Max dimension of a subspace of singular $ntimes n$ matrices
Clash Royale CLAN TAG#URR8PPP
up vote
24
down vote
favorite
I am sure the answer to this is (kind of) well known. I've searched the web and the site for a proof and found nothing, and if this is a duplicate, I'm sorry.
The following question was given in a contest I took part. I had an approach but it didn't solve the problem.
Consider $V$ a linear subspace of the real vector space $mathcalM_n(BbbR)$ ($ntimes n$ real entries matrices) such that $V$ contains only singular matrices (i.e matrices with determinant equal to $0$). What is the maximal dimension of $V$?
A quick guess would be $n^2-n$ since if we consider $W$ the set of $ntimes n$ real matrices with last line equal to $0$ then this space has dimension $n^2-n$ and it is a linear space of singular matrices.
Now the only thing there is to prove is that if $V$ is a subspace of $mathcalM_n(BbbR)$ of dimension $k > n^2-n$ then $V$ contains a non-singular matrix. The official proof was unsatisfactory for me, because it was a combinatorial one, and seemed to have few things in common with linear algebra. I was hoping for a pure linear algebra proof.
My approach was to search for a permutation matrix in $V$, but I used some 'false theorem' in between, which I am ashamed to post here.
linear-algebra matrices
asked Sep 23 '11 at 5:28
Beni Bogosel
17.2k345110
add a comment |Â
up vote
24
down vote
favorite
I am sure the answer to this is (kind of) well known. I've searched the web and the site for a proof and found nothing, and if this is a duplicate, I'm sorry.
The following question was given in a contest I took part. I had an approach but it didn't solve the problem.
Consider $V$ a linear subspace of the real vector space $mathcalM_n(BbbR)$ ($ntimes n$ real entries matrices) such that $V$ contains only singular matrices (i.e matrices with determinant equal to $0$). What is the maximal dimension of $V$?
A quick guess would be $n^2-n$ since if we consider $W$ the set of $ntimes n$ real matrices with last line equal to $0$ then this space has dimension $n^2-n$ and it is a linear space of singular matrices.
Now the only thing there is to prove is that if $V$ is a subspace of $mathcalM_n(BbbR)$ of dimension $k > n^2-n$ then $V$ contains a non-singular matrix. The official proof was unsatisfactory for me, because it was a combinatorial one, and seemed to have few things in common with linear algebra. I was hoping for a pure linear algebra proof.
My approach was to search for a permutation matrix in $V$, but I used some 'false theorem' in between, which I am ashamed to post here.
linear-algebra matrices
asked Sep 23 '11 at 5:28
Beni Bogosel
17.2k345110
3
I'm a little curious about the combinatorial proof myself.
– anon
Sep 23 '11 at 5:42
I'll try and write it once I remember it. :)
– Beni Bogosel
Sep 23 '11 at 5:52
I am looking forward to it too. What's the basis for singular matrices then?
– user13838
Sep 23 '11 at 8:30
1
Some relevant papers: arxiv.org/pdf/1004.0298 ; citeseerx.ist.psu.edu/viewdoc/… ; math.technion.ac.il/~meshulam/eprints/maxrank.pdf. The first published proofs seem to be in J. Dieudonné, Sur une généralisation du groupe orthogonal à quatre variables, Arch. Math., 1 (1949) 282-287 and H. Flanders, On spaces of linear transformations with bounded rank, J. Lond. Math. Soc., 37 (1962) 10-16; I couldn't find either of those two with free access.
– joriki
Sep 23 '11 at 11:43
add a comment |Â
up vote
24
down vote
favorite
up vote
24
down vote
favorite
I am sure the answer to this is (kind of) well known. I've searched the web and the site for a proof and found nothing, and if this is a duplicate, I'm sorry.
The following question was given in a contest I took part. I had an approach but it didn't solve the problem.
Consider $V$ a linear subspace of the real vector space $mathcalM_n(BbbR)$ ($ntimes n$ real entries matrices) such that $V$ contains only singular matrices (i.e matrices with determinant equal to $0$). What is the maximal dimension of $V$?
A quick guess would be $n^2-n$ since if we consider $W$ the set of $ntimes n$ real matrices with last line equal to $0$ then this space has dimension $n^2-n$ and it is a linear space of singular matrices.
