathbfr_n)=cdet(mathbfr_1,dots,mathbfr_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbfx+mathbfr_1,mathbfr_2,dots,mathbfr_n)=det(mathbfx,dots,mathbfr_n)+det(mathbfr_1,dots,mathbfr_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"
answered 1 hour ago
TomGrubbTomGrubb
11.2k11639
$endgroup$
add a comment |
Your Answer
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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%2f3189406%2fdeterminant-is-linear-as-a-function-of-each-of-the-rows-of-the-matrix%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have
$$detbeginpmatrixr_1 \ vdots \r_i \ vdots \ r_nendpmatrix = detbeginpmatrixr_1 \ vdots \ sa+tb \ vdots \ r_nendpmatrix = sdetbeginpmatrixr_1 \ vdots \ a \ vdots \ r_nendpmatrix + tdetbeginpmatrixr_1 \ vdots \ b \ vdots \ r_nendpmatrix$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
$endgroup$
add a comment |
$begingroup$
If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have
$$detbeginpmatrixr_1 \ vdots \r_i \ vdots \ r_nendpmatrix = detbeginpmatrixr_1 \ vdots \ sa+tb \ vdots \ r_nendpmatrix = sdetbeginpmatrixr_1 \ vdots \ a \ vdots \ r_nendpmatrix + tdetbeginpmatrixr_1 \ vdots \ b \ vdots \ r_nendpmatrix$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
$endgroup$
add a comment |
$begingroup$
If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have
$$detbeginpmatrixr_1 \ vdots \r_i \ vdots \ r_nendpmatrix = detbeginpmatrixr_1 \ vdots \ sa+tb \ vdots \ r_nendpmatrix = sdetbeginpmatrixr_1 \ vdots \ a \ vdots \ r_nendpmatrix + tdetbeginpmatrixr_1 \ vdots \ b \ vdots \ r_nendpmatrix$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
$endgroup$
If $r_1, ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have
$$detbeginpmatrixr_1 \ vdots \r_i \ vdots \ r_nendpmatrix = detbeginpmatrixr_1 \ vdots \ sa+tb \ vdots \ r_nendpmatrix = sdetbeginpmatrixr_1 \ vdots \ a \ vdots \ r_nendpmatrix + tdetbeginpmatrixr_1 \ vdots \ b \ vdots \ r_nendpmatrix$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
answered 1 hour ago
trancelocationtrancelocation
14.1k1829
14.1k1829
add a comment |
add a comment |
$begingroup$
Let $M$ be an $ntimes n$ matrix with rows $mathbfr_1,dots,mathbfr_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbfr_1,dots,mathbfr_n).
$$
To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
$$
det(mathbfr_1,dots,mathbfr_i-1,cmathbfr_i,mathbfr_i+1dotsmathbfr_n)=cdet(mathbfr_1,dots,mathbfr_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbfx+mathbfr_1,mathbfr_2,dots,mathbfr_n)=det(mathbfx,dots,mathbfr_n)+det(mathbfr_1,dots,mathbfr_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"
answered 1 hour ago
TomGrubbTomGrubb
11.2k11639
$endgroup$
add a comment |
$begingroup$
Let $M$ be an $ntimes n$ matrix with rows $mathbfr_1,dots,mathbfr_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbfr_1,dots,mathbfr_n).
$$
To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
$$
det(mathbfr_1,dots,mathbfr_i-1,cmathbfr_i,mathbfr_i+1dotsmathbfr_n)=cdet(mathbfr_1,dots,mathbfr_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbfx+mathbfr_1,mathbfr_2,dots,mathbfr_n)=det(mathbfx,dots,mathbfr_n)+det(mathbfr_1,dots,mathbfr_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"
answered 1 hour ago
TomGrubbTomGrubb
11.2k11639
$endgroup$
add a comment |
$begingroup$
Let $M$ be an $ntimes n$ matrix with rows $mathbfr_1,dots,mathbfr_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbfr_1,dots,mathbfr_n).
$$
To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
$$
det(mathbfr_1,dots,mathbfr_i-1,cmathbfr_i,mathbfr_i+1dotsmathbfr_n)=cdet(mathbfr_1,dots,mathbfr_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbfx+mathbfr_1,mathbfr_2,dots,mathbfr_n)=det(mathbfx,dots,mathbfr_n)+det(mathbfr_1,dots,mathbfr_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"
answered 1 hour ago
TomGrubbTomGrubb
11.2k11639
$endgroup$
Let $M$ be an $ntimes n$ matrix with rows $mathbfr_1,dots,mathbfr_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbfr_1,dots,mathbfr_n).
$$
To say that $det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is,
$$
det(mathbfr_1,dots,mathbfr_i-1,cmathbfr_i,mathbfr_i+1dotsmathbfr_n)=cdet(mathbfr_1,dots,mathbfr_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbfx+mathbfr_1,mathbfr_2,dots,mathbfr_n)=det(mathbfx,dots,mathbfr_n)+det(mathbfr_1,dots,mathbfr_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"
answered 1 hour ago
TomGrubbTomGrubb
11.2k11639
answered 1 hour ago
TomGrubbTomGrubb
11.2k11639
answered 1 hour ago
TomGrubbTomGrubb
11.2k11639
answered 1 hour ago
TomGrubbTomGrubb
11.2k11639
11.2k11639
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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%2f3189406%2fdeterminant-is-linear-as-a-function-of-each-of-the-rows-of-the-matrix%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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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
$begingroup$
I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right?
$endgroup$
– StammeringMathematician
1 hour ago
$begingroup$
Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant.
$endgroup$
– copper.hat
1 hour ago