Ggthjy

How do I stop a creek from eroding my steep embankment?

G-Code for resetting to 100% speed

Stars Make Stars

How do I mention the quality of my school without bragging

Sorting numerically

Bonus calculation: Am I making a mountain out of a molehill?

How to bypass password on Windows XP account?

Single word antonym of "flightless"

Why are there no cargo aircraft with "flying wing" design?

Can Pao de Queijo, and similar foods, be kosher for Passover?

When to stop saving and start investing?

Why was the term "discrete" used in discrete logarithm?

Why aren't air breathing engines used as small first stages

Diagram with tikz

Storing hydrofluoric acid before the invention of plastics

ListPlot join points by nearest neighbor rather than order

How to recreate this effect in Photoshop?

How to motivate offshore teams and trust them to deliver?

Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?

What are the motives behind Cersei's orders given to Bronn?

How can I fade player when goes inside or outside of the area?

Does accepting a pardon have any bearing on trying that person for the same crime in a sovereign jurisdiction?

Why is black pepper both grey and black?

What is the longest distance a 13th-level monk can jump while attacking on the same turn?

Determinant is linear as a function of each of the rows of the matrix.

2

$begingroup$

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

share|cite|improve this question

edited 57 mins 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
  49 mins 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
  48 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|cite|improve this question

edited 57 mins 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
  49 mins 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
  48 mins 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

share|cite|improve this question

edited 57 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

$endgroup$

share|cite|improve this question

edited 57 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

share|cite|improve this question

edited 57 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

edited 57 mins ago

Rodrigo de Azevedo

13.2k41961

edited 57 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

2 Answers
2

active

oldest

votes

3

$begingroup$

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

$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$

This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.

share|cite|improve this answer

answered 52 mins ago

trancelocationtrancelocation

14.1k1829

$endgroup$

add a comment | 

2

$begingroup$

Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
$$
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,
$$
det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"

share|cite|improve this answer

answered 51 mins 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
});


}
});

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%2f3189406%2fdeterminant-is-linear-as-a-function-of-each-of-the-rows-of-the-matrix%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

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

$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$

This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.

share|cite|improve this answer

answered 52 mins ago

trancelocationtrancelocation

14.1k1829

$endgroup$

add a comment | 

3

$begingroup$

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

$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$

This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.

share|cite|improve this answer

answered 52 mins 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

$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$

This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.

share|cite|improve this answer

answered 52 mins ago

trancelocationtrancelocation

14.1k1829

$endgroup$

share|cite|improve this answer

answered 52 mins ago

trancelocationtrancelocation

14.1k1829

answered 52 mins ago

trancelocationtrancelocation

14.1k1829

answered 52 mins ago

trancelocationtrancelocation

14.1k1829

2

$begingroup$

Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
$$
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,
$$
det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"

share|cite|improve this answer

answered 51 mins ago

TomGrubbTomGrubb

11.2k11639

$endgroup$

add a comment | 

2

$begingroup$

Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
$$
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,
$$
det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"

share|cite|improve this answer

answered 51 mins ago

TomGrubbTomGrubb

11.2k11639

$endgroup$

add a comment | 

$begingroup$

Let $M$ be an $ntimes n$ matrix with rows $mathbf{r}_1,dots,mathbf{r}_n$. Then we may think of the determinant as a function of the rows
$$
det(M)=det(mathbf{r}_1,dots,mathbf{r}_n).
$$
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,
$$
det(mathbf{r}_1,dots,mathbf{r}_{i-1},cmathbf{r}_i,mathbf{r}_{i+1}dotsmathbf{r}_n)=cdet(mathbf{r}_1,dots,mathbf{r}_n).
$$
Similarly if we fix all but one row (say the first), we obtain
$$
det(mathbf{x}+mathbf{r}_1,mathbf{r}_2,dots,mathbf{r}_n)=det(mathbf{x},dots,mathbf{r}_n)+det(mathbf{r}_1,dots,mathbf{r}_n).
$$
Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"

share|cite|improve this answer

answered 51 mins ago

TomGrubbTomGrubb

11.2k11639

$endgroup$

share|cite|improve this answer

answered 51 mins ago

TomGrubbTomGrubb

11.2k11639

answered 51 mins ago

TomGrubbTomGrubb

11.2k11639

answered 51 mins ago

TomGrubbTomGrubb

11.2k11639