Mixing
Determinant of block matrix with singular blocks on the diagonal
$begingroup$
Let $A$ and $D$ be square matrices, and let $B$ and $C$ be matrices of valid shapes to allow the formation of
$$
M =
begin{bmatrix}
A & B \
C & D
end{bmatrix}.
$$
If $det{A}neq0$, we may use the Schur complement to express $det{M}$ in terms of its constituent blocks as
$$
det{M} = det{A}cdotdet(D-CA^{-1}B),
$$
and if $det{D}neq0$ we have in a similar fashion that
$$
det{M} = det(A-BD^{-1}C)cdotdet{D}.
$$
My question: Does there exist a similar formula expressing $det{M}$ in terms of its constituent blocks, that is valid in case $det{A}=det{D}=0$?
matrices determinant block-matrices schur-complement
edited Mar 7 at 22:13
Rodrigo de Azevedo
13.2k41960
asked Sep 11 '18 at 11:26
Mårten WMårten W
2,56241837
$endgroup$
add a comment |
$begingroup$
Let $A$ and $D$ be square matrices, and let $B$ and $C$ be matrices of valid shapes to allow the formation of
$$
M =
begin{bmatrix}
A & B \
C & D
end{bmatrix}.
$$
If $det{A}neq0$, we may use the Schur complement to express $det{M}$ in terms of its constituent blocks as
$$
det{M} = det{A}cdotdet(D-CA^{-1}B),
$$
and if $det{D}neq0$ we have in a similar fashion that
$$
det{M} = det(A-BD^{-1}C)cdotdet{D}.
$$
My question: Does there exist a similar formula expressing $det{M}$ in terms of its constituent blocks, that is valid in case $det{A}=det{D}=0$?
matrices determinant block-matrices schur-complement
edited Mar 7 at 22:13
Rodrigo de Azevedo
13.2k41960
asked Sep 11 '18 at 11:26
Mårten WMårten W
2,56241837
$endgroup$
add a comment |
$begingroup$
Let $A$ and $D$ be square matrices, and let $B$ and $C$ be matrices of valid shapes to allow the formation of
$$
M =
begin{bmatrix}
A & B \
C & D
end{bmatrix}.
$$
If $det{A}neq0$, we may use the Schur complement to express $det{M}$ in terms of its constituent blocks as
$$
det{M} = det{A}cdotdet(D-CA^{-1}B),
$$
and if $det{D}neq0$ we have in a similar fashion that
$$
det{M} = det(A-BD^{-1}C)cdotdet{D}.
$$
My question: Does there exist a similar formula expressing $det{M}$ in terms of its constituent blocks, that is valid in case $det{A}=det{D}=0$?
matrices determinant block-matrices schur-complement
edited Mar 7 at 22:13
Rodrigo de Azevedo
13.2k41960
asked Sep 11 '18 at 11:26
Mårten WMårten W
2,56241837
$endgroup$
Let $A$ and $D$ be square matrices, and let $B$ and $C$ be matrices of valid shapes to allow the formation of
$$
M =
begin{bmatrix}
A & B \
C & D
end{bmatrix}.
$$
If $det{A}neq0$, we may use the Schur complement to express $det{M}$ in terms of its constituent blocks as
$$
det{M} = det{A}cdotdet(D-CA^{-1}B),
$$
and if $det{D}neq0$ we have in a similar fashion that
$$
det{M} = det(A-BD^{-1}C)cdotdet{D}.
$$
My question: Does there exist a similar formula expressing $det{M}$ in terms of its constituent blocks, that is valid in case $det{A}=det{D}=0$?
matrices determinant block-matrices schur-complement
matrices determinant block-matrices schur-complement
edited Mar 7 at 22:13
Rodrigo de Azevedo
13.2k41960
asked Sep 11 '18 at 11:26
Mårten WMårten W
2,56241837
edited Mar 7 at 22:13
Rodrigo de Azevedo
13.2k41960
asked Sep 11 '18 at 11:26
Mårten WMårten W
2,56241837
edited Mar 7 at 22:13
Rodrigo de Azevedo
13.2k41960
edited Mar 7 at 22:13
Rodrigo de Azevedo
13.2k41960
edited Mar 7 at 22:13
Rodrigo de Azevedo
13.2k41960
13.2k41960
asked Sep 11 '18 at 11:26
Mårten WMårten W
2,56241837
asked Sep 11 '18 at 11:26
Mårten WMårten W
2,56241837
asked Sep 11 '18 at 11:26
Mårten WMårten W
2,56241837
2,56241837
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Without loss of generality, let $A$ be $m$-by-$m$ and $D$ be $n$-by-$n$, with $mge n$. Consider
$$
f(t)=detleft(
begin{array}{cc}
A+tI_m&B\
C&D
end{array}
right).
