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
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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 |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
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 |
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$begingroup$
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