Show that there is an orthogonal matrix $O$ such that $OA_1=A_2O$.












1












$begingroup$


Let $A_1,A_2$ be two real $n times n$ matrix. And suppose that they are two orthogonal and anti-symmetric matrices. Show that there is an orthogonal matrix $O$ such that $OA_1=A_2O$.



I have no idea how to find such $O$.






linear-algebra matrices matrix-equations orthogonal-matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    So, are both $A_1,A_2$ orthogonal and antisymmetric at once?
    $endgroup$
    – Berci
    Dec 7 '18 at 21:53










  • $begingroup$
    @Berci Yes, $A_1,A_2$ are orthogonal and anti-symmetric matrix.
    $endgroup$
    – whereamI
    Dec 7 '18 at 21:57
















1












$begingroup$


Let $A_1,A_2$ be two real $n times n$ matrix. And suppose that they are two orthogonal and anti-symmetric matrices. Show that there is an orthogonal matrix $O$ such that $OA_1=A_2O$.



I have no idea how to find such $O$.






linear-algebra matrices matrix-equations orthogonal-matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    So, are both $A_1,A_2$ orthogonal and antisymmetric at once?
    $endgroup$
    – Berci
    Dec 7 '18 at 21:53










  • $begingroup$
    @Berci Yes, $A_1,A_2$ are orthogonal and anti-symmetric matrix.
    $endgroup$
    – whereamI
    Dec 7 '18 at 21:57














1












1








1


1



$begingroup$


Let $A_1,A_2$ be two real $n times n$ matrix. And suppose that they are two orthogonal and anti-symmetric matrices. Show that there is an orthogonal matrix $O$ such that $OA_1=A_2O$.



I have no idea how to find such $O$.






linear-algebra matrices matrix-equations orthogonal-matrices







share|cite|improve this question











$endgroup$




Let $A_1,A_2$ be two real $n times n$ matrix. And suppose that they are two orthogonal and anti-symmetric matrices. Show that there is an orthogonal matrix $O$ such that $OA_1=A_2O$.



I have no idea how to find such $O$.






linear-algebra matrices matrix-equations orthogonal-matrices




linear-algebra matrices matrix-equations orthogonal-matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 22:31









Rodrigo de Azevedo

13k41960




13k41960










asked Dec 7 '18 at 21:42









whereamIwhereamI

327115




327115












  • $begingroup$
    So, are both $A_1,A_2$ orthogonal and antisymmetric at once?
    $endgroup$
    – Berci
    Dec 7 '18 at 21:53










  • $begingroup$
    @Berci Yes, $A_1,A_2$ are orthogonal and anti-symmetric matrix.
    $endgroup$
    – whereamI
    Dec 7 '18 at 21:57


















  • $begingroup$
    So, are both $A_1,A_2$ orthogonal and antisymmetric at once?
    $endgroup$
    – Berci
    Dec 7 '18 at 21:53










  • $begingroup$
    @Berci Yes, $A_1,A_2$ are orthogonal and anti-symmetric matrix.
    $endgroup$
    – whereamI
    Dec 7 '18 at 21:57
















$begingroup$
So, are both $A_1,A_2$ orthogonal and antisymmetric at once?
$endgroup$
– Berci
Dec 7 '18 at 21:53




$begingroup$
So, are both $A_1,A_2$ orthogonal and antisymmetric at once?
$endgroup$
– Berci
Dec 7 '18 at 21:53












$begingroup$
@Berci Yes, $A_1,A_2$ are orthogonal and anti-symmetric matrix.
$endgroup$
– whereamI
Dec 7 '18 at 21:57




$begingroup$
@Berci Yes, $A_1,A_2$ are orthogonal and anti-symmetric matrix.
$endgroup$
– whereamI
Dec 7 '18 at 21:57










1 Answer
1






active

oldest

votes


















2












$begingroup$

By assumption we have $A_i^2=-I_n$ for $i=1,2$, where $I_n$ is the identity matrix. In particular, $A_i$ is invertible.



But any invertible anti-symmetric matrix is a symplectic form, that is, it must be similar to the canonical form
$$begin{pmatrix} 0 & I_{n/2} \ -I_{n/2} & 0 end{pmatrix}.$$
Thus, $A_1$ and $A_2$ are similar. Note that $A_i$ is orthogonal, so we can choose the transition matrix to be orthogonal too.