$$
Obviously,
$f(t)$ is a polynomial of $t$, for which it is analytic on $mathbb{R}$;
$f(0)$ returns the desired determinant;
$g(t)=detleft(A+tI_mright)$ is also a polynomial of $t$, for which it has isolated zeros;
$g(0)=0$ provided that $A$ is singular, yet since $t=0$ is an isolated zero of $g(t)$, $g(t)ne 0$ for all $tinleft(-delta,deltaright)setminusleft{0right}$ for some $delta>0$.
Now, since $g(t)ne 0$ on some $left(-delta,deltaright)setminusleft{0right}$, it follows that $A+tI_m$ is invertible on this domain. Therefore,
$$
f(t)=detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright).
$$
Consequently, the continuity of $f(t)$ yields
$$
f(0)=lim_{tto 0}left(detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright)right).
$$
Now, let us focus on two distinct cases. First, if $A=O_m$, i.e., $A$ is a zero matrix of order $m$. In this case, the above formula gives
begin{align}
f(0)&=lim_{tto 0}left(detleft(tI_mright)detleft(D-Cleft(tI_mright)^{-1}Bright)right)\
&=lim_{tto 0}left(t^mdetleft(D-frac{1}{t}CBright)right)\
&=lim_{tto 0}left(t^mdetleft(frac{1}{t}left(tD-CBright)right)right)\
&=lim_{tto 0}left(t^{m-n}detleft(tD-CBright)right).
end{align}
Recall that $mge n$. We therefore obtain
- If $m>n$, it is obvious that $f(0)=0$;
- If $m=n$, it follows that $f(0)=detleft(-CBright)=left(-1right)^ndetleft(CBright)$.
Second, consider $Ane O_m$. This case is more complicated, and there is no elegant form for $f(0)$ with only $A$, $B$, $C$, and $D$ involved. However, since $Ane O_m$, we may perform some elementary operations of the second type, i.e., row- and column-switching transformations, such that after the operations, we obtain
$$
f(0)=detleft(
begin{array}{cc}
A'&B'\
C'&D'
end{array}
right),
$$
where, e.g., $A'$ results from $A$ by switching $A$'s rows and columns, such that $A''$, the $k$-by-$k$ square matrix made up of the first $k$ rows and columns of $A'$, is invertible. The existence of such an $A''$ is guaranteed by the fact that $Ane O_m$. In this way,
$$
f(0)=detleft(
begin{array}{cc}
A''&B''\
C''&D''
end{array}
right).
$$
Thanks to the invertibility of $A''$,
$$
f(0)=detleft(A''right)detleft(D''-C''left(A''right)^{-1}B''right).
$$
This result is much less elegant. $A''$ is only part of $A$. $B''$ contains part of both $A$ and $B$, and so does $C''$. $D''$ contains the whole $D$, and part of $A$, $B$, and $C$. Besides, there are switches of rows and columns as well.
answered Jan 29 at 3:16
hypernovahypernova
4,989414
$endgroup$
$begingroup$
While this answer did not provide a formula for the general case, it did provide a very nice idea which might be possible to use in some important special cases. Congratulations for the first (smaller) bonus!
$endgroup$
– Mårten W
Feb 4 at 22:50
$begingroup$
@MårtenW: Wow! Thank you for your acknowledgement and your bonus! It is so beyond the generous of you! I hope this partial answer would be somewhat helpful for you. By the way, I think it quite promising that its results for the special cases would help clarify the sign ambiguity in polfosol's answer. Thanks :-)
$endgroup$
– hypernova
Feb 5 at 0:49
add a comment |
$begingroup$
For $M=begin{bmatrix}A & B\C & Dend{bmatrix}$ let $N=M^TM$. So that
$$N:=begin{bmatrix}E & F\F^T & Gend{bmatrix}$$
where
$$begin{align}
E&=A^T A+C^T C\F&=A^T B+C^T D\G&=B^T B+D^T D
end{align}$$
Now, if $M$ is non-singular then $N$ is a positive definite matrix. Hence according to this other post, $E$ is a positive definite (i.e. non-singular) block matrix.
Since $E$ is non-singular, we can use Schur complement to obtain $det N$.
$$det{N} = det{E}cdotdet(G-F^T E^{-1}F)=(det M)^2$$
The only remaining part in this case is to determine the sign of $det M$, which apparently, there is no easy way to do that in general.