Now the only thing there is to prove is that if $V$ is a subspace of $mathcalM_n(BbbR)$ of dimension $k > n^2-n$ then $V$ contains a non-singular matrix. The official proof was unsatisfactory for me, because it was a combinatorial one, and seemed to have few things in common with linear algebra. I was hoping for a pure linear algebra proof.
My approach was to search for a permutation matrix in $V$, but I used some 'false theorem' in between, which I am ashamed to post here.
linear-algebra matrices
asked Sep 23 '11 at 5:28
Beni Bogosel
17.2k345110
I am sure the answer to this is (kind of) well known. I've searched the web and the site for a proof and found nothing, and if this is a duplicate, I'm sorry.
The following question was given in a contest I took part. I had an approach but it didn't solve the problem.
Consider $V$ a linear subspace of the real vector space $mathcalM_n(BbbR)$ ($ntimes n$ real entries matrices) such that $V$ contains only singular matrices (i.e matrices with determinant equal to $0$). What is the maximal dimension of $V$?
A quick guess would be $n^2-n$ since if we consider $W$ the set of $ntimes n$ real matrices with last line equal to $0$ then this space has dimension $n^2-n$ and it is a linear space of singular matrices.
Now the only thing there is to prove is that if $V$ is a subspace of $mathcalM_n(BbbR)$ of dimension $k > n^2-n$ then $V$ contains a non-singular matrix. The official proof was unsatisfactory for me, because it was a combinatorial one, and seemed to have few things in common with linear algebra. I was hoping for a pure linear algebra proof.
My approach was to search for a permutation matrix in $V$, but I used some 'false theorem' in between, which I am ashamed to post here.
linear-algebra matrices
asked Sep 23 '11 at 5:28
Beni Bogosel
17.2k345110
asked Sep 23 '11 at 5:28
Beni Bogosel
17.2k345110
asked Sep 23 '11 at 5:28
Beni Bogosel
17.2k345110
asked Sep 23 '11 at 5:28
Beni Bogosel
17.2k345110
17.2k345110
3
I'm a little curious about the combinatorial proof myself.
– anon
Sep 23 '11 at 5:42
I'll try and write it once I remember it. :)
– Beni Bogosel
Sep 23 '11 at 5:52
I am looking forward to it too. What's the basis for singular matrices then?
– user13838
Sep 23 '11 at 8:30
1
Some relevant papers: arxiv.org/pdf/1004.0298 ; citeseerx.ist.psu.edu/viewdoc/… ; math.technion.ac.il/~meshulam/eprints/maxrank.pdf. The first published proofs seem to be in J. Dieudonné, Sur une généralisation du groupe orthogonal à quatre variables, Arch. Math., 1 (1949) 282-287 and H. Flanders, On spaces of linear transformations with bounded rank, J. Lond. Math. Soc., 37 (1962) 10-16; I couldn't find either of those two with free access.
– joriki
Sep 23 '11 at 11:43
add a comment |Â
3
I'm a little curious about the combinatorial proof myself.
– anon
Sep 23 '11 at 5:42
I'll try and write it once I remember it. :)
– Beni Bogosel
Sep 23 '11 at 5:52
I am looking forward to it too. What's the basis for singular matrices then?
– user13838
Sep 23 '11 at 8:30
1
Some relevant papers: arxiv.org/pdf/1004.0298 ; citeseerx.ist.psu.edu/viewdoc/… ; math.technion.ac.il/~meshulam/eprints/maxrank.pdf. The first published proofs seem to be in J. Dieudonné, Sur une généralisation du groupe orthogonal à quatre variables, Arch. Math., 1 (1949) 282-287 and H. Flanders, On spaces of linear transformations with bounded rank, J. Lond. Math. Soc., 37 (1962) 10-16; I couldn't find either of those two with free access.
– joriki
Sep 23 '11 at 11:43
3
3
I'm a little curious about the combinatorial proof myself.
– anon
Sep 23 '11 at 5:42
I'm a little curious about the combinatorial proof myself.
– anon
Sep 23 '11 at 5:42
I'll try and write it once I remember it. :)
– Beni Bogosel
Sep 23 '11 at 5:52
I'll try and write it once I remember it. :)
– Beni Bogosel
Sep 23 '11 at 5:52
I am looking forward to it too. What's the basis for singular matrices then?
– user13838
Sep 23 '11 at 8:30
I am looking forward to it too. What's the basis for singular matrices then?
