Determinant is linear as a function of each of the rows of the matrix.
$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
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
2 hours 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
2 hours 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
$endgroup$
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 3 hours ago
Rodrigo de Azevedo
13.2k41961
13.2k41961
asked 3 hours ago
StammeringMathematicianStammeringMathematician
2,8171324
2,8171324
$begingroup$
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
2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours 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
$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
$endgroup$
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"
$endgroup$
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');
});
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
$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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
$$detbegin{pmatrix}r_1 \ vdots \r_i \ vdots \ r_nend{pmatrix} = detbegin{pmatrix}r_1 \ vdots \ sa+tb \ vdots \ r_nend{pmatrix} = sdetbegin{pmatrix}r_1 \ vdots \ a \ vdots \ r_nend{pmatrix} + tdetbegin{pmatrix}r_1 \ vdots \ b \ vdots \ r_nend{pmatrix}$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
answered 2 hours ago
trancelocationtrancelocation
14.2k1829
14.2k1829
add a comment |
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"
$endgroup$
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"
$endgroup$
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"
$endgroup$
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"
answered 2 hours 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');
});
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');
});
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
$begingroup$
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
2 hours 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
2 hours ago