We might be able to get the sign of $det M$ by a trick similar to the one discussed at the end of @hypernova's answer. I'll update this answer if anything comes up.
answered Jan 29 at 17:55
polfosolpolfosol
5,93931945
$endgroup$
1
$begingroup$
Nice answer (+1) and nicer nickname by the way :)
$endgroup$
– Mostafa Ayaz
Feb 1 at 21:44
1
$begingroup$
Except for leaving that sign ambiguity, this answer does precisely what I was asking for, and for that reason I'm going to award the bigger bounty as soon as the system lets me.
$endgroup$
– Mårten W
Feb 4 at 22:47
add a comment |
$begingroup$
Hint:
$$
M =
begin{bmatrix}
A & B \
C & D
end{bmatrix}.
=
begin{bmatrix}
0 & I \
I & 0
end{bmatrix}.
begin{bmatrix}
C & D \
A & B
end{bmatrix}
$$
Where $I$ is the idenity matrix of appropriate size.
answered Jan 29 at 3:47
zimbra314zimbra314
630312
$endgroup$
2
$begingroup$
I thought about the same hint, but C may not even be square, so this is a dead end in general.
$endgroup$
– Dirk
Jan 29 at 6:33
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Without loss of generality, let $A$ be $m$-by-$m$ and $D$ be $n$-by-$n$, with $mge n$. Consider
$$
f(t)=detleft(
begin{array}{cc}
A+tI_m&B\
C&D
end{array}
right).
$$
Obviously,
$f(t)$ is a polynomial of $t$, for which it is analytic on $mathbb{R}$;
$f(0)$ returns the desired determinant;
$g(t)=detleft(A+tI_mright)$ is also a polynomial of $t$, for which it has isolated zeros;
$g(0)=0$ provided that $A$ is singular, yet since $t=0$ is an isolated zero of $g(t)$, $g(t)ne 0$ for all $tinleft(-delta,deltaright)setminusleft{0right}$ for some $delta>0$.
Now, since $g(t)ne 0$ on some $left(-delta,deltaright)setminusleft{0right}$, it follows that $A+tI_m$ is invertible on this domain. Therefore,
$$
f(t)=detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright).
$$
Consequently, the continuity of $f(t)$ yields
$$
f(0)=lim_{tto 0}left(detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright)right).
$$
Now, let us focus on two distinct cases. First, if $A=O_m$, i.e., $A$ is a zero matrix of order $m$. In this case, the above formula gives
begin{align}
f(0)&=lim_{tto 0}left(detleft(tI_mright)detleft(D-Cleft(tI_mright)^{-1}Bright)right)\
&=lim_{tto 0}left(t^mdetleft(D-frac{1}{t}CBright)right)\
&=lim_{tto 0}left(t^mdetleft(frac{1}{t}left(tD-CBright)right)right)\
&=lim_{tto 0}left(t^{m-n}detleft(tD-CBright)right).
end{align}
Recall that $mge n$. We therefore obtain
- If $m>n$, it is obvious that $f(0)=0$;
- If $m=n$, it follows that $f(0)=detleft(-CBright)=left(-1right)^ndetleft(CBright)$.
Second, consider $Ane O_m$. This case is more complicated, and there is no elegant form for $f(0)$ with only $A$, $B$, $C$, and $D$ involved. However, since $Ane O_m$, we may perform some elementary operations of the second type, i.e., row- and column-switching transformations, such that after the operations, we obtain
$$
f(0)=detleft(
begin{array}{cc}
A'&B'\
C'&D'
end{array}
right),
$$
where, e.g., $A'$ results from $A$ by switching $A$'s rows and columns, such that $A''$, the $k$-by-$k$ square matrix made up of the first $k$ rows and columns of $A'$, is invertible. The existence of such an $A''$ is guaranteed by the fact that $Ane O_m$. In this way,
$$
f(0)=detleft(
begin{array}{cc}
A''&B''\
C''&D''
end{array}
right).
$$
Thanks to the invertibility of $A''$,
$$
f(0)=detleft(A''right)detleft(D''-C''left(A''right)^{-1}B''right).
$$
This result is much less elegant. $A''$ is only part of $A$. $B''$ contains part of both $A$ and $B$, and so does $C''$. $D''$ contains the whole $D$, and part of $A$, $B$, and $C$. Besides, there are switches of rows and columns as well.
answered Jan 29 at 3:16
hypernovahypernova
4,989414
$endgroup$
$begingroup$
While this answer did not provide a formula for the general case, it did provide a very nice idea which might be possible to use in some important special cases. Congratulations for the first (smaller) bonus!