– user13838
Sep 23 '11 at 8:30
1
1
Some relevant papers: arxiv.org/pdf/1004.0298 ; citeseerx.ist.psu.edu/viewdoc/… ; math.technion.ac.il/~meshulam/eprints/maxrank.pdf. The first published proofs seem to be in J. Dieudonné, Sur une généralisation du groupe orthogonal à quatre variables, Arch. Math., 1 (1949) 282-287 and H. Flanders, On spaces of linear transformations with bounded rank, J. Lond. Math. Soc., 37 (1962) 10-16; I couldn't find either of those two with free access.
– joriki
Sep 23 '11 at 11:43
Some relevant papers: arxiv.org/pdf/1004.0298 ; citeseerx.ist.psu.edu/viewdoc/… ; math.technion.ac.il/~meshulam/eprints/maxrank.pdf. The first published proofs seem to be in J. Dieudonné, Sur une généralisation du groupe orthogonal à quatre variables, Arch. Math., 1 (1949) 282-287 and H. Flanders, On spaces of linear transformations with bounded rank, J. Lond. Math. Soc., 37 (1962) 10-16; I couldn't find either of those two with free access.
– joriki
Sep 23 '11 at 11:43
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
15
down vote
accepted
We can show more generally that if $mathcal M$ is a linear subspace of $mathcal M_n(mathbb R)$ such that all its element have a rank less than or equal to $p$, where $1leq p<n$, then the dimension of $mathcal M$ is less than or equal to $np$. To see that, consider the subspace $mathcal E:=leftbeginpmatrix0&B\^tB&Aendpmatrix, Ainmathcal M_n-p(mathbb R),Binmathcal M_p,n-p(mathbb R) right$. Its dimension is $p(n-p)+(n-p)^2=(n-p)(p+n-p)=n(n-p)$. Let $mathcal M$ a linear subspace of $mathcal M_n(mathbb R)$ such that $displaystylemax_Minmathcal Moperatornamerank(M)=p$. We can assume that this space contains the matrix $J:=beginpmatrixI_p&0\0&0endpmatrixinmathcal M_n(mathbb R)$. Indeed, if $M_0inmathcal M$ is such that $operatornamerankM_0=p$, we can find $P,Qinmathcal M_n(mathbb R)$ invertible matrices such that $J=PM_0Q$, and the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)=PMQ$ is a rank-preserving bijective linear map.
If we take $Minmathcal Mcap mathcal E$, then we can show, considering $M+lambda Jinmathcal M$, that $M=0$. Therefore, since
$$dim (mathcal M+mathcal E)=dim(mathcal M)+dim(mathcal E)leq dim(mathcal M_n(mathbb R))=n^2, $$
we have
$$dim (mathcal M)leq n^2-n(n-p)=np.$$
edited Aug 16 at 8:36
Jose Brox
1,9151920
answered Sep 23 '11 at 12:01
Davide Giraudo
121k15147250
I have one question: Why can we assume that $J in mathcalM$?
– Beni Bogosel
Sep 23 '11 at 18:30
@beni: we can find a matrix $M_0inmathcal M$ whose rank is $p$. Now, we take $Pinmathcal M_n(mathbb R)$ invertible such that $J=PM_0P^-1$, and we consider the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)= PMP^-1$. It's an isomorphism, hence $varphi(mathcal M)$ and $mathcal M$ have the same dimension.
– Davide Giraudo
Sep 23 '11 at 18:34
@DavideGiraudo: Ok. Now is clear.
– Beni Bogosel
Sep 23 '11 at 18:54
@DavideGiraudo How can we prove easily that $rank(M+λJ)leq p$ for all $lambda$ implies $M=0$? I see a proof computing the $p+1$ minors via the Schur complement, is there some cleaner argument?
– Jose Brox
Aug 16 at 8:41
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
We can show more generally that if $mathcal M$ is a linear subspace of $mathcal M_n(mathbb R)$ such that all its element have a rank less than or equal to $p$, where $1leq p<n$, then the dimension of $mathcal M$ is less than or equal to $np$. To see that, consider the subspace $mathcal E:=leftbeginpmatrix0&B\^tB&Aendpmatrix, Ainmathcal M_n-p(mathbb R),Binmathcal M_p,n-p(mathbb R) right$. Its dimension is $p(n-p)+(n-p)^2=(n-p)(p+n-p)=n(n-p)$. Let $mathcal M$ a linear subspace of $mathcal M_n(mathbb R)$ such that $displaystylemax_Minmathcal Moperatornamerank(M)=p$. We can assume that this space contains the matrix $J:=beginpmatrixI_p&0\0&0endpmatrixinmathcal M_n(mathbb R)$. Indeed, if $M_0inmathcal M$ is such that $operatornamerankM_0=p$, we can find $P,Qinmathcal M_n(mathbb R)$ invertible matrices such that $J=PM_0Q$, and the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)=PMQ$ is a rank-preserving bijective linear map.
