Mixing
Decomposition of Determinant of Sub-Matrices of a Matrix
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Consider an $n times n$ matrix $bf A$ over a field. Let $bf A$ is constructed by the product of $n times n$ matrices $B_i$, for $1leq i leq m$ which means
$$
bf A=prod_i=1^m, bf B_i, .
$$
From Linear Algebra we know that the relation between determinant of $bf A$ and $bf B_i$'s are as follows:
$$
det(bf A)=prod_i=1^m, det(bf B_i), .
$$
Suppose that an $k times k$ sub-matrix of $bf A$ for $1leq k leq n$ is denoted with $bf A^(k)$.
My question:
Is there a condition that can be used to check $bf A^(k)$ is singular or non-singular matrix based on the determinants of $bf B_i$'s or sub-matrices of $bf B_i$'s.
Thanks for your assistance in advance.
linear-algebra matrices determinant matrix-decomposition
asked Jul 15 at 20:44
user0410
1287
 |Â
show 2 more comments
up vote
0
down vote
favorite
Consider an $n times n$ matrix $bf A$ over a field. Let $bf A$ is constructed by the product of $n times n$ matrices $B_i$, for $1leq i leq m$ which means
$$
bf A=prod_i=1^m, bf B_i, .
$$
From Linear Algebra we know that the relation between determinant of $bf A$ and $bf B_i$'s are as follows:
$$
det(bf A)=prod_i=1^m, det(bf B_i), .
$$
Suppose that an $k times k$ sub-matrix of $bf A$ for $1leq k leq n$ is denoted with $bf A^(k)$.
My question:
Is there a condition that can be used to check $bf A^(k)$ is singular or non-singular matrix based on the determinants of $bf B_i$'s or sub-matrices of $bf B_i$'s.
Thanks for your assistance in advance.
linear-algebra matrices determinant matrix-decomposition
asked Jul 15 at 20:44
user0410
1287
Every entry of $B_i$ is a submatrix of $B_i$.
– user1551
Jul 16 at 8:47
@user1551 I can not understand what did you mean by your comment. Is it possible to ask you to explain more. Thanks
– user0410
Jul 16 at 8:50
I mean your question is vague. $det A^(k)$ is a polynomial in the entries of $B_1,B_2,ldots,B_m$. Since every entry of $B_i$ is a submatrix of $B_i$, the answer to your question is undoubtedly yes, because "$det A^(k)=0$" is already such a condition.
– user1551
Jul 16 at 9:14
@user1551 Maybe the first comment of this question in mathoverflow be helpful to make clean my question.
– user0410
Jul 16 at 9:21
1
In general no, but even in the case $k=1$ and $m=2$ I'd have to multiply it out to decide. If you know a better way in that case then Gauss Bonnet tells you how to do the general case.
– ancientmathematician
Jul 17 at 6:35
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider an $n times n$ matrix $bf A$ over a field. Let $bf A$ is constructed by the product of $n times n$ matrices $B_i$, for $1leq i leq m$ which means
$$
bf A=prod_i=1^m, bf B_i, .
$$
From Linear Algebra we know that the relation between determinant of $bf A$ and $bf B_i$'s are as follows:
$$
det(bf A)=prod_i=1^m, det(bf B_i), .
$$
Suppose that an $k times k$ sub-matrix of $bf A$ for $1leq k leq n$ is denoted with $bf A^(k)$.
My question:
Is there a condition that can be used to check $bf A^(k)$ is singular or non-singular matrix based on the determinants of $bf B_i$'s or sub-matrices of $bf B_i$'s.
Thanks for your assistance in advance.
linear-algebra matrices determinant matrix-decomposition
asked Jul 15 at 20:44
user0410
1287
Consider an $n times n$ matrix $bf A$ over a field. Let $bf A$ is constructed by the product of $n times n$ matrices $B_i$, for $1leq i leq m$ which means
$$
bf A=prod_i=1^m, bf B_i, .
$$
From Linear Algebra we know that the relation between determinant of $bf A$ and $bf B_i$'s are as follows:
$$
det(bf A)=prod_i=1^m, det(bf B_i), .
$$
Suppose that an $k times k$ sub-matrix of $bf A$ for $1leq k leq n$ is denoted with $bf A^(k)$.
My question:
Is there a condition that can be used to check $bf A^(k)$ is singular or non-singular matrix based on the determinants of $bf B_i$'s or sub-matrices of $bf B_i$'s.
Thanks for your assistance in advance.
linear-algebra matrices determinant matrix-decomposition
asked Jul 15 at 20:44
user0410
1287
edited Jul 16 at 7:46
asked Jul 15 at 20:44
user0410
1287
asked Jul 15 at 20:44
user0410
1287
asked Jul 15 at 20:44
user0410
1287
1287
Every entry of $B_i$ is a submatrix of $B_i$.
