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Iwasawa Matrix Decomposition Proof
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Iwasawa Decomposition (special case): Let $G=SL_n(BbbR)$, $K=$ real unitary matrices, $U=$ upper triangular matrices with $1$'s on the diagonal (called unipotent), and $A=$ diagonal matrices with positive elements ($0$ everywhere else). Then, the product map $UtimesAtimesKrightarrowG$ given by $(u,a,k)mapstouak$ is a bijection.
Here is the proof by Serge Lang in his book Undergraduate Algebra Section 6 Chapter 4 pg 246 (Read below picture for question(s) please :)
I have a few questions about this proof (understand it roughly as a whole):
1) How does one get that $B=au$, following $g^-1=Bk^-1$?
2) Why does it follow that A has positive diagonal elements - that is $a_i=b_ii>0$? (My guess is that the QR decomposition guarantees this for R and note $B=R$)
3) I can't see any point in the proof where Lang makes a reference to the fact that $g$ has determinant $1$, in fact it seems that $g$ could have any non-zero determinant, hence $ginGL_n(BbbR)$. Why is this not so?
Thank you.
linear-algebra abstract-algebra matrices matrix-decomposition linear-groups
asked yesterday
Daniele1234
716214
add a comment |Â
up vote
2
down vote
favorite
Iwasawa Decomposition (special case): Let $G=SL_n(BbbR)$, $K=$ real unitary matrices, $U=$ upper triangular matrices with $1$'s on the diagonal (called unipotent), and $A=$ diagonal matrices with positive elements ($0$ everywhere else). Then, the product map $UtimesAtimesKrightarrowG$ given by $(u,a,k)mapstouak$ is a bijection.
Here is the proof by Serge Lang in his book Undergraduate Algebra Section 6 Chapter 4 pg 246 (Read below picture for question(s) please :)
I have a few questions about this proof (understand it roughly as a whole):
1) How does one get that $B=au$, following $g^-1=Bk^-1$?
2) Why does it follow that A has positive diagonal elements - that is $a_i=b_ii>0$? (My guess is that the QR decomposition guarantees this for R and note $B=R$)
3) I can't see any point in the proof where Lang makes a reference to the fact that $g$ has determinant $1$, in fact it seems that $g$ could have any non-zero determinant, hence $ginGL_n(BbbR)$. Why is this not so?
Thank you.
linear-algebra abstract-algebra matrices matrix-decomposition linear-groups
asked yesterday
Daniele1234
716214
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Iwasawa Decomposition (special case): Let $G=SL_n(BbbR)$, $K=$ real unitary matrices, $U=$ upper triangular matrices with $1$'s on the diagonal (called unipotent), and $A=$ diagonal matrices with positive elements ($0$ everywhere else). Then, the product map $UtimesAtimesKrightarrowG$ given by $(u,a,k)mapstouak$ is a bijection.
Here is the proof by Serge Lang in his book Undergraduate Algebra Section 6 Chapter 4 pg 246 (Read below picture for question(s) please :)
I have a few questions about this proof (understand it roughly as a whole):
1) How does one get that $B=au$, following $g^-1=Bk^-1$?
2) Why does it follow that A has positive diagonal elements - that is $a_i=b_ii>0$? (My guess is that the QR decomposition guarantees this for R and note $B=R$)
3) I can't see any point in the proof where Lang makes a reference to the fact that $g$ has determinant $1$, in fact it seems that $g$ could have any non-zero determinant, hence $ginGL_n(BbbR)$. Why is this not so?
Thank you.
linear-algebra abstract-algebra matrices matrix-decomposition linear-groups
asked yesterday
Daniele1234
716214
Iwasawa Decomposition (special case): Let $G=SL_n(BbbR)$, $K=$ real unitary matrices, $U=$ upper triangular matrices with $1$'s on the diagonal (called unipotent), and $A=$ diagonal matrices with positive elements ($0$ everywhere else). Then, the product map $UtimesAtimesKrightarrowG$ given by $(u,a,k)mapstouak$ is a bijection.
Here is the proof by Serge Lang in his book Undergraduate Algebra Section 6 Chapter 4 pg 246 (Read below picture for question(s) please :)
I have a few questions about this proof (understand it roughly as a whole):
1) How does one get that $B=au$, following $g^-1=Bk^-1$?