$endgroup$
– Mårten W
Feb 4 at 22:50
$begingroup$
@MårtenW: Wow! Thank you for your acknowledgement and your bonus! It is so beyond the generous of you! I hope this partial answer would be somewhat helpful for you. By the way, I think it quite promising that its results for the special cases would help clarify the sign ambiguity in polfosol's answer. Thanks :-)
$endgroup$
– hypernova
Feb 5 at 0:49
add a comment |
$begingroup$
Without loss of generality, let $A$ be $m$-by-$m$ and $D$ be $n$-by-$n$, with $mge n$. Consider
$$
f(t)=detleft(
begin{array}{cc}
A+tI_m&B\
C&D
end{array}
right).
$$
Obviously,
$f(t)$ is a polynomial of $t$, for which it is analytic on $mathbb{R}$;
$f(0)$ returns the desired determinant;
$g(t)=detleft(A+tI_mright)$ is also a polynomial of $t$, for which it has isolated zeros;
$g(0)=0$ provided that $A$ is singular, yet since $t=0$ is an isolated zero of $g(t)$, $g(t)ne 0$ for all $tinleft(-delta,deltaright)setminusleft{0right}$ for some $delta>0$.
Now, since $g(t)ne 0$ on some $left(-delta,deltaright)setminusleft{0right}$, it follows that $A+tI_m$ is invertible on this domain. Therefore,
$$
f(t)=detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright).
$$
Consequently, the continuity of $f(t)$ yields
$$
f(0)=lim_{tto 0}left(detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright)right).
$$
Now, let us focus on two distinct cases. First, if $A=O_m$, i.e., $A$ is a zero matrix of order $m$. In this case, the above formula gives
begin{align}
f(0)&=lim_{tto 0}left(detleft(tI_mright)detleft(D-Cleft(tI_mright)^{-1}Bright)right)\
&=lim_{tto 0}left(t^mdetleft(D-frac{1}{t}CBright)right)\
&=lim_{tto 0}left(t^mdetleft(frac{1}{t}left(tD-CBright)right)right)\
&=lim_{tto 0}left(t^{m-n}detleft(tD-CBright)right).
end{align}
Recall that $mge n$. We therefore obtain
- If $m>n$, it is obvious that $f(0)=0$;
- If $m=n$, it follows that $f(0)=detleft(-CBright)=left(-1right)^ndetleft(CBright)$.
Second, consider $Ane O_m$. This case is more complicated, and there is no elegant form for $f(0)$ with only $A$, $B$, $C$, and $D$ involved. However, since $Ane O_m$, we may perform some elementary operations of the second type, i.e., row- and column-switching transformations, such that after the operations, we obtain
$$
f(0)=detleft(
begin{array}{cc}
A'&B'\
C'&D'
end{array}
right),
$$
where, e.g., $A'$ results from $A$ by switching $A$'s rows and columns, such that $A''$, the $k$-by-$k$ square matrix made up of the first $k$ rows and columns of $A'$, is invertible. The existence of such an $A''$ is guaranteed by the fact that $Ane O_m$. In this way,
$$
f(0)=detleft(
begin{array}{cc}
A''&B''\
C''&D''
end{array}
right).
$$
Thanks to the invertibility of $A''$,
$$
f(0)=detleft(A''right)detleft(D''-C''left(A''right)^{-1}B''right).
$$
This result is much less elegant. $A''$ is only part of $A$. $B''$ contains part of both $A$ and $B$, and so does $C''$. $D''$ contains the whole $D$, and part of $A$, $B$, and $C$. Besides, there are switches of rows and columns as well.
answered Jan 29 at 3:16
hypernovahypernova
4,989414
$endgroup$
$begingroup$
While this answer did not provide a formula for the general case, it did provide a very nice idea which might be possible to use in some important special cases. Congratulations for the first (smaller) bonus!
$endgroup$
– Mårten W
Feb 4 at 22:50
$begingroup$
@MårtenW: Wow! Thank you for your acknowledgement and your bonus! It is so beyond the generous of you! I hope this partial answer would be somewhat helpful for you. By the way, I think it quite promising that its results for the special cases would help clarify the sign ambiguity in polfosol's answer. Thanks :-)
$endgroup$
– hypernova
Feb 5 at 0:49
add a comment |
$begingroup$
Without loss of generality, let $A$ be $m$-by-$m$ and $D$ be $n$-by-$n$, with $mge n$. Consider
$$
f(t)=detleft(
begin{array}{cc}
A+tI_m&B\
C&D
end{array}
right).
$$
Obviously,
$f(t)$ is a polynomial of $t$, for which it is analytic on $mathbb{R}$;
$f(0)$ returns the desired determinant;
$g(t)=detleft(A+tI_mright)$ is also a polynomial of $t$, for which it has isolated zeros;
$g(0)=0$ provided that $A$ is singular, yet since $t=0$ is an isolated zero of $g(t)$, $g(t)ne 0$ for all $tinleft(-delta,deltaright)setminusleft{0right}$ for some $delta>0$.
