Mrthfx

How do I mention the quality of my school without bragging

How widely used is the term Treppenwitz? Is it something that most Germans know?

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

"Seemed to had" is it correct?

Is the address of a local variable a constexpr?

I am not a queen, who am I?

Why is "Consequences inflicted." not a sentence?

Is there a "higher Segal conjecture"?

How can I make names more distinctive without making them longer?

What are 'alternative tunings' of a guitar and why would you use them? Doesn't it make it more difficult to play?

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

Is 1 ppb equal to 1 μg/kg?

Why is black pepper both grey and black?

Why don't the Weasley twins use magic outside of school if the Trace can only find the location of spells cast?

How discoverable are IPv6 addresses and AAAA names by potential attackers?

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

What happens to sewage if there is no river near by?

List *all* the tuples!

Output the ŋarâþ crîþ alphabet song without using (m)any letters

Stars Make Stars

What would be the ideal power source for a cybernetic eye?

Proof involving the spectral radius and the Jordan canonical form

Is there a documented rationale why the House Ways and Means chairman can demand tax info?

Do you forfeit tax refunds/credits if you aren't required to and don't file by April 15?

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 59 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
  51 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
  51 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 59 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
  51 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
  51 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 59 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

$endgroup$

share|cite|improve this question

edited 59 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

share|cite|improve this question

edited 59 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

StammeringMathematicianStammeringMathematician

2,8121324

edited 59 mins ago

Rodrigo de Azevedo

13.2k41961

edited 59 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 54 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 53 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 54 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 54 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 54 mins ago

trancelocationtrancelocation

14.1k1829

$endgroup$

share|cite|improve this answer

answered 54 mins ago

trancelocationtrancelocation

14.1k1829

answered 54 mins ago

trancelocationtrancelocation

14.1k1829

answered 54 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 53 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 53 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 53 mins ago

TomGrubbTomGrubb

11.2k11639

$endgroup$

share|cite|improve this answer

answered 53 mins ago

TomGrubbTomGrubb

11.2k11639

answered 53 mins ago

TomGrubbTomGrubb

11.2k11639

answered 53 mins ago

TomGrubbTomGrubb

11.2k11639