Mixing
Every invertible matrix can be written as a finite composition of elementary matrices with real eigenvalues?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Suppose $Psi$ is an invertible $ntimes n$ matrix that does not have only real eigenvalues. I read that such a matrix can be written as a finite composition of elementary matrices with real eigenvalues. Specifically:
Every automorphism of $mathbbR^n$ is a finite composition of
automorphisms with real eigenvalues (elementary matrices).
I have not been able to find a proof of this, does anyone know how to show it?
matrices group-theory matrix-decomposition
edited Jul 29 at 11:25
William Elliot
5,0722414
asked Jul 29 at 9:58
csss
1,22811221
add a comment |Â
up vote
0
down vote
favorite
Suppose $Psi$ is an invertible $ntimes n$ matrix that does not have only real eigenvalues. I read that such a matrix can be written as a finite composition of elementary matrices with real eigenvalues. Specifically:
Every automorphism of $mathbbR^n$ is a finite composition of
automorphisms with real eigenvalues (elementary matrices).
I have not been able to find a proof of this, does anyone know how to show it?
matrices group-theory matrix-decomposition
edited Jul 29 at 11:25
William Elliot
5,0722414
asked Jul 29 at 9:58
csss
1,22811221
3
The title claim is already wrong for $n=1$. Take the $1times 1$-matrix $A=(i)$ with $i^2=-1$.
– Dietrich Burde
Jul 29 at 10:03
@DietrichBurde This statement is used in Step 6. on page 70 of these notes: people.math.ethz.ch/~salamon/PREPRINTS/measure.pdf - are these notes (which have now been released as a book) wrong?
– csss
Jul 31 at 15:15
Dear csss, the statement with automorphisms is correct, but the statement in your title is not correct, as you see already for the case $n=1$.
– Dietrich Burde
Jul 31 at 16:58
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $Psi$ is an invertible $ntimes n$ matrix that does not have only real eigenvalues. I read that such a matrix can be written as a finite composition of elementary matrices with real eigenvalues. Specifically:
Every automorphism of $mathbbR^n$ is a finite composition of
automorphisms with real eigenvalues (elementary matrices).
I have not been able to find a proof of this, does anyone know how to show it?
matrices group-theory matrix-decomposition
edited Jul 29 at 11:25
William Elliot
5,0722414
asked Jul 29 at 9:58
csss
1,22811221
Suppose $Psi$ is an invertible $ntimes n$ matrix that does not have only real eigenvalues. I read that such a matrix can be written as a finite composition of elementary matrices with real eigenvalues. Specifically:
Every automorphism of $mathbbR^n$ is a finite composition of
automorphisms with real eigenvalues (elementary matrices).
I have not been able to find a proof of this, does anyone know how to show it?
matrices group-theory matrix-decomposition
edited Jul 29 at 11:25
William Elliot
5,0722414
asked Jul 29 at 9:58
csss
1,22811221
edited Jul 29 at 11:25
William Elliot
5,0722414
edited Jul 29 at 11:25
William Elliot
5,0722414
edited Jul 29 at 11:25
William Elliot
5,0722414
5,0722414
asked Jul 29 at 9:58
csss
1,22811221
asked Jul 29 at 9:58
csss
1,22811221
asked Jul 29 at 9:58
csss
1,22811221
1,22811221
3
The title claim is already wrong for $n=1$. Take the $1times 1$-matrix $A=(i)$ with $i^2=-1$.
– Dietrich Burde
Jul 29 at 10:03
@DietrichBurde This statement is used in Step 6. on page 70 of these notes: people.math.ethz.ch/~salamon/PREPRINTS/measure.pdf - are these notes (which have now been released as a book) wrong?
– csss
Jul 31 at 15:15
Dear csss, the statement with automorphisms is correct, but the statement in your title is not correct, as you see already for the case $n=1$.
– Dietrich Burde
Jul 31 at 16:58
add a comment |Â
3
The title claim is already wrong for $n=1$. Take the $1times 1$-matrix $A=(i)$ with $i^2=-1$.
– Dietrich Burde
Jul 29 at 10:03
@DietrichBurde This statement is used in Step 6. on page 70 of these notes: people.math.ethz.ch/~salamon/PREPRINTS/measure.pdf - are these notes (which have now been released as a book) wrong?
– csss
Jul 31 at 15:15
Dear csss, the statement with automorphisms is correct, but the statement in your title is not correct, as you see already for the case $n=1$.
– Dietrich Burde
Jul 31 at 16:58
3
3
The title claim is already wrong for $n=1$. Take the $1times 1$-matrix $A=(i)$ with $i^2=-1$.
– Dietrich Burde
Jul 29 at 10:03
The title claim is already wrong for $n=1$. Take the $1times 1$-matrix $A=(i)$ with $i^2=-1$.
– Dietrich Burde
Jul 29 at 10:03
@DietrichBurde This statement is used in Step 6. on page 70 of these notes: people.math.ethz.ch/~salamon/PREPRINTS/measure.pdf - are these notes (which have now been released as a book) wrong?