Now, since $g(t)ne 0$ on some $left(-delta,deltaright)setminusleft{0right}$, it follows that $A+tI_m$ is invertible on this domain. Therefore,
$$
f(t)=detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright).
$$
Consequently, the continuity of $f(t)$ yields
$$
f(0)=lim_{tto 0}left(detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright)right).
$$
Now, let us focus on two distinct cases. First, if $A=O_m$, i.e., $A$ is a zero matrix of order $m$. In this case, the above formula gives
begin{align}
f(0)&=lim_{tto 0}left(detleft(tI_mright)detleft(D-Cleft(tI_mright)^{-1}Bright)right)\
&=lim_{tto 0}left(t^mdetleft(D-frac{1}{t}CBright)right)\
&=lim_{tto 0}left(t^mdetleft(frac{1}{t}left(tD-CBright)right)right)\
&=lim_{tto 0}left(t^{m-n}detleft(tD-CBright)right).
end{align}
Recall that $mge n$. We therefore obtain
- If $m>n$, it is obvious that $f(0)=0$;
- If $m=n$, it follows that $f(0)=detleft(-CBright)=left(-1right)^ndetleft(CBright)$.
Second, consider $Ane O_m$. This case is more complicated, and there is no elegant form for $f(0)$ with only $A$, $B$, $C$, and $D$ involved. However, since $Ane O_m$, we may perform some elementary operations of the second type, i.e., row- and column-switching transformations, such that after the operations, we obtain
$$
f(0)=detleft(
begin{array}{cc}
A'&B'\
C'&D'
end{array}
right),
$$
where, e.g., $A'$ results from $A$ by switching $A$'s rows and columns, such that $A''$, the $k$-by-$k$ square matrix made up of the first $k$ rows and columns of $A'$, is invertible. The existence of such an $A''$ is guaranteed by the fact that $Ane O_m$. In this way,
$$
f(0)=detleft(
begin{array}{cc}
A''&B''\
C''&D''
end{array}
right).
$$
Thanks to the invertibility of $A''$,
$$
f(0)=detleft(A''right)detleft(D''-C''left(A''right)^{-1}B''right).
$$
This result is much less elegant. $A''$ is only part of $A$. $B''$ contains part of both $A$ and $B$, and so does $C''$. $D''$ contains the whole $D$, and part of $A$, $B$, and $C$. Besides, there are switches of rows and columns as well.
answered Jan 29 at 3:16
hypernovahypernova
4,989414
$endgroup$
Without loss of generality, let $A$ be $m$-by-$m$ and $D$ be $n$-by-$n$, with $mge n$. Consider
$$
f(t)=detleft(
begin{array}{cc}
A+tI_m&B\
C&D
end{array}
right).
$$
Obviously,
$f(t)$ is a polynomial of $t$, for which it is analytic on $mathbb{R}$;
$f(0)$ returns the desired determinant;
$g(t)=detleft(A+tI_mright)$ is also a polynomial of $t$, for which it has isolated zeros;
$g(0)=0$ provided that $A$ is singular, yet since $t=0$ is an isolated zero of $g(t)$, $g(t)ne 0$ for all $tinleft(-delta,deltaright)setminusleft{0right}$ for some $delta>0$.
Now, since $g(t)ne 0$ on some $left(-delta,deltaright)setminusleft{0right}$, it follows that $A+tI_m$ is invertible on this domain. Therefore,
$$
f(t)=detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright).
$$
Consequently, the continuity of $f(t)$ yields
$$
f(0)=lim_{tto 0}left(detleft(A+tI_mright)detleft(D-Cleft(A+tI_mright)^{-1}Bright)right).
$$
Now, let us focus on two distinct cases. First, if $A=O_m$, i.e., $A$ is a zero matrix of order $m$. In this case, the above formula gives
begin{align}
f(0)&=lim_{tto 0}left(detleft(tI_mright)detleft(D-Cleft(tI_mright)^{-1}Bright)right)\
&=lim_{tto 0}left(t^mdetleft(D-frac{1}{t}CBright)right)\
&=lim_{tto 0}left(t^mdetleft(frac{1}{t}left(tD-CBright)right)right)\
&=lim_{tto 0}left(t^{m-n}detleft(tD-CBright)right).
end{align}
Recall that $mge n$. We therefore obtain
- If $m>n$, it is obvious that $f(0)=0$;
- If $m=n$, it follows that $f(0)=detleft(-CBright)=left(-1right)^ndetleft(CBright)$.
