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Iwasawa Decomposition $SL_n(BbbR)$ Proof
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A special case of the Iwasawa Decomposition for matrices is the following: 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.
I am looking at the proof of this in Serge Lang's Undergraduate Algebra, Section 6 Chapter 4 pg 246, however, I need some help in understanding the beginning of the proof:
"Let $e_1,..,e_n$ be the standard vertical unit vectors of $BbbR^n$. Let $g=(g_ij)inG$. Then we have $ge_i=beginpmatrix g_1i \ vdots \ g_ni endpmatrix=g^(i)=sum_q=1^n g_qie_q$. There exists an upper triangular matrix $B=(b_ij)$, so with $b_ij=0$ if $i>j$ such that $b_11g^(1)=e'_1$,..........,$b_1jg^(1)+b_2jg^(2)+...+b_jjg^(j)=e'_j$,.............., $b_1ng^(1)+b_2ng^(2)+...+b_nng^(n)=e'_n$, such that the diagonal elements are positive, that is $b_11,...,b_nn>0$, and such that the vectors $e'_1,...,e'_n$ are mutually perpendicular unit vectors. Getting such a matrix B is merely applying the usual Gram Schmidt orthogonalisation process, subtracting a linear combination of previous vectors to get orthogonality, and then dividing by the norms to get unit vectors."
My problem here is understanding the matrix $B$ and how it is obtained. Having studied the Gram Schmidt Orthogonalisation process in a linear algebra course, I simply learnt it as a way to construct an orthogonal (orthonormal by diving by norms) basis for a finite dimensional inner product space (in this case $BbbR$) given any basis. Could someone please explain how Lang then guarantees the existence of this matrix $B$.
Thank you.
ADDITIONAL (carrying on where Lang's proof left off): Let $gB=kinK$. Then $ke_i=e'_i$, so the columns of $k$ are orthonormal, so $k$ is real unitary, and $g=kB^-1$. Then $g^-1=Bk^-1$ and $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$.
linear-algebra abstract-algebra matrices orthogonality matrix-decomposition
asked 2 days ago
Daniele1234
716214
add a comment |Â
up vote
3
down vote
favorite
A special case of the Iwasawa Decomposition for matrices is the following: 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.
I am looking at the proof of this in Serge Lang's Undergraduate Algebra, Section 6 Chapter 4 pg 246, however, I need some help in understanding the beginning of the proof:
"Let $e_1,..,e_n$ be the standard vertical unit vectors of $BbbR^n$. Let $g=(g_ij)inG$. Then we have $ge_i=beginpmatrix g_1i \ vdots \ g_ni endpmatrix=g^(i)=sum_q=1^n g_qie_q$. There exists an upper triangular matrix $B=(b_ij)$, so with $b_ij=0$ if $i>j$ such that $b_11g^(1)=e'_1$,..........,$b_1jg^(1)+b_2jg^(2)+...+b_jjg^(j)=e'_j$,.............., $b_1ng^(1)+b_2ng^(2)+...+b_nng^(n)=e'_n$, such that the diagonal elements are positive, that is $b_11,...,b_nn>0$, and such that the vectors $e'_1,...,e'_n$ are mutually perpendicular unit vectors. Getting such a matrix B is merely applying the usual Gram Schmidt orthogonalisation process, subtracting a linear combination of previous vectors to get orthogonality, and then dividing by the norms to get unit vectors."
My problem here is understanding the matrix $B$ and how it is obtained. Having studied the Gram Schmidt Orthogonalisation process in a linear algebra course, I simply learnt it as a way to construct an orthogonal (orthonormal by diving by norms) basis for a finite dimensional inner product space (in this case $BbbR$) given any basis. Could someone please explain how Lang then guarantees the existence of this matrix $B$.
Thank you.
ADDITIONAL (carrying on where Lang's proof left off): Let $gB=kinK$. Then $ke_i=e'_i$, so the columns of $k$ are orthonormal, so $k$ is real unitary, and $g=kB^-1$. Then $g^-1=Bk^-1$ and $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$.
linear-algebra abstract-algebra matrices orthogonality matrix-decomposition
asked 2 days ago
Daniele1234
716214
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
A special case of the Iwasawa Decomposition for matrices is the following: 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.