For example, choose any symplectic basis $e_1,ldots,e_n$. (This is a standard result which can be done by induction.) Let's write $n=2m$. Then we have
$$A_i(e_1,ldots,e_m)=(e_{m+1},ldots,e_{2m}), quad A_i(e_{m+1},ldots,e_{2m})=-(e_1,ldots,e_m).$$
Apply Gram-Schmidt to $e_1,ldots,e_m$. We have
$$(e_1,ldots,e_m)T=(e_1',ldots,e_m'),quad (e_{m+1},ldots,e_{2m})T=(e_{m+1}',ldots,e_{2m}').$$
Here $e_1',ldots,e_m'$ are mutually orthogonal and of unit length. It is easy to show that $e_{m+1}',ldots,e_{2m}'$ are also mutually orthogonal and of unit length, and that $e_1',ldots,e_{2m}'$ is also a symplectic basis. Now under this new basis, which is orthonormal, $A_i$ has the desired matrix, and the transition matrix is orthogonal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot! I don't know the invertible anti-symmetric matrix has such a property. And would you mind explain a bit more about how to choose the transition matrix to be orthogonal? It is not so obvious to me.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:12












  • $begingroup$
    @whereamI I added some details. Hope it is clearer now.
    $endgroup$
    – Eclipse Sun
    Dec 7 '18 at 22:26










  • $begingroup$
    Thank you very much! I got it now.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:31











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

By assumption we have $A_i^2=-I_n$ for $i=1,2$, where $I_n$ is the identity matrix. In particular, $A_i$ is invertible.



But any invertible anti-symmetric matrix is a symplectic form, that is, it must be similar to the canonical form
$$begin{pmatrix} 0 & I_{n/2} \ -I_{n/2} & 0 end{pmatrix}.$$
Thus, $A_1$ and $A_2$ are similar. Note that $A_i$ is orthogonal, so we can choose the transition matrix to be orthogonal too.



For example, choose any symplectic basis $e_1,ldots,e_n$. (This is a standard result which can be done by induction.) Let's write $n=2m$. Then we have
$$A_i(e_1,ldots,e_m)=(e_{m+1},ldots,e_{2m}), quad A_i(e_{m+1},ldots,e_{2m})=-(e_1,ldots,e_m).$$
Apply Gram-Schmidt to $e_1,ldots,e_m$. We have
$$(e_1,ldots,e_m)T=(e_1',ldots,e_m'),quad (e_{m+1},ldots,e_{2m})T=(e_{m+1}',ldots,e_{2m}').$$
Here $e_1',ldots,e_m'$ are mutually orthogonal and of unit length. It is easy to show that $e_{m+1}',ldots,e_{2m}'$ are also mutually orthogonal and of unit length, and that $e_1',ldots,e_{2m}'$ is also a symplectic basis. Now under this new basis, which is orthonormal, $A_i$ has the desired matrix, and the transition matrix is orthogonal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot! I don't know the invertible anti-symmetric matrix has such a property. And would you mind explain a bit more about how to choose the transition matrix to be orthogonal? It is not so obvious to me.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:12












  • $begingroup$
    @whereamI I added some details. Hope it is clearer now.
    $endgroup$
    – Eclipse Sun
    Dec 7 '18 at 22:26










  • $begingroup$
    Thank you very much! I got it now.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:31
















2












$begingroup$

By assumption we have $A_i^2=-I_n$ for $i=1,2$, where $I_n$ is the identity matrix. In particular, $A_i$ is invertible.



But any invertible anti-symmetric matrix is a symplectic form, that is, it must be similar to the canonical form
$$begin{pmatrix} 0 & I_{n/2} \ -I_{n/2} & 0 end{pmatrix}.$$
Thus, $A_1$ and $A_2$ are similar. Note that $A_i$ is orthogonal, so we can choose the transition matrix to be orthogonal too.



For example, choose any symplectic basis $e_1,ldots,e_n$. (This is a standard result which can be done by induction.) Let's write $n=2m$. Then we have
$$A_i(e_1,ldots,e_m)=(e_{m+1},ldots,e_{2m}), quad A_i(e_{m+1},ldots,e_{2m})=-(e_1,ldots,e_m).$$
Apply Gram-Schmidt to $e_1,ldots,e_m$. We have
$$(e_1,ldots,e_m)T=(e_1',ldots,e_m'),quad (e_{m+1},ldots,e_{2m})T=(e_{m+1}',ldots,e_{2m}').$$
Here $e_1',ldots,e_m'$ are mutually orthogonal and of unit length. It is easy to show that $e_{m+1}',ldots,e_{2m}'$ are also mutually orthogonal and of unit length, and that $e_1',ldots,e_{2m}'$ is also a symplectic basis. Now under this new basis, which is orthonormal, $A_i$ has the desired matrix, and the transition matrix is orthogonal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot! I don't know the invertible anti-symmetric matrix has such a property. And would you mind explain a bit more about how to choose the transition matrix to be orthogonal? It is not so obvious to me.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:12












  • $begingroup$
    @whereamI I added some details. Hope it is clearer now.
    $endgroup$
    – Eclipse Sun
    Dec 7 '18 at 22:26










  • $begingroup$
    Thank you very much! I got it now.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:31














2












2








2





$begingroup$

By assumption we have $A_i^2=-I_n$ for $i=1,2$, where $I_n$ is the identity matrix. In particular, $A_i$ is invertible.