Second, consider $Ane O_m$. This case is more complicated, and there is no elegant form for $f(0)$ with only $A$, $B$, $C$, and $D$ involved. However, since $Ane O_m$, we may perform some elementary operations of the second type, i.e., row- and column-switching transformations, such that after the operations, we obtain
$$
f(0)=detleft(
begin{array}{cc}
A'&B'\
C'&D'
end{array}
right),
$$
where, e.g., $A'$ results from $A$ by switching $A$'s rows and columns, such that $A''$, the $k$-by-$k$ square matrix made up of the first $k$ rows and columns of $A'$, is invertible. The existence of such an $A''$ is guaranteed by the fact that $Ane O_m$. In this way,
$$
f(0)=detleft(
begin{array}{cc}
A''&B''\
C''&D''
end{array}
right).
$$
Thanks to the invertibility of $A''$,
$$
f(0)=detleft(A''right)detleft(D''-C''left(A''right)^{-1}B''right).
$$
This result is much less elegant. $A''$ is only part of $A$. $B''$ contains part of both $A$ and $B$, and so does $C''$. $D''$ contains the whole $D$, and part of $A$, $B$, and $C$. Besides, there are switches of rows and columns as well.
answered Jan 29 at 3:16
hypernovahypernova
4,989414
answered Jan 29 at 3:16
hypernovahypernova
4,989414
answered Jan 29 at 3:16
hypernovahypernova
4,989414
answered Jan 29 at 3:16
hypernovahypernova
4,989414
4,989414
$begingroup$
While this answer did not provide a formula for the general case, it did provide a very nice idea which might be possible to use in some important special cases. Congratulations for the first (smaller) bonus!
$endgroup$
– Mårten W
Feb 4 at 22:50
$begingroup$
@MårtenW: Wow! Thank you for your acknowledgement and your bonus! It is so beyond the generous of you! I hope this partial answer would be somewhat helpful for you. By the way, I think it quite promising that its results for the special cases would help clarify the sign ambiguity in polfosol's answer. Thanks :-)
$endgroup$
– hypernova
Feb 5 at 0:49
add a comment |
$begingroup$
While this answer did not provide a formula for the general case, it did provide a very nice idea which might be possible to use in some important special cases. Congratulations for the first (smaller) bonus!
$endgroup$
– Mårten W
Feb 4 at 22:50
$begingroup$
@MårtenW: Wow! Thank you for your acknowledgement and your bonus! It is so beyond the generous of you! I hope this partial answer would be somewhat helpful for you. By the way, I think it quite promising that its results for the special cases would help clarify the sign ambiguity in polfosol's answer. Thanks :-)
$endgroup$
– hypernova
Feb 5 at 0:49
$begingroup$
While this answer did not provide a formula for the general case, it did provide a very nice idea which might be possible to use in some important special cases. Congratulations for the first (smaller) bonus!
$endgroup$
– Mårten W
Feb 4 at 22:50
$begingroup$
While this answer did not provide a formula for the general case, it did provide a very nice idea which might be possible to use in some important special cases. Congratulations for the first (smaller) bonus!
$endgroup$
– Mårten W
Feb 4 at 22:50
$begingroup$
@MårtenW: Wow! Thank you for your acknowledgement and your bonus! It is so beyond the generous of you! I hope this partial answer would be somewhat helpful for you. By the way, I think it quite promising that its results for the special cases would help clarify the sign ambiguity in polfosol's answer. Thanks :-)
$endgroup$
– hypernova
Feb 5 at 0:49
$begingroup$
@MårtenW: Wow! Thank you for your acknowledgement and your bonus! It is so beyond the generous of you! I hope this partial answer would be somewhat helpful for you. By the way, I think it quite promising that its results for the special cases would help clarify the sign ambiguity in polfosol's answer. Thanks :-)
$endgroup$
– hypernova
Feb 5 at 0:49
add a comment |
$begingroup$
For $M=begin{bmatrix}A & B\C & Dend{bmatrix}$ let $N=M^TM$. So that
$$N:=begin{bmatrix}E & F\F^T & Gend{bmatrix}$$
where
$$begin{align}
E&=A^T A+C^T C\F&=A^T B+C^T D\G&=B^T B+D^T D
end{align}$$
Now, if $M$ is non-singular then $N$ is a positive definite matrix. Hence according to this other post, $E$ is a positive definite (i.e. non-singular) block matrix.