I am looking at the proof of this in Serge Lang's Undergraduate Algebra, Section 6 Chapter 4 pg 246, however, I need some help in understanding the beginning of the proof:
"Let $e_1,..,e_n$ be the standard vertical unit vectors of $BbbR^n$. Let $g=(g_ij)inG$. Then we have $ge_i=beginpmatrix g_1i \ vdots \ g_ni endpmatrix=g^(i)=sum_q=1^n g_qie_q$. There exists an upper triangular matrix $B=(b_ij)$, so with $b_ij=0$ if $i>j$ such that $b_11g^(1)=e'_1$,..........,$b_1jg^(1)+b_2jg^(2)+...+b_jjg^(j)=e'_j$,.............., $b_1ng^(1)+b_2ng^(2)+...+b_nng^(n)=e'_n$, such that the diagonal elements are positive, that is $b_11,...,b_nn>0$, and such that the vectors $e'_1,...,e'_n$ are mutually perpendicular unit vectors. Getting such a matrix B is merely applying the usual Gram Schmidt orthogonalisation process, subtracting a linear combination of previous vectors to get orthogonality, and then dividing by the norms to get unit vectors."
My problem here is understanding the matrix $B$ and how it is obtained. Having studied the Gram Schmidt Orthogonalisation process in a linear algebra course, I simply learnt it as a way to construct an orthogonal (orthonormal by diving by norms) basis for a finite dimensional inner product space (in this case $BbbR$) given any basis. Could someone please explain how Lang then guarantees the existence of this matrix $B$.
Thank you.
ADDITIONAL (carrying on where Lang's proof left off): Let $gB=kinK$. Then $ke_i=e'_i$, so the columns of $k$ are orthonormal, so $k$ is real unitary, and $g=kB^-1$. Then $g^-1=Bk^-1$ and $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$.
linear-algebra abstract-algebra matrices orthogonality matrix-decomposition
asked 2 days ago
Daniele1234
716214
A special case of the Iwasawa Decomposition for matrices is the following: 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.
I am looking at the proof of this in Serge Lang's Undergraduate Algebra, Section 6 Chapter 4 pg 246, however, I need some help in understanding the beginning of the proof:
"Let $e_1,..,e_n$ be the standard vertical unit vectors of $BbbR^n$. Let $g=(g_ij)inG$. Then we have $ge_i=beginpmatrix g_1i \ vdots \ g_ni endpmatrix=g^(i)=sum_q=1^n g_qie_q$. There exists an upper triangular matrix $B=(b_ij)$, so with $b_ij=0$ if $i>j$ such that $b_11g^(1)=e'_1$,..........,$b_1jg^(1)+b_2jg^(2)+...+b_jjg^(j)=e'_j$,.............., $b_1ng^(1)+b_2ng^(2)+...+b_nng^(n)=e'_n$, such that the diagonal elements are positive, that is $b_11,...,b_nn>0$, and such that the vectors $e'_1,...,e'_n$ are mutually perpendicular unit vectors. Getting such a matrix B is merely applying the usual Gram Schmidt orthogonalisation process, subtracting a linear combination of previous vectors to get orthogonality, and then dividing by the norms to get unit vectors."
My problem here is understanding the matrix $B$ and how it is obtained. Having studied the Gram Schmidt Orthogonalisation process in a linear algebra course, I simply learnt it as a way to construct an orthogonal (orthonormal by diving by norms) basis for a finite dimensional inner product space (in this case $BbbR$) given any basis. Could someone please explain how Lang then guarantees the existence of this matrix $B$.
Thank you.
