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Question about Segre embedding proof in Karen's Smith book.
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The main problems that I am having with the proof is linear algebra. First, why is the Segre mapping contained in the set defined by 2x2 minors of the matrix $(z_ij)$ why does it follow from the fact that 2x2 sub- determinants vanish ? In the converse why is it the case that when all 2x2 subdeterminants vanish, then rank of the matrix at most 1 ? Finally, why (m + 1)(n + 1) of rank k factors according to what the book mentioned ? 
algebraic-geometry
asked Jul 27 at 15:54
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384111
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up vote
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The main problems that I am having with the proof is linear algebra. First, why is the Segre mapping contained in the set defined by 2x2 minors of the matrix $(z_ij)$ why does it follow from the fact that 2x2 sub- determinants vanish ? In the converse why is it the case that when all 2x2 subdeterminants vanish, then rank of the matrix at most 1 ? Finally, why (m + 1)(n + 1) of rank k factors according to what the book mentioned ?
algebraic-geometry
asked Jul 27 at 15:54
Newbie
384111
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The main problems that I am having with the proof is linear algebra. First, why is the Segre mapping contained in the set defined by 2x2 minors of the matrix $(z_ij)$ why does it follow from the fact that 2x2 sub- determinants vanish ? In the converse why is it the case that when all 2x2 subdeterminants vanish, then rank of the matrix at most 1 ? Finally, why (m + 1)(n + 1) of rank k factors according to what the book mentioned ?
algebraic-geometry
asked Jul 27 at 15:54
Newbie
384111
The main problems that I am having with the proof is linear algebra. First, why is the Segre mapping contained in the set defined by 2x2 minors of the matrix $(z_ij)$ why does it follow from the fact that 2x2 sub- determinants vanish ? In the converse why is it the case that when all 2x2 subdeterminants vanish, then rank of the matrix at most 1 ? Finally, why (m + 1)(n + 1) of rank k factors according to what the book mentioned ?
algebraic-geometry
asked Jul 27 at 15:54
Newbie
384111
asked Jul 27 at 15:54
Newbie
384111
asked Jul 27 at 15:54
Newbie
384111
asked Jul 27 at 15:54
Newbie
384111
384111
add a comment |Â
add a comment |Â
1 Answer
1
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To answer your first question, assume for contradiction that some $2times2$ minor does not vanish. After reordering the rows and columns of your matrix (which does not change its rank), you may assume that it is the top left $2times 2$ submatrix. Now note that the second column can not be a scalar multiple of the first, because otherwise the top left $2times 2$ block would have the form
$$
beginpmatrix a & lambda a \ b & lambda b endpmatrix
$$
for some $lambdain Bbbk$; and the determinant of this matrix vanishes.Remark. Of course, this works more generally. If you have a non-vanishing $ktimes k$ minor, then you can conclude that there are at least $k$ linearly independent rows, which means that your matrix has rank at least $k$.
- To answer your second question, assume that all $2times 2$ minors of the matrix $z=(z_ij)$ vanish. We may assume that the matrix is nonzero, otherwise we are done (it has rank less than one). Without loss of generality, assume that the entry $z_11$ is nonzero. Now, for any $i$ and $j$, you have that the matrix
$$
beginpmatrix z_11 & z_1j \ z_i1 & z_ij endpmatrix
$$
has vanishing determinant, so the second column must be a multiple of the first, i.e. there is some $lambdainBbbk$ with $z_1j=lambda z_11$ and $z_ij=lambda z_i1$. Since this holds for all $i$, it follows that the $j$-th column of $z$ is a scalar multiple of the first column. Since that holds for all $j$, all columns are scalar multiples of the first column. Hence, the matrix has rank one. - To see that any rank $k$ matrix factors like this, recall that the rank of a linear map is the dimension of its image. Hence, your $(m+1)times(n+1)$ matrix describes a linear map $f:Bbbk^n+1toBbbk^m+1$ such that $V:=f(Bbbk^n+1)congBbbk^k$. In other words, the map factors as $Bbbk^n+1to VtoBbbk^m+1$, where the first map is the co-restriction of $f$ and the second is just the inclusion. Choosing appropriate bases, you may describe this as $Bbbk^n+1toBbbk^ktoBbbk^m+1$, i.e. the first map is given by a $ktimes(n+1)$ matrix and the second one by an $(m+1)times k$ matrix.
answered Jul 27 at 22:25
Jesko Hüttenhain
9,96912253
One final thing why is the image of the Segre map contained in the 2x2 minors ?