Since $E$ is non-singular, we can use Schur complement to obtain $det N$.
$$det{N} = det{E}cdotdet(G-F^T E^{-1}F)=(det M)^2$$
The only remaining part in this case is to determine the sign of $det M$, which apparently, there is no easy way to do that in general.
We might be able to get the sign of $det M$ by a trick similar to the one discussed at the end of @hypernova's answer. I'll update this answer if anything comes up.
answered Jan 29 at 17:55
polfosolpolfosol
5,93931945
$endgroup$
1
$begingroup$
Nice answer (+1) and nicer nickname by the way :)
$endgroup$
– Mostafa Ayaz
Feb 1 at 21:44
1
$begingroup$
Except for leaving that sign ambiguity, this answer does precisely what I was asking for, and for that reason I'm going to award the bigger bounty as soon as the system lets me.
$endgroup$
– Mårten W
Feb 4 at 22:47
add a comment |
$begingroup$
For $M=begin{bmatrix}A & B\C & Dend{bmatrix}$ let $N=M^TM$. So that
$$N:=begin{bmatrix}E & F\F^T & Gend{bmatrix}$$
where
$$begin{align}
E&=A^T A+C^T C\F&=A^T B+C^T D\G&=B^T B+D^T D
end{align}$$
Now, if $M$ is non-singular then $N$ is a positive definite matrix. Hence according to this other post, $E$ is a positive definite (i.e. non-singular) block matrix.
Since $E$ is non-singular, we can use Schur complement to obtain $det N$.
$$det{N} = det{E}cdotdet(G-F^T E^{-1}F)=(det M)^2$$
The only remaining part in this case is to determine the sign of $det M$, which apparently, there is no easy way to do that in general.
We might be able to get the sign of $det M$ by a trick similar to the one discussed at the end of @hypernova's answer. I'll update this answer if anything comes up.
answered Jan 29 at 17:55
polfosolpolfosol
5,93931945
$endgroup$
1
$begingroup$
Nice answer (+1) and nicer nickname by the way :)
$endgroup$
– Mostafa Ayaz
Feb 1 at 21:44
1
$begingroup$
Except for leaving that sign ambiguity, this answer does precisely what I was asking for, and for that reason I'm going to award the bigger bounty as soon as the system lets me.
$endgroup$
– Mårten W
Feb 4 at 22:47
add a comment |
$begingroup$
For $M=begin{bmatrix}A & B\C & Dend{bmatrix}$ let $N=M^TM$. So that
$$N:=begin{bmatrix}E & F\F^T & Gend{bmatrix}$$
where
$$begin{align}
E&=A^T A+C^T C\F&=A^T B+C^T D\G&=B^T B+D^T D
end{align}$$
Now, if $M$ is non-singular then $N$ is a positive definite matrix. Hence according to this other post, $E$ is a positive definite (i.e. non-singular) block matrix.
Since $E$ is non-singular, we can use Schur complement to obtain $det N$.
$$det{N} = det{E}cdotdet(G-F^T E^{-1}F)=(det M)^2$$
The only remaining part in this case is to determine the sign of $det M$, which apparently, there is no easy way to do that in general.
We might be able to get the sign of $det M$ by a trick similar to the one discussed at the end of @hypernova's answer. I'll update this answer if anything comes up.
answered Jan 29 at 17:55
polfosolpolfosol
5,93931945
$endgroup$
For $M=begin{bmatrix}A & B\C & Dend{bmatrix}$ let $N=M^TM$. So that
$$N:=begin{bmatrix}E & F\F^T & Gend{bmatrix}$$
where
$$begin{align}
E&=A^T A+C^T C\F&=A^T B+C^T D\G&=B^T B+D^T D
end{align}$$
Now, if $M$ is non-singular then $N$ is a positive definite matrix. Hence according to this other post, $E$ is a positive definite (i.e. non-singular) block matrix.
Since $E$ is non-singular, we can use Schur complement to obtain $det N$.
$$det{N} = det{E}cdotdet(G-F^T E^{-1}F)=(det M)^2$$
The only remaining part in this case is to determine the sign of $det M$, which apparently, there is no easy way to do that in general.