ADDITIONAL (carrying on where Lang's proof left off): Let $gB=kinK$. Then $ke_i=e'_i$, so the columns of $k$ are orthonormal, so $k$ is real unitary, and $g=kB^-1$. Then $g^-1=Bk^-1$ and $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$.
linear-algebra abstract-algebra matrices orthogonality matrix-decomposition
asked 2 days ago
Daniele1234
716214
edited 2 days ago
asked 2 days ago
Daniele1234
716214
asked 2 days ago
Daniele1234
716214
asked 2 days ago
Daniele1234
716214
716214
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Gram Schmidt is as follows, depicted visually 
such that
$$a_1 = r_11 q_1 \a_2 = r_12q_1 + r_22q_2\ a_3= r_13q_1 +r_23q_2 + r_33 q_3 \ vdots \ a_n = r_1n q_1 + r_2nq_2 + cdots + r_nn q_n $$
giving
$$A = hatQhatR $$
The matrix $B$ appears to be the matrix $R$.
How it is found
$$ q_1 = fraca_1r_11\ q_2 = fraca_2-r_12q_1r_22\q_3 = fraca_3-r_13q_1-r_23q_2r_33\ vdots \ q_n = fraca_n - sum_i=1^n r_inq_ir_nn$$
where
$$ |r_ij | = q_i^*a_j \
| r_jj| = | a_j = sum_i=1^j-1 r_ijq_i|_2$$
seen here
Giving your $B$ from that
answered 2 days ago
Geronimo
809715
So this is just a reference to the QR decomposition, which uses the Gram-Schmidt process
– Daniele1234
2 days ago
1
The QR decomposition is Gram-Schmidt. It decomposes a matrix into an orthogonal matrix and an upper triangular matrix of coefficients. Your question was constructing the matrix B which was an upper triangular matrix with vectors that $e_1, e_2, cdots, e_n$ that are orthogonal...these $q_1, q_2, cdots q_n $ I would imagine...you can't construct the orthogonal vectors without the coefficients...you're subtracting them off.
– Geronimo
2 days ago
Lang then states that one can write $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$. I understand why $u$ is thus unipotent, but could you please explain how he guarantees $B=au$. Thank you
– Daniele1234
2 days ago
This is after having stated, $gB=k$, where $k$ is unitary , so $g=kB^-1$, so $g^-1=Bk^-1$ (I understand this) and (thus?) $B=au$.
– Daniele1234
2 days ago
@Daniele1234 I'll have to look at Langs book
– Geronimo
2 days ago
 |Â
show 21 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Gram Schmidt is as follows, depicted visually
such that
$$a_1 = r_11 q_1 \a_2 = r_12q_1 + r_22q_2\ a_3= r_13q_1 +r_23q_2 + r_33 q_3 \ vdots \ a_n = r_1n q_1 + r_2nq_2 + cdots + r_nn q_n $$
giving
$$A = hatQhatR $$
The matrix $B$ appears to be the matrix $R$.
How it is found
$$ q_1 = fraca_1r_11\ q_2 = fraca_2-r_12q_1r_22\q_3 = fraca_3-r_13q_1-r_23q_2r_33\ vdots \ q_n = fraca_n - sum_i=1^n r_inq_ir_nn$$
where
$$ |r_ij | = q_i^*a_j \
| r_jj| = | a_j = sum_i=1^j-1 r_ijq_i|_2$$
seen here
Giving your $B$ from that
answered 2 days ago
Geronimo
809715
So this is just a reference to the QR decomposition, which uses the Gram-Schmidt process
– Daniele1234
2 days ago
1
The QR decomposition is Gram-Schmidt. It decomposes a matrix into an orthogonal matrix and an upper triangular matrix of coefficients. Your question was constructing the matrix B which was an upper triangular matrix with vectors that $e_1, e_2, cdots, e_n$ that are orthogonal...these $q_1, q_2, cdots q_n $ I would imagine...you can't construct the orthogonal vectors without the coefficients...you're subtracting them off.
– Geronimo
2 days ago
Lang then states that one can write $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$. I understand why $u$ is thus unipotent, but could you please explain how he guarantees $B=au$. Thank you
– Daniele1234
2 days ago
This is after having stated, $gB=k$, where $k$ is unitary , so $g=kB^-1$, so $g^-1=Bk^-1$ (I understand this) and (thus?) $B=au$.