2) Why does it follow that A has positive diagonal elements - that is $a_i=b_ii>0$? (My guess is that the QR decomposition guarantees this for R and note $B=R$)
3) I can't see any point in the proof where Lang makes a reference to the fact that $g$ has determinant $1$, in fact it seems that $g$ could have any non-zero determinant, hence $ginGL_n(BbbR)$. Why is this not so?
Thank you.
linear-algebra abstract-algebra matrices matrix-decomposition linear-groups
asked yesterday
Daniele1234
716214
edited yesterday
asked yesterday
Daniele1234
716214
asked yesterday
Daniele1234
716214
asked yesterday
Daniele1234
716214
716214
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add a comment |Â
1 Answer
1
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2
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The diagonal entries of $B$ are positive, because they are the lengths of the vectors in Gram-Schmidt. To write $B=au$, will do it in $2times 2$ so that you see what happens:
$$
B=beginbmatrix b_11 & b_12 \ 0& b_22endbmatrix
=beginbmatrix b_11 &0 \ 0& b_22 endbmatrix
beginbmatrix 1&b_12/b_11\ 0&1endbmatrix.
$$
As for the determinant, the way the decomposition is phrased it is not restricted to the case $det g=1$, since $a$ is allowed arbitrary diagonal entries so the determinant can be arbitrary. Note that $det u=1$, $det k=pm1$, so $$det g=pmdet a.$$
answered yesterday
Martin Argerami
115k1070164
Thank you for this. So this decomposition is in fact for $GL_n(BbbR)$ as $g$ can be any non-zero determinant?
– Daniele1234
yesterday
Yes. $ $
– Martin Argerami
yesterday
Why would Lang then so heavily restrict himself to special linear when he could use the much more general General Linear group - perhaps an aesthetic thing?
– Daniele1234
23 hours ago
I have no idea. To my taste, the whole writing of the proof is horrible.
– Martin Argerami
23 hours ago
No wonder I was confused! Thank you for your input.
– Daniele1234
22 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The diagonal entries of $B$ are positive, because they are the lengths of the vectors in Gram-Schmidt. To write $B=au$, will do it in $2times 2$ so that you see what happens:
$$
B=beginbmatrix b_11 & b_12 \ 0& b_22endbmatrix
=beginbmatrix b_11 &0 \ 0& b_22 endbmatrix
beginbmatrix 1&b_12/b_11\ 0&1endbmatrix.
$$
As for the determinant, the way the decomposition is phrased it is not restricted to the case $det g=1$, since $a$ is allowed arbitrary diagonal entries so the determinant can be arbitrary. Note that $det u=1$, $det k=pm1$, so $$det g=pmdet a.$$
answered yesterday
Martin Argerami
115k1070164
Thank you for this. So this decomposition is in fact for $GL_n(BbbR)$ as $g$ can be any non-zero determinant?
– Daniele1234
yesterday
Yes. $ $
– Martin Argerami
yesterday
Why would Lang then so heavily restrict himself to special linear when he could use the much more general General Linear group - perhaps an aesthetic thing?
– Daniele1234
23 hours ago
I have no idea. To my taste, the whole writing of the proof is horrible.
– Martin Argerami
23 hours ago
No wonder I was confused! Thank you for your input.
– Daniele1234
22 hours ago
add a comment |Â
up vote
2
down vote
accepted
The diagonal entries of $B$ are positive, because they are the lengths of the vectors in Gram-Schmidt. To write $B=au$, will do it in $2times 2$ so that you see what happens:
$$
B=beginbmatrix b_11 & b_12 \ 0& b_22endbmatrix
=beginbmatrix b_11 &0 \ 0& b_22 endbmatrix
beginbmatrix 1&b_12/b_11\ 0&1endbmatrix.
$$
As for the determinant, the way the decomposition is phrased it is not restricted to the case $det g=1$, since $a$ is allowed arbitrary diagonal entries so the determinant can be arbitrary. Note that $det u=1$, $det k=pm1$, so $$det g=pmdet a.$$
answered yesterday
Martin Argerami
115k1070164
Thank you for this. So this decomposition is in fact for $GL_n(BbbR)$ as $g$ can be any non-zero determinant?