But any invertible anti-symmetric matrix is a symplectic form, that is, it must be similar to the canonical form
$$begin{pmatrix} 0 & I_{n/2} \ -I_{n/2} & 0 end{pmatrix}.$$
Thus, $A_1$ and $A_2$ are similar. Note that $A_i$ is orthogonal, so we can choose the transition matrix to be orthogonal too.



For example, choose any symplectic basis $e_1,ldots,e_n$. (This is a standard result which can be done by induction.) Let's write $n=2m$. Then we have
$$A_i(e_1,ldots,e_m)=(e_{m+1},ldots,e_{2m}), quad A_i(e_{m+1},ldots,e_{2m})=-(e_1,ldots,e_m).$$
Apply Gram-Schmidt to $e_1,ldots,e_m$. We have
$$(e_1,ldots,e_m)T=(e_1',ldots,e_m'),quad (e_{m+1},ldots,e_{2m})T=(e_{m+1}',ldots,e_{2m}').$$
Here $e_1',ldots,e_m'$ are mutually orthogonal and of unit length. It is easy to show that $e_{m+1}',ldots,e_{2m}'$ are also mutually orthogonal and of unit length, and that $e_1',ldots,e_{2m}'$ is also a symplectic basis. Now under this new basis, which is orthonormal, $A_i$ has the desired matrix, and the transition matrix is orthogonal.






share|cite|improve this answer











$endgroup$



By assumption we have $A_i^2=-I_n$ for $i=1,2$, where $I_n$ is the identity matrix. In particular, $A_i$ is invertible.



But any invertible anti-symmetric matrix is a symplectic form, that is, it must be similar to the canonical form
$$begin{pmatrix} 0 & I_{n/2} \ -I_{n/2} & 0 end{pmatrix}.$$
Thus, $A_1$ and $A_2$ are similar. Note that $A_i$ is orthogonal, so we can choose the transition matrix to be orthogonal too.



For example, choose any symplectic basis $e_1,ldots,e_n$. (This is a standard result which can be done by induction.) Let's write $n=2m$. Then we have
$$A_i(e_1,ldots,e_m)=(e_{m+1},ldots,e_{2m}), quad A_i(e_{m+1},ldots,e_{2m})=-(e_1,ldots,e_m).$$
Apply Gram-Schmidt to $e_1,ldots,e_m$. We have
$$(e_1,ldots,e_m)T=(e_1',ldots,e_m'),quad (e_{m+1},ldots,e_{2m})T=(e_{m+1}',ldots,e_{2m}').$$
Here $e_1',ldots,e_m'$ are mutually orthogonal and of unit length. It is easy to show that $e_{m+1}',ldots,e_{2m}'$ are also mutually orthogonal and of unit length, and that $e_1',ldots,e_{2m}'$ is also a symplectic basis. Now under this new basis, which is orthonormal, $A_i$ has the desired matrix, and the transition matrix is orthogonal.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '18 at 22:25

























answered Dec 7 '18 at 22:00









Eclipse SunEclipse Sun

7,8801438




7,8801438












  • $begingroup$
    Thanks a lot! I don't know the invertible anti-symmetric matrix has such a property. And would you mind explain a bit more about how to choose the transition matrix to be orthogonal? It is not so obvious to me.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:12












  • $begingroup$
    @whereamI I added some details. Hope it is clearer now.
    $endgroup$
    – Eclipse Sun
    Dec 7 '18 at 22:26










  • $begingroup$
    Thank you very much! I got it now.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:31


















  • $begingroup$
    Thanks a lot! I don't know the invertible anti-symmetric matrix has such a property. And would you mind explain a bit more about how to choose the transition matrix to be orthogonal? It is not so obvious to me.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:12












  • $begingroup$
    @whereamI I added some details. Hope it is clearer now.
    $endgroup$
    – Eclipse Sun
    Dec 7 '18 at 22:26










  • $begingroup$
    Thank you very much! I got it now.
    $endgroup$
    – whereamI
    Dec 7 '18 at 22:31
















$begingroup$
Thanks a lot! I don't know the invertible anti-symmetric matrix has such a property. And would you mind explain a bit more about how to choose the transition matrix to be orthogonal? It is not so obvious to me.
$endgroup$
– whereamI
Dec 7 '18 at 22:12






$begingroup$
Thanks a lot! I don't know the invertible anti-symmetric matrix has such a property. And would you mind explain a bit more about how to choose the transition matrix to be orthogonal? It is not so obvious to me.
$endgroup$
– whereamI
Dec 7 '18 at 22:12














$begingroup$
@whereamI I added some details. Hope it is clearer now.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 22:26




$begingroup$
@whereamI I added some details. Hope it is clearer now.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 22:26












$begingroup$
Thank you very much! I got it now.
$endgroup$
– whereamI
Dec 7 '18 at 22:31




$begingroup$
Thank you very much! I got it now.
$endgroup$
– whereamI
Dec 7 '18 at 22:31


















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