– Daniele1234
2 days ago
@Daniele1234 I'll have to look at Langs book
– Geronimo
2 days ago
 |Â
show 21 more comments
up vote
1
down vote
accepted
Gram Schmidt is as follows, depicted visually
such that
$$a_1 = r_11 q_1 \a_2 = r_12q_1 + r_22q_2\ a_3= r_13q_1 +r_23q_2 + r_33 q_3 \ vdots \ a_n = r_1n q_1 + r_2nq_2 + cdots + r_nn q_n $$
giving
$$A = hatQhatR $$
The matrix $B$ appears to be the matrix $R$.
How it is found
$$ q_1 = fraca_1r_11\ q_2 = fraca_2-r_12q_1r_22\q_3 = fraca_3-r_13q_1-r_23q_2r_33\ vdots \ q_n = fraca_n - sum_i=1^n r_inq_ir_nn$$
where
$$ |r_ij | = q_i^*a_j \
| r_jj| = | a_j = sum_i=1^j-1 r_ijq_i|_2$$
seen here
Giving your $B$ from that
answered 2 days ago
Geronimo
809715
So this is just a reference to the QR decomposition, which uses the Gram-Schmidt process
– Daniele1234
2 days ago
1
The QR decomposition is Gram-Schmidt. It decomposes a matrix into an orthogonal matrix and an upper triangular matrix of coefficients. Your question was constructing the matrix B which was an upper triangular matrix with vectors that $e_1, e_2, cdots, e_n$ that are orthogonal...these $q_1, q_2, cdots q_n $ I would imagine...you can't construct the orthogonal vectors without the coefficients...you're subtracting them off.
– Geronimo
2 days ago
Lang then states that one can write $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$. I understand why $u$ is thus unipotent, but could you please explain how he guarantees $B=au$. Thank you
– Daniele1234
2 days ago
This is after having stated, $gB=k$, where $k$ is unitary , so $g=kB^-1$, so $g^-1=Bk^-1$ (I understand this) and (thus?) $B=au$.
– Daniele1234
2 days ago
@Daniele1234 I'll have to look at Langs book
– Geronimo
2 days ago
 |Â
show 21 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Gram Schmidt is as follows, depicted visually
such that
$$a_1 = r_11 q_1 \a_2 = r_12q_1 + r_22q_2\ a_3= r_13q_1 +r_23q_2 + r_33 q_3 \ vdots \ a_n = r_1n q_1 + r_2nq_2 + cdots + r_nn q_n $$
giving
$$A = hatQhatR $$
The matrix $B$ appears to be the matrix $R$.
How it is found
$$ q_1 = fraca_1r_11\ q_2 = fraca_2-r_12q_1r_22\q_3 = fraca_3-r_13q_1-r_23q_2r_33\ vdots \ q_n = fraca_n - sum_i=1^n r_inq_ir_nn$$
where
$$ |r_ij | = q_i^*a_j \
| r_jj| = | a_j = sum_i=1^j-1 r_ijq_i|_2$$
seen here
Giving your $B$ from that
answered 2 days ago
Geronimo
809715
Gram Schmidt is as follows, depicted visually
such that
$$a_1 = r_11 q_1 \a_2 = r_12q_1 + r_22q_2\ a_3= r_13q_1 +r_23q_2 + r_33 q_3 \ vdots \ a_n = r_1n q_1 + r_2nq_2 + cdots + r_nn q_n $$
giving
$$A = hatQhatR $$
The matrix $B$ appears to be the matrix $R$.
How it is found
$$ q_1 = fraca_1r_11\ q_2 = fraca_2-r_12q_1r_22\q_3 = fraca_3-r_13q_1-r_23q_2r_33\ vdots \ q_n = fraca_n - sum_i=1^n r_inq_ir_nn$$
where
$$ |r_ij | = q_i^*a_j \
| r_jj| = | a_j = sum_i=1^j-1 r_ijq_i|_2$$
seen here
Giving your $B$ from that
answered 2 days ago
Geronimo
809715
edited 2 days ago
answered 2 days ago
Geronimo
809715
answered 2 days ago
Geronimo
809715
answered 2 days ago
Geronimo
809715
809715
So this is just a reference to the QR decomposition, which uses the Gram-Schmidt process
– Daniele1234
2 days ago
1
The QR decomposition is Gram-Schmidt. It decomposes a matrix into an orthogonal matrix and an upper triangular matrix of coefficients. Your question was constructing the matrix B which was an upper triangular matrix with vectors that $e_1, e_2, cdots, e_n$ that are orthogonal...these $q_1, q_2, cdots q_n $ I would imagine...you can't construct the orthogonal vectors without the coefficients...you're subtracting them off.