If we take $Minmathcal Mcap mathcal E$, then we can show, considering $M+lambda Jinmathcal M$, that $M=0$. Therefore, since
$$dim (mathcal M+mathcal E)=dim(mathcal M)+dim(mathcal E)leq dim(mathcal M_n(mathbb R))=n^2, $$
we have
$$dim (mathcal M)leq n^2-n(n-p)=np.$$
edited Aug 16 at 8:36
Jose Brox
1,9151920
answered Sep 23 '11 at 12:01
Davide Giraudo
121k15147250
I have one question: Why can we assume that $J in mathcalM$?
– Beni Bogosel
Sep 23 '11 at 18:30
@beni: we can find a matrix $M_0inmathcal M$ whose rank is $p$. Now, we take $Pinmathcal M_n(mathbb R)$ invertible such that $J=PM_0P^-1$, and we consider the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)= PMP^-1$. It's an isomorphism, hence $varphi(mathcal M)$ and $mathcal M$ have the same dimension.
– Davide Giraudo
Sep 23 '11 at 18:34
@DavideGiraudo: Ok. Now is clear.
– Beni Bogosel
Sep 23 '11 at 18:54
@DavideGiraudo How can we prove easily that $rank(M+λJ)leq p$ for all $lambda$ implies $M=0$? I see a proof computing the $p+1$ minors via the Schur complement, is there some cleaner argument?
– Jose Brox
Aug 16 at 8:41
add a comment |Â
up vote
15
down vote
accepted
We can show more generally that if $mathcal M$ is a linear subspace of $mathcal M_n(mathbb R)$ such that all its element have a rank less than or equal to $p$, where $1leq p<n$, then the dimension of $mathcal M$ is less than or equal to $np$. To see that, consider the subspace $mathcal E:=leftbeginpmatrix0&B\^tB&Aendpmatrix, Ainmathcal M_n-p(mathbb R),Binmathcal M_p,n-p(mathbb R) right$. Its dimension is $p(n-p)+(n-p)^2=(n-p)(p+n-p)=n(n-p)$. Let $mathcal M$ a linear subspace of $mathcal M_n(mathbb R)$ such that $displaystylemax_Minmathcal Moperatornamerank(M)=p$. We can assume that this space contains the matrix $J:=beginpmatrixI_p&0\0&0endpmatrixinmathcal M_n(mathbb R)$. Indeed, if $M_0inmathcal M$ is such that $operatornamerankM_0=p$, we can find $P,Qinmathcal M_n(mathbb R)$ invertible matrices such that $J=PM_0Q$, and the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)=PMQ$ is a rank-preserving bijective linear map.
If we take $Minmathcal Mcap mathcal E$, then we can show, considering $M+lambda Jinmathcal M$, that $M=0$. Therefore, since
$$dim (mathcal M+mathcal E)=dim(mathcal M)+dim(mathcal E)leq dim(mathcal M_n(mathbb R))=n^2, $$
we have
$$dim (mathcal M)leq n^2-n(n-p)=np.$$
edited Aug 16 at 8:36
Jose Brox
1,9151920
answered Sep 23 '11 at 12:01
Davide Giraudo
121k15147250
I have one question: Why can we assume that $J in mathcalM$?
– Beni Bogosel
Sep 23 '11 at 18:30
@beni: we can find a matrix $M_0inmathcal M$ whose rank is $p$. Now, we take $Pinmathcal M_n(mathbb R)$ invertible such that $J=PM_0P^-1$, and we consider the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)= PMP^-1$. It's an isomorphism, hence $varphi(mathcal M)$ and $mathcal M$ have the same dimension.
– Davide Giraudo
Sep 23 '11 at 18:34
@DavideGiraudo: Ok. Now is clear.
– Beni Bogosel
Sep 23 '11 at 18:54
@DavideGiraudo How can we prove easily that $rank(M+λJ)leq p$ for all $lambda$ implies $M=0$? I see a proof computing the $p+1$ minors via the Schur complement, is there some cleaner argument?