– Newbie
Jul 28 at 3:27
well, because every point in the image has rank one and we just established that a matrix has rank one if and only if it is nonzero and all of its $2times2$ minors vanish.
– Jesko Hüttenhain
Jul 28 at 8:11
Yeah, that is good. Thanks.
– Newbie
Jul 28 at 15:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To answer your first question, assume for contradiction that some $2times2$ minor does not vanish. After reordering the rows and columns of your matrix (which does not change its rank), you may assume that it is the top left $2times 2$ submatrix. Now note that the second column can not be a scalar multiple of the first, because otherwise the top left $2times 2$ block would have the form
$$
beginpmatrix a & lambda a \ b & lambda b endpmatrix
$$
for some $lambdain Bbbk$; and the determinant of this matrix vanishes.Remark. Of course, this works more generally. If you have a non-vanishing $ktimes k$ minor, then you can conclude that there are at least $k$ linearly independent rows, which means that your matrix has rank at least $k$.
- To answer your second question, assume that all $2times 2$ minors of the matrix $z=(z_ij)$ vanish. We may assume that the matrix is nonzero, otherwise we are done (it has rank less than one). Without loss of generality, assume that the entry $z_11$ is nonzero. Now, for any $i$ and $j$, you have that the matrix
$$
beginpmatrix z_11 & z_1j \ z_i1 & z_ij endpmatrix
$$
has vanishing determinant, so the second column must be a multiple of the first, i.e. there is some $lambdainBbbk$ with $z_1j=lambda z_11$ and $z_ij=lambda z_i1$. Since this holds for all $i$, it follows that the $j$-th column of $z$ is a scalar multiple of the first column. Since that holds for all $j$, all columns are scalar multiples of the first column. Hence, the matrix has rank one. - To see that any rank $k$ matrix factors like this, recall that the rank of a linear map is the dimension of its image. Hence, your $(m+1)times(n+1)$ matrix describes a linear map $f:Bbbk^n+1toBbbk^m+1$ such that $V:=f(Bbbk^n+1)congBbbk^k$. In other words, the map factors as $Bbbk^n+1to VtoBbbk^m+1$, where the first map is the co-restriction of $f$ and the second is just the inclusion. Choosing appropriate bases, you may describe this as $Bbbk^n+1toBbbk^ktoBbbk^m+1$, i.e. the first map is given by a $ktimes(n+1)$ matrix and the second one by an $(m+1)times k$ matrix.
answered Jul 27 at 22:25
Jesko Hüttenhain
9,96912253
One final thing why is the image of the Segre map contained in the 2x2 minors ?
– Newbie
Jul 28 at 3:27
well, because every point in the image has rank one and we just established that a matrix has rank one if and only if it is nonzero and all of its $2times2$ minors vanish.
– Jesko Hüttenhain
Jul 28 at 8:11
Yeah, that is good. Thanks.
– Newbie
Jul 28 at 15:49
add a comment |Â
up vote
1
down vote
accepted
To answer your first question, assume for contradiction that some $2times2$ minor does not vanish. After reordering the rows and columns of your matrix (which does not change its rank), you may assume that it is the top left $2times 2$ submatrix. Now note that the second column can not be a scalar multiple of the first, because otherwise the top left $2times 2$ block would have the form
$$
beginpmatrix a & lambda a \ b & lambda b endpmatrix
$$
for some $lambdain Bbbk$; and the determinant of this matrix vanishes.Remark. Of course, this works more generally. If you have a non-vanishing $ktimes k$ minor, then you can conclude that there are at least $k$ linearly independent rows, which means that your matrix has rank at least $k$.
