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 issueThe determinant function is the only one satisfying the conditionsLinear Algebra solution when determinant is zeroIf a NxN matrix has two identical columns will its determinant be zero?Finding the determinant of a block matrix (Linear Algebra)Determinant of an elementary matrixDeterminant when rows reversedDeterminant and determinant function(explanation)Determinant of a matrix and linear independence (explanation needed)Operations upon the determinant of a matrix
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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 issueThe determinant function is the only one satisfying the conditionsLinear Algebra solution when determinant is zeroIf a NxN matrix has two identical columns will its determinant be zero?Finding the determinant of a block matrix (Linear Algebra)Determinant of an elementary matrixDeterminant when rows reversedDeterminant and determinant function(explanation)Determinant of a matrix and linear independence (explanation needed)Operations upon the determinant of a 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
edited 57 mins 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 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
edited 57 mins 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 57 mins ago
Rodrigo de Azevedo
13.2k41961
asked 1 hour ago
StammeringMathematicianStammeringMathematician
2,8121324
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
edited 57 mins 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
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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 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 52 mins 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 50 mins ago
TomGrubbTomGrubb
11.2k11639
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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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 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
$$detbeginpmatrixr_1 \ vdots \r_i \ vdots \ r_nendpmatrix = detbeginpmatrixr_1 \ vdots \ sa+tb \ vdots \ r_nendpmatrix = sdetbeginpmatrixr_1 \ vdots \ a \ vdots \ r_nendpmatrix + tdetbeginpmatrixr_1 \ vdots \ b \ vdots \ r_nendpmatrix$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
answered 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
$$detbeginpmatrixr_1 \ vdots \r_i \ vdots \ r_nendpmatrix = detbeginpmatrixr_1 \ vdots \ sa+tb \ vdots \ r_nendpmatrix = sdetbeginpmatrixr_1 \ vdots \ a \ vdots \ r_nendpmatrix + tdetbeginpmatrixr_1 \ vdots \ b \ vdots \ r_nendpmatrix$$
This holds for any row $i=1,ldots , n$. And similarly this also applies to columns.
answered 52 mins 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 52 mins ago
trancelocationtrancelocation
14.1k1829
answered 52 mins ago
trancelocationtrancelocation
14.1k1829
answered 52 mins ago
trancelocationtrancelocation
14.1k1829
answered 52 mins 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 50 mins 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 50 mins 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 50 mins 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 50 mins ago
TomGrubbTomGrubb
11.2k11639
answered 50 mins ago
TomGrubbTomGrubb
11.2k11639
answered 50 mins ago
TomGrubbTomGrubb
11.2k11639
answered 50 mins 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
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