– Jose Brox
Aug 16 at 8:41
add a comment |Â
up vote
15
down vote
accepted
up vote
15
down vote
accepted
We can show more generally that if $mathcal M$ is a linear subspace of $mathcal M_n(mathbb R)$ such that all its element have a rank less than or equal to $p$, where $1leq p<n$, then the dimension of $mathcal M$ is less than or equal to $np$. To see that, consider the subspace $mathcal E:=leftbeginpmatrix0&B\^tB&Aendpmatrix, Ainmathcal M_n-p(mathbb R),Binmathcal M_p,n-p(mathbb R) right$. Its dimension is $p(n-p)+(n-p)^2=(n-p)(p+n-p)=n(n-p)$. Let $mathcal M$ a linear subspace of $mathcal M_n(mathbb R)$ such that $displaystylemax_Minmathcal Moperatornamerank(M)=p$. We can assume that this space contains the matrix $J:=beginpmatrixI_p&0\0&0endpmatrixinmathcal M_n(mathbb R)$. Indeed, if $M_0inmathcal M$ is such that $operatornamerankM_0=p$, we can find $P,Qinmathcal M_n(mathbb R)$ invertible matrices such that $J=PM_0Q$, and the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)=PMQ$ is a rank-preserving bijective linear map.
If we take $Minmathcal Mcap mathcal E$, then we can show, considering $M+lambda Jinmathcal M$, that $M=0$. Therefore, since
$$dim (mathcal M+mathcal E)=dim(mathcal M)+dim(mathcal E)leq dim(mathcal M_n(mathbb R))=n^2, $$
we have
$$dim (mathcal M)leq n^2-n(n-p)=np.$$
edited Aug 16 at 8:36
Jose Brox
1,9151920
answered Sep 23 '11 at 12:01
Davide Giraudo
121k15147250
We can show more generally that if $mathcal M$ is a linear subspace of $mathcal M_n(mathbb R)$ such that all its element have a rank less than or equal to $p$, where $1leq p<n$, then the dimension of $mathcal M$ is less than or equal to $np$. To see that, consider the subspace $mathcal E:=leftbeginpmatrix0&B\^tB&Aendpmatrix, Ainmathcal M_n-p(mathbb R),Binmathcal M_p,n-p(mathbb R) right$. Its dimension is $p(n-p)+(n-p)^2=(n-p)(p+n-p)=n(n-p)$. Let $mathcal M$ a linear subspace of $mathcal M_n(mathbb R)$ such that $displaystylemax_Minmathcal Moperatornamerank(M)=p$. We can assume that this space contains the matrix $J:=beginpmatrixI_p&0\0&0endpmatrixinmathcal M_n(mathbb R)$. Indeed, if $M_0inmathcal M$ is such that $operatornamerankM_0=p$, we can find $P,Qinmathcal M_n(mathbb R)$ invertible matrices such that $J=PM_0Q$, and the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)=PMQ$ is a rank-preserving bijective linear map.
If we take $Minmathcal Mcap mathcal E$, then we can show, considering $M+lambda Jinmathcal M$, that $M=0$. Therefore, since
$$dim (mathcal M+mathcal E)=dim(mathcal M)+dim(mathcal E)leq dim(mathcal M_n(mathbb R))=n^2, $$
we have
$$dim (mathcal M)leq n^2-n(n-p)=np.$$
edited Aug 16 at 8:36
Jose Brox
1,9151920
answered Sep 23 '11 at 12:01
Davide Giraudo
121k15147250
edited Aug 16 at 8:36
Jose Brox
1,9151920
edited Aug 16 at 8:36
Jose Brox
1,9151920
edited Aug 16 at 8:36
Jose Brox
1,9151920
1,9151920
answered Sep 23 '11 at 12:01
Davide Giraudo
121k15147250
answered Sep 23 '11 at 12:01
Davide Giraudo
121k15147250
answered Sep 23 '11 at 12:01
Davide Giraudo
121k15147250
121k15147250
I have one question: Why can we assume that $J in mathcalM$?
– Beni Bogosel
Sep 23 '11 at 18:30
@beni: we can find a matrix $M_0inmathcal M$ whose rank is $p$. Now, we take $Pinmathcal M_n(mathbb R)$ invertible such that $J=PM_0P^-1$, and we consider the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)= PMP^-1$. It's an isomorphism, hence $varphi(mathcal M)$ and $mathcal M$ have the same dimension.