- To answer your second question, assume that all $2times 2$ minors of the matrix $z=(z_ij)$ vanish. We may assume that the matrix is nonzero, otherwise we are done (it has rank less than one). Without loss of generality, assume that the entry $z_11$ is nonzero. Now, for any $i$ and $j$, you have that the matrix
$$
beginpmatrix z_11 & z_1j \ z_i1 & z_ij endpmatrix
$$
has vanishing determinant, so the second column must be a multiple of the first, i.e. there is some $lambdainBbbk$ with $z_1j=lambda z_11$ and $z_ij=lambda z_i1$. Since this holds for all $i$, it follows that the $j$-th column of $z$ is a scalar multiple of the first column. Since that holds for all $j$, all columns are scalar multiples of the first column. Hence, the matrix has rank one. - To see that any rank $k$ matrix factors like this, recall that the rank of a linear map is the dimension of its image. Hence, your $(m+1)times(n+1)$ matrix describes a linear map $f:Bbbk^n+1toBbbk^m+1$ such that $V:=f(Bbbk^n+1)congBbbk^k$. In other words, the map factors as $Bbbk^n+1to VtoBbbk^m+1$, where the first map is the co-restriction of $f$ and the second is just the inclusion. Choosing appropriate bases, you may describe this as $Bbbk^n+1toBbbk^ktoBbbk^m+1$, i.e. the first map is given by a $ktimes(n+1)$ matrix and the second one by an $(m+1)times k$ matrix.
answered Jul 27 at 22:25
Jesko Hüttenhain
9,96912253
One final thing why is the image of the Segre map contained in the 2x2 minors ?
– Newbie
Jul 28 at 3:27
well, because every point in the image has rank one and we just established that a matrix has rank one if and only if it is nonzero and all of its $2times2$ minors vanish.
– Jesko Hüttenhain
Jul 28 at 8:11
Yeah, that is good. Thanks.
– Newbie
Jul 28 at 15:49
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To answer your first question, assume for contradiction that some $2times2$ minor does not vanish. After reordering the rows and columns of your matrix (which does not change its rank), you may assume that it is the top left $2times 2$ submatrix. Now note that the second column can not be a scalar multiple of the first, because otherwise the top left $2times 2$ block would have the form
$$
beginpmatrix a & lambda a \ b & lambda b endpmatrix
$$
for some $lambdain Bbbk$; and the determinant of this matrix vanishes.Remark. Of course, this works more generally. If you have a non-vanishing $ktimes k$ minor, then you can conclude that there are at least $k$ linearly independent rows, which means that your matrix has rank at least $k$.
- To answer your second question, assume that all $2times 2$ minors of the matrix $z=(z_ij)$ vanish. We may assume that the matrix is nonzero, otherwise we are done (it has rank less than one). Without loss of generality, assume that the entry $z_11$ is nonzero. Now, for any $i$ and $j$, you have that the matrix
$$
beginpmatrix z_11 & z_1j \ z_i1 & z_ij endpmatrix
$$
has vanishing determinant, so the second column must be a multiple of the first, i.e. there is some $lambdainBbbk$ with $z_1j=lambda z_11$ and $z_ij=lambda z_i1$. Since this holds for all $i$, it follows that the $j$-th column of $z$ is a scalar multiple of the first column. Since that holds for all $j$, all columns are scalar multiples of the first column. Hence, the matrix has rank one. - To see that any rank $k$ matrix factors like this, recall that the rank of a linear map is the dimension of its image. Hence, your $(m+1)times(n+1)$ matrix describes a linear map $f:Bbbk^n+1toBbbk^m+1$ such that $V:=f(Bbbk^n+1)congBbbk^k$. In other words, the map factors as $Bbbk^n+1to VtoBbbk^m+1$, where the first map is the co-restriction of $f$ and the second is just the inclusion. Choosing appropriate bases, you may describe this as $Bbbk^n+1toBbbk^ktoBbbk^m+1$, i.e. the first map is given by a $ktimes(n+1)$ matrix and the second one by an $(m+1)times k$ matrix.