– user1551
Jul 16 at 8:47
@user1551 I can not understand what did you mean by your comment. Is it possible to ask you to explain more. Thanks
– user0410
Jul 16 at 8:50
I mean your question is vague. $det A^(k)$ is a polynomial in the entries of $B_1,B_2,ldots,B_m$. Since every entry of $B_i$ is a submatrix of $B_i$, the answer to your question is undoubtedly yes, because "$det A^(k)=0$" is already such a condition.
– user1551
Jul 16 at 9:14
@user1551 Maybe the first comment of this question in mathoverflow be helpful to make clean my question.
– user0410
Jul 16 at 9:21
1
In general no, but even in the case $k=1$ and $m=2$ I'd have to multiply it out to decide. If you know a better way in that case then Gauss Bonnet tells you how to do the general case.
– ancientmathematician
Jul 17 at 6:35
 |Â
show 2 more comments
Every entry of $B_i$ is a submatrix of $B_i$.
– user1551
Jul 16 at 8:47
@user1551 I can not understand what did you mean by your comment. Is it possible to ask you to explain more. Thanks
– user0410
Jul 16 at 8:50
I mean your question is vague. $det A^(k)$ is a polynomial in the entries of $B_1,B_2,ldots,B_m$. Since every entry of $B_i$ is a submatrix of $B_i$, the answer to your question is undoubtedly yes, because "$det A^(k)=0$" is already such a condition.
– user1551
Jul 16 at 9:14
@user1551 Maybe the first comment of this question in mathoverflow be helpful to make clean my question.
– user0410
Jul 16 at 9:21
1
In general no, but even in the case $k=1$ and $m=2$ I'd have to multiply it out to decide. If you know a better way in that case then Gauss Bonnet tells you how to do the general case.
– ancientmathematician
Jul 17 at 6:35
Every entry of $B_i$ is a submatrix of $B_i$.
– user1551
Jul 16 at 8:47
Every entry of $B_i$ is a submatrix of $B_i$.
– user1551
Jul 16 at 8:47
@user1551 I can not understand what did you mean by your comment. Is it possible to ask you to explain more. Thanks
– user0410
Jul 16 at 8:50
@user1551 I can not understand what did you mean by your comment. Is it possible to ask you to explain more. Thanks
– user0410
Jul 16 at 8:50
I mean your question is vague. $det A^(k)$ is a polynomial in the entries of $B_1,B_2,ldots,B_m$. Since every entry of $B_i$ is a submatrix of $B_i$, the answer to your question is undoubtedly yes, because "$det A^(k)=0$" is already such a condition.
– user1551
Jul 16 at 9:14
I mean your question is vague. $det A^(k)$ is a polynomial in the entries of $B_1,B_2,ldots,B_m$. Since every entry of $B_i$ is a submatrix of $B_i$, the answer to your question is undoubtedly yes, because "$det A^(k)=0$" is already such a condition.
– user1551
Jul 16 at 9:14
@user1551 Maybe the first comment of this question in mathoverflow be helpful to make clean my question.
– user0410
Jul 16 at 9:21
@user1551 Maybe the first comment of this question in mathoverflow be helpful to make clean my question.
– user0410
Jul 16 at 9:21
1
1
In general no, but even in the case $k=1$ and $m=2$ I'd have to multiply it out to decide. If you know a better way in that case then Gauss Bonnet tells you how to do the general case.
– ancientmathematician
Jul 17 at 6:35
In general no, but even in the case $k=1$ and $m=2$ I'd have to multiply it out to decide. If you know a better way in that case then Gauss Bonnet tells you how to do the general case.
– ancientmathematician
Jul 17 at 6:35
 |Â
show 2 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2852830%2fdecomposition-of-determinant-of-sub-matrices-of-a-matrix%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
Every entry of $B_i$ is a submatrix of $B_i$.
– user1551
Jul 16 at 8:47
@user1551 I can not understand what did you mean by your comment. Is it possible to ask you to explain more. Thanks
– user0410
Jul 16 at 8:50
I mean your question is vague. $det A^(k)$ is a polynomial in the entries of $B_1,B_2,ldots,B_m$. Since every entry of $B_i$ is a submatrix of $B_i$, the answer to your question is undoubtedly yes, because "$det A^(k)=0$" is already such a condition.
– user1551
Jul 16 at 9:14
@user1551 Maybe the first comment of this question in mathoverflow be helpful to make clean my question.
– user0410
Jul 16 at 9:21
1
In general no, but even in the case $k=1$ and $m=2$ I'd have to multiply it out to decide. If you know a better way in that case then Gauss Bonnet tells you how to do the general case.
– ancientmathematician
Jul 17 at 6:35