– csss
Jul 31 at 15:15
@DietrichBurde This statement is used in Step 6. on page 70 of these notes: people.math.ethz.ch/~salamon/PREPRINTS/measure.pdf - are these notes (which have now been released as a book) wrong?
– csss
Jul 31 at 15:15
Dear csss, the statement with automorphisms is correct, but the statement in your title is not correct, as you see already for the case $n=1$.
– Dietrich Burde
Jul 31 at 16:58
Dear csss, the statement with automorphisms is correct, but the statement in your title is not correct, as you see already for the case $n=1$.
– Dietrich Burde
Jul 31 at 16:58
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
It's a standard result from introductory linear algebra that, by multiplying by elementary matrices, you can row reduce any matrix into row reduced echelon form.
In the case of an invertible matrix, the row reduced echelon form is the identity matrix, which is also an elementary matrix, and so you get the theorem.
answered Jul 29 at 10:08
Hurkyl
107k9112253
I don't understand. Say we have an invertible matrix $M$. How can the fact that $M$ has the identity as a row-reduced echelon form be used to show that $M$ is a finite composition of matrices with real eigenvalues?
– csss
Jul 31 at 15:22
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
It's a standard result from introductory linear algebra that, by multiplying by elementary matrices, you can row reduce any matrix into row reduced echelon form.
In the case of an invertible matrix, the row reduced echelon form is the identity matrix, which is also an elementary matrix, and so you get the theorem.
answered Jul 29 at 10:08
Hurkyl
107k9112253
I don't understand. Say we have an invertible matrix $M$. How can the fact that $M$ has the identity as a row-reduced echelon form be used to show that $M$ is a finite composition of matrices with real eigenvalues?
– csss
Jul 31 at 15:22
add a comment |Â
up vote
2
down vote
It's a standard result from introductory linear algebra that, by multiplying by elementary matrices, you can row reduce any matrix into row reduced echelon form.
In the case of an invertible matrix, the row reduced echelon form is the identity matrix, which is also an elementary matrix, and so you get the theorem.
answered Jul 29 at 10:08
Hurkyl
107k9112253
I don't understand. Say we have an invertible matrix $M$. How can the fact that $M$ has the identity as a row-reduced echelon form be used to show that $M$ is a finite composition of matrices with real eigenvalues?
– csss
Jul 31 at 15:22
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It's a standard result from introductory linear algebra that, by multiplying by elementary matrices, you can row reduce any matrix into row reduced echelon form.
In the case of an invertible matrix, the row reduced echelon form is the identity matrix, which is also an elementary matrix, and so you get the theorem.
answered Jul 29 at 10:08
Hurkyl
107k9112253
It's a standard result from introductory linear algebra that, by multiplying by elementary matrices, you can row reduce any matrix into row reduced echelon form.
In the case of an invertible matrix, the row reduced echelon form is the identity matrix, which is also an elementary matrix, and so you get the theorem.
answered Jul 29 at 10:08
Hurkyl
107k9112253
answered Jul 29 at 10:08
Hurkyl
107k9112253
answered Jul 29 at 10:08
Hurkyl
107k9112253
answered Jul 29 at 10:08
Hurkyl
107k9112253
107k9112253
I don't understand. Say we have an invertible matrix $M$. How can the fact that $M$ has the identity as a row-reduced echelon form be used to show that $M$ is a finite composition of matrices with real eigenvalues?
– csss
Jul 31 at 15:22
add a comment |Â
I don't understand. Say we have an invertible matrix $M$. How can the fact that $M$ has the identity as a row-reduced echelon form be used to show that $M$ is a finite composition of matrices with real eigenvalues?
– csss
Jul 31 at 15:22
I don't understand. Say we have an invertible matrix $M$. How can the fact that $M$ has the identity as a row-reduced echelon form be used to show that $M$ is a finite composition of matrices with real eigenvalues?
– csss
Jul 31 at 15:22
I don't understand. Say we have an invertible matrix $M$. How can the fact that $M$ has the identity as a row-reduced echelon form be used to show that $M$ is a finite composition of matrices with real eigenvalues?
– csss
Jul 31 at 15:22
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%2f2865942%2fevery-invertible-matrix-can-be-written-as-a-finite-composition-of-elementary-mat%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
The title claim is already wrong for $n=1$. Take the $1times 1$-matrix $A=(i)$ with $i^2=-1$.
– Dietrich Burde
Jul 29 at 10:03
@DietrichBurde This statement is used in Step 6. on page 70 of these notes: people.math.ethz.ch/~salamon/PREPRINTS/measure.pdf - are these notes (which have now been released as a book) wrong?
– csss
Jul 31 at 15:15
Dear csss, the statement with automorphisms is correct, but the statement in your title is not correct, as you see already for the case $n=1$.
– Dietrich Burde
Jul 31 at 16:58