We might be able to get the sign of $det M$ by a trick similar to the one discussed at the end of @hypernova's answer. I'll update this answer if anything comes up.
answered Jan 29 at 17:55
polfosolpolfosol
5,93931945
edited Jan 30 at 11:17
answered Jan 29 at 17:55
polfosolpolfosol
5,93931945
answered Jan 29 at 17:55
polfosolpolfosol
5,93931945
answered Jan 29 at 17:55
polfosolpolfosol
5,93931945
5,93931945
1
$begingroup$
Nice answer (+1) and nicer nickname by the way :)
$endgroup$
– Mostafa Ayaz
Feb 1 at 21:44
1
$begingroup$
Except for leaving that sign ambiguity, this answer does precisely what I was asking for, and for that reason I'm going to award the bigger bounty as soon as the system lets me.
$endgroup$
– Mårten W
Feb 4 at 22:47
add a comment |
1
$begingroup$
Nice answer (+1) and nicer nickname by the way :)
$endgroup$
– Mostafa Ayaz
Feb 1 at 21:44
1
$begingroup$
Except for leaving that sign ambiguity, this answer does precisely what I was asking for, and for that reason I'm going to award the bigger bounty as soon as the system lets me.
$endgroup$
– Mårten W
Feb 4 at 22:47
1
1
$begingroup$
Nice answer (+1) and nicer nickname by the way :)
$endgroup$
– Mostafa Ayaz
Feb 1 at 21:44
$begingroup$
Nice answer (+1) and nicer nickname by the way :)
$endgroup$
– Mostafa Ayaz
Feb 1 at 21:44
1
1
$begingroup$
Except for leaving that sign ambiguity, this answer does precisely what I was asking for, and for that reason I'm going to award the bigger bounty as soon as the system lets me.
$endgroup$
– Mårten W
Feb 4 at 22:47
$begingroup$
Except for leaving that sign ambiguity, this answer does precisely what I was asking for, and for that reason I'm going to award the bigger bounty as soon as the system lets me.
$endgroup$
– Mårten W
Feb 4 at 22:47
add a comment |
$begingroup$
Hint:
$$
M =
begin{bmatrix}
A & B \
C & D
end{bmatrix}.
=
begin{bmatrix}
0 & I \
I & 0
end{bmatrix}.
begin{bmatrix}
C & D \
A & B
end{bmatrix}
$$
Where $I$ is the idenity matrix of appropriate size.
answered Jan 29 at 3:47
zimbra314zimbra314
630312
$endgroup$
2
$begingroup$
I thought about the same hint, but C may not even be square, so this is a dead end in general.
$endgroup$
– Dirk
Jan 29 at 6:33
add a comment |
$begingroup$
Hint:
$$
M =
begin{bmatrix}
A & B \
C & D
end{bmatrix}.
=
begin{bmatrix}
0 & I \
I & 0
end{bmatrix}.
begin{bmatrix}
C & D \
A & B
end{bmatrix}
$$
Where $I$ is the idenity matrix of appropriate size.
answered Jan 29 at 3:47
zimbra314zimbra314
630312
$endgroup$
2
$begingroup$
I thought about the same hint, but C may not even be square, so this is a dead end in general.
$endgroup$
– Dirk
Jan 29 at 6:33
add a comment |
$begingroup$
Hint:
$$
M =
begin{bmatrix}
A & B \
C & D
end{bmatrix}.
=
begin{bmatrix}
0 & I \
I & 0
end{bmatrix}.
begin{bmatrix}
C & D \
A & B
end{bmatrix}
$$
Where $I$ is the idenity matrix of appropriate size.
answered Jan 29 at 3:47
zimbra314zimbra314
630312
$endgroup$
Hint:
$$
M =
begin{bmatrix}
A & B \
C & D
end{bmatrix}.
=
begin{bmatrix}
0 & I \
I & 0
end{bmatrix}.
begin{bmatrix}
C & D \
A & B
end{bmatrix}
$$
Where $I$ is the idenity matrix of appropriate size.
answered Jan 29 at 3:47
zimbra314zimbra314
630312
answered Jan 29 at 3:47
zimbra314zimbra314
630312
answered Jan 29 at 3:47
zimbra314zimbra314
630312
answered Jan 29 at 3:47
zimbra314zimbra314
630312
630312
2
$begingroup$
I thought about the same hint, but C may not even be square, so this is a dead end in general.
$endgroup$
– Dirk
Jan 29 at 6:33
add a comment |
2
$begingroup$
I thought about the same hint, but C may not even be square, so this is a dead end in general.
$endgroup$
– Dirk
Jan 29 at 6:33
2
2
$begingroup$
I thought about the same hint, but C may not even be square, so this is a dead end in general.
$endgroup$
– Dirk
Jan 29 at 6:33
$begingroup$
I thought about the same hint, but C may not even be square, so this is a dead end in general.
$endgroup$
– Dirk
Jan 29 at 6:33
add a comment |
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