answered Jul 27 at 22:25
Jesko Hüttenhain
9,96912253
To answer your first question, assume for contradiction that some $2times2$ minor does not vanish. After reordering the rows and columns of your matrix (which does not change its rank), you may assume that it is the top left $2times 2$ submatrix. Now note that the second column can not be a scalar multiple of the first, because otherwise the top left $2times 2$ block would have the form
$$
beginpmatrix a & lambda a \ b & lambda b endpmatrix
$$
for some $lambdain Bbbk$; and the determinant of this matrix vanishes.Remark. Of course, this works more generally. If you have a non-vanishing $ktimes k$ minor, then you can conclude that there are at least $k$ linearly independent rows, which means that your matrix has rank at least $k$.
- To answer your second question, assume that all $2times 2$ minors of the matrix $z=(z_ij)$ vanish. We may assume that the matrix is nonzero, otherwise we are done (it has rank less than one). Without loss of generality, assume that the entry $z_11$ is nonzero. Now, for any $i$ and $j$, you have that the matrix
$$
beginpmatrix z_11 & z_1j \ z_i1 & z_ij endpmatrix
$$
has vanishing determinant, so the second column must be a multiple of the first, i.e. there is some $lambdainBbbk$ with $z_1j=lambda z_11$ and $z_ij=lambda z_i1$. Since this holds for all $i$, it follows that the $j$-th column of $z$ is a scalar multiple of the first column. Since that holds for all $j$, all columns are scalar multiples of the first column. Hence, the matrix has rank one. - To see that any rank $k$ matrix factors like this, recall that the rank of a linear map is the dimension of its image. Hence, your $(m+1)times(n+1)$ matrix describes a linear map $f:Bbbk^n+1toBbbk^m+1$ such that $V:=f(Bbbk^n+1)congBbbk^k$. In other words, the map factors as $Bbbk^n+1to VtoBbbk^m+1$, where the first map is the co-restriction of $f$ and the second is just the inclusion. Choosing appropriate bases, you may describe this as $Bbbk^n+1toBbbk^ktoBbbk^m+1$, i.e. the first map is given by a $ktimes(n+1)$ matrix and the second one by an $(m+1)times k$ matrix.
answered Jul 27 at 22:25
Jesko Hüttenhain
9,96912253
answered Jul 27 at 22:25
Jesko Hüttenhain
9,96912253
answered Jul 27 at 22:25
Jesko Hüttenhain
9,96912253
answered Jul 27 at 22:25
Jesko Hüttenhain
9,96912253
9,96912253
One final thing why is the image of the Segre map contained in the 2x2 minors ?
– Newbie
Jul 28 at 3:27
well, because every point in the image has rank one and we just established that a matrix has rank one if and only if it is nonzero and all of its $2times2$ minors vanish.
– Jesko Hüttenhain
Jul 28 at 8:11
Yeah, that is good. Thanks.
– Newbie
Jul 28 at 15:49
add a comment |Â
One final thing why is the image of the Segre map contained in the 2x2 minors ?
– Newbie
Jul 28 at 3:27
well, because every point in the image has rank one and we just established that a matrix has rank one if and only if it is nonzero and all of its $2times2$ minors vanish.
– Jesko Hüttenhain
Jul 28 at 8:11
Yeah, that is good. Thanks.
– Newbie
Jul 28 at 15:49
One final thing why is the image of the Segre map contained in the 2x2 minors ?
– Newbie
Jul 28 at 3:27
One final thing why is the image of the Segre map contained in the 2x2 minors ?
– Newbie
Jul 28 at 3:27
well, because every point in the image has rank one and we just established that a matrix has rank one if and only if it is nonzero and all of its $2times2$ minors vanish.
– Jesko Hüttenhain
Jul 28 at 8:11
well, because every point in the image has rank one and we just established that a matrix has rank one if and only if it is nonzero and all of its $2times2$ minors vanish.
– Jesko Hüttenhain
Jul 28 at 8:11
Yeah, that is good. Thanks.
– Newbie
Jul 28 at 15:49
Yeah, that is good. Thanks.
– Newbie
Jul 28 at 15:49
add a comment |Â
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