– Geronimo
2 days ago
Lang then states that one can write $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$. I understand why $u$ is thus unipotent, but could you please explain how he guarantees $B=au$. Thank you
– Daniele1234
2 days ago
This is after having stated, $gB=k$, where $k$ is unitary , so $g=kB^-1$, so $g^-1=Bk^-1$ (I understand this) and (thus?) $B=au$.
– Daniele1234
2 days ago
@Daniele1234 I'll have to look at Langs book
– Geronimo
2 days ago
 |Â
show 21 more comments
So this is just a reference to the QR decomposition, which uses the Gram-Schmidt process
– Daniele1234
2 days ago
1
The QR decomposition is Gram-Schmidt. It decomposes a matrix into an orthogonal matrix and an upper triangular matrix of coefficients. Your question was constructing the matrix B which was an upper triangular matrix with vectors that $e_1, e_2, cdots, e_n$ that are orthogonal...these $q_1, q_2, cdots q_n $ I would imagine...you can't construct the orthogonal vectors without the coefficients...you're subtracting them off.
– Geronimo
2 days ago
Lang then states that one can write $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$. I understand why $u$ is thus unipotent, but could you please explain how he guarantees $B=au$. Thank you
– Daniele1234
2 days ago
This is after having stated, $gB=k$, where $k$ is unitary , so $g=kB^-1$, so $g^-1=Bk^-1$ (I understand this) and (thus?) $B=au$.
– Daniele1234
2 days ago
@Daniele1234 I'll have to look at Langs book
– Geronimo
2 days ago
So this is just a reference to the QR decomposition, which uses the Gram-Schmidt process
– Daniele1234
2 days ago
So this is just a reference to the QR decomposition, which uses the Gram-Schmidt process
– Daniele1234
2 days ago
1
1
The QR decomposition is Gram-Schmidt. It decomposes a matrix into an orthogonal matrix and an upper triangular matrix of coefficients. Your question was constructing the matrix B which was an upper triangular matrix with vectors that $e_1, e_2, cdots, e_n$ that are orthogonal...these $q_1, q_2, cdots q_n $ I would imagine...you can't construct the orthogonal vectors without the coefficients...you're subtracting them off.
– Geronimo
2 days ago
The QR decomposition is Gram-Schmidt. It decomposes a matrix into an orthogonal matrix and an upper triangular matrix of coefficients. Your question was constructing the matrix B which was an upper triangular matrix with vectors that $e_1, e_2, cdots, e_n$ that are orthogonal...these $q_1, q_2, cdots q_n $ I would imagine...you can't construct the orthogonal vectors without the coefficients...you're subtracting them off.
– Geronimo
2 days ago
Lang then states that one can write $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$. I understand why $u$ is thus unipotent, but could you please explain how he guarantees $B=au$. Thank you
– Daniele1234
2 days ago
Lang then states that one can write $B=au$, where $a$ is the diagonal matrix with $a_i=b_ii$ and $u$ is unipotent, $u=a^-1B$. I understand why $u$ is thus unipotent, but could you please explain how he guarantees $B=au$. Thank you
– Daniele1234
2 days ago
This is after having stated, $gB=k$, where $k$ is unitary , so $g=kB^-1$, so $g^-1=Bk^-1$ (I understand this) and (thus?) $B=au$.
– Daniele1234
2 days ago
This is after having stated, $gB=k$, where $k$ is unitary , so $g=kB^-1$, so $g^-1=Bk^-1$ (I understand this) and (thus?) $B=au$.
– Daniele1234
2 days ago
@Daniele1234 I'll have to look at Langs book
– Geronimo
2 days ago
@Daniele1234 I'll have to look at Langs book
– Geronimo
2 days ago
 |Â
show 21 more comments
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