– Davide Giraudo
Sep 23 '11 at 18:34
@DavideGiraudo: Ok. Now is clear.
– Beni Bogosel
Sep 23 '11 at 18:54
@DavideGiraudo How can we prove easily that $rank(M+λJ)leq p$ for all $lambda$ implies $M=0$? I see a proof computing the $p+1$ minors via the Schur complement, is there some cleaner argument?
– Jose Brox
Aug 16 at 8:41
add a comment |Â
I have one question: Why can we assume that $J in mathcalM$?
– Beni Bogosel
Sep 23 '11 at 18:30
@beni: we can find a matrix $M_0inmathcal M$ whose rank is $p$. Now, we take $Pinmathcal M_n(mathbb R)$ invertible such that $J=PM_0P^-1$, and we consider the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)= PMP^-1$. It's an isomorphism, hence $varphi(mathcal M)$ and $mathcal M$ have the same dimension.
– Davide Giraudo
Sep 23 '11 at 18:34
@DavideGiraudo: Ok. Now is clear.
– Beni Bogosel
Sep 23 '11 at 18:54
@DavideGiraudo How can we prove easily that $rank(M+λJ)leq p$ for all $lambda$ implies $M=0$? I see a proof computing the $p+1$ minors via the Schur complement, is there some cleaner argument?
– Jose Brox
Aug 16 at 8:41
I have one question: Why can we assume that $J in mathcalM$?
– Beni Bogosel
Sep 23 '11 at 18:30
I have one question: Why can we assume that $J in mathcalM$?
– Beni Bogosel
Sep 23 '11 at 18:30
@beni: we can find a matrix $M_0inmathcal M$ whose rank is $p$. Now, we take $Pinmathcal M_n(mathbb R)$ invertible such that $J=PM_0P^-1$, and we consider the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)= PMP^-1$. It's an isomorphism, hence $varphi(mathcal M)$ and $mathcal M$ have the same dimension.
– Davide Giraudo
Sep 23 '11 at 18:34
@beni: we can find a matrix $M_0inmathcal M$ whose rank is $p$. Now, we take $Pinmathcal M_n(mathbb R)$ invertible such that $J=PM_0P^-1$, and we consider the map $varphicolon mathcal Mtovarphi(mathcal M)$ defined by $varphi(M)= PMP^-1$. It's an isomorphism, hence $varphi(mathcal M)$ and $mathcal M$ have the same dimension.
– Davide Giraudo
Sep 23 '11 at 18:34
@DavideGiraudo: Ok. Now is clear.
– Beni Bogosel
Sep 23 '11 at 18:54
@DavideGiraudo: Ok. Now is clear.
– Beni Bogosel
Sep 23 '11 at 18:54
@DavideGiraudo How can we prove easily that $rank(M+λJ)leq p$ for all $lambda$ implies $M=0$? I see a proof computing the $p+1$ minors via the Schur complement, is there some cleaner argument?
– Jose Brox
Aug 16 at 8:41
@DavideGiraudo How can we prove easily that $rank(M+λJ)leq p$ for all $lambda$ implies $M=0$? I see a proof computing the $p+1$ minors via the Schur complement, is there some cleaner argument?
– Jose Brox
Aug 16 at 8:41
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f66877%2fmax-dimension-of-a-subspace-of-singular-n-times-n-matrices%23new-answer', 'question_page');
);
Post as a guest
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
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
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
3
I'm a little curious about the combinatorial proof myself.
– anon
Sep 23 '11 at 5:42
I'll try and write it once I remember it. :)
– Beni Bogosel
Sep 23 '11 at 5:52
I am looking forward to it too. What's the basis for singular matrices then?
– user13838
Sep 23 '11 at 8:30
1
Some relevant papers: arxiv.org/pdf/1004.0298 ; citeseerx.ist.psu.edu/viewdoc/… ; math.technion.ac.il/~meshulam/eprints/maxrank.pdf. The first published proofs seem to be in J. Dieudonné, Sur une généralisation du groupe orthogonal à quatre variables, Arch. Math., 1 (1949) 282-287 and H. Flanders, On spaces of linear transformations with bounded rank, J. Lond. Math. Soc., 37 (1962) 10-16; I couldn't find either of those two with free access.
– joriki
Sep 23 '11 at 11:43