– Daniele1234
yesterday
Yes. $ $
– Martin Argerami
yesterday
Why would Lang then so heavily restrict himself to special linear when he could use the much more general General Linear group - perhaps an aesthetic thing?
– Daniele1234
23 hours ago
I have no idea. To my taste, the whole writing of the proof is horrible.
– Martin Argerami
23 hours ago
No wonder I was confused! Thank you for your input.
– Daniele1234
22 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The diagonal entries of $B$ are positive, because they are the lengths of the vectors in Gram-Schmidt. To write $B=au$, will do it in $2times 2$ so that you see what happens:
$$
B=beginbmatrix b_11 & b_12 \ 0& b_22endbmatrix
=beginbmatrix b_11 &0 \ 0& b_22 endbmatrix
beginbmatrix 1&b_12/b_11\ 0&1endbmatrix.
$$
As for the determinant, the way the decomposition is phrased it is not restricted to the case $det g=1$, since $a$ is allowed arbitrary diagonal entries so the determinant can be arbitrary. Note that $det u=1$, $det k=pm1$, so $$det g=pmdet a.$$
answered yesterday
Martin Argerami
115k1070164
The diagonal entries of $B$ are positive, because they are the lengths of the vectors in Gram-Schmidt. To write $B=au$, will do it in $2times 2$ so that you see what happens:
$$
B=beginbmatrix b_11 & b_12 \ 0& b_22endbmatrix
=beginbmatrix b_11 &0 \ 0& b_22 endbmatrix
beginbmatrix 1&b_12/b_11\ 0&1endbmatrix.
$$
As for the determinant, the way the decomposition is phrased it is not restricted to the case $det g=1$, since $a$ is allowed arbitrary diagonal entries so the determinant can be arbitrary. Note that $det u=1$, $det k=pm1$, so $$det g=pmdet a.$$
answered yesterday
Martin Argerami
115k1070164
answered yesterday
Martin Argerami
115k1070164
answered yesterday
Martin Argerami
115k1070164
answered yesterday
Martin Argerami
115k1070164
115k1070164
Thank you for this. So this decomposition is in fact for $GL_n(BbbR)$ as $g$ can be any non-zero determinant?
– Daniele1234
yesterday
Yes. $ $
– Martin Argerami
yesterday
Why would Lang then so heavily restrict himself to special linear when he could use the much more general General Linear group - perhaps an aesthetic thing?
– Daniele1234
23 hours ago
I have no idea. To my taste, the whole writing of the proof is horrible.
– Martin Argerami
23 hours ago
No wonder I was confused! Thank you for your input.
– Daniele1234
22 hours ago
add a comment |Â
Thank you for this. So this decomposition is in fact for $GL_n(BbbR)$ as $g$ can be any non-zero determinant?
– Daniele1234
yesterday
Yes. $ $
– Martin Argerami
yesterday
Why would Lang then so heavily restrict himself to special linear when he could use the much more general General Linear group - perhaps an aesthetic thing?
– Daniele1234
23 hours ago
I have no idea. To my taste, the whole writing of the proof is horrible.
– Martin Argerami
23 hours ago
No wonder I was confused! Thank you for your input.
– Daniele1234
22 hours ago
Thank you for this. So this decomposition is in fact for $GL_n(BbbR)$ as $g$ can be any non-zero determinant?
– Daniele1234
yesterday
Thank you for this. So this decomposition is in fact for $GL_n(BbbR)$ as $g$ can be any non-zero determinant?
– Daniele1234
yesterday
Yes. $ $
– Martin Argerami
yesterday
Yes. $ $
– Martin Argerami
yesterday
Why would Lang then so heavily restrict himself to special linear when he could use the much more general General Linear group - perhaps an aesthetic thing?
– Daniele1234
23 hours ago
Why would Lang then so heavily restrict himself to special linear when he could use the much more general General Linear group - perhaps an aesthetic thing?
– Daniele1234
23 hours ago
I have no idea. To my taste, the whole writing of the proof is horrible.
– Martin Argerami
23 hours ago
I have no idea. To my taste, the whole writing of the proof is horrible.
– Martin Argerami
23 hours ago
No wonder I was confused! Thank you for your input.
– Daniele1234
22 hours ago
No wonder I was confused! Thank you for your input.
– Daniele1234
22 hours ago
add a comment |Â
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