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A “unique” solution to an equation over the orthogonal matrices?

0

$begingroup$

Set $D=textdiag(-1,1,1,dots ,1)$ be an $n times n$ real diagonal matrix (where $D_11=-1$ and $D_ii=1$ for $i>1$).

Let $R,Q$ be special orthogonal matrices, satisfying $RDQ=D$. Is it true that $R=Q^-1$?

I know that $Q^TDR^T=D$, so $Q^TDR^T=RDQ Rightarrow UDU=D$, where $U=R^TQ^T$.

Is there a slick way to continue from here (to shows that $U=textId$? $UDU=D$ is equivalent to $UD=DU^T$, i.e. $U_ijD_j=D_iU_ji$.

matrix-equations matrix-calculus orthogonality orthogonal-matrices

share|cite|improve this question

edited Mar 26 at 10:01

Asaf Shachar

asked Mar 26 at 9:43

Asaf ShacharAsaf Shachar

5,79931145

$endgroup$

add a comment | 

0

$begingroup$

Set $D=textdiag(-1,1,1,dots ,1)$ be an $n times n$ real diagonal matrix (where $D_11=-1$ and $D_ii=1$ for $i>1$).

Let $R,Q$ be special orthogonal matrices, satisfying $RDQ=D$. Is it true that $R=Q^-1$?

I know that $Q^TDR^T=D$, so $Q^TDR^T=RDQ Rightarrow UDU=D$, where $U=R^TQ^T$.

Is there a slick way to continue from here (to shows that $U=textId$? $UDU=D$ is equivalent to $UD=DU^T$, i.e. $U_ijD_j=D_iU_ji$.

matrix-equations matrix-calculus orthogonality orthogonal-matrices

share|cite|improve this question

edited Mar 26 at 10:01

Asaf Shachar

asked Mar 26 at 9:43

Asaf ShacharAsaf Shachar

5,79931145

$endgroup$

add a comment | 

$begingroup$

Set $D=textdiag(-1,1,1,dots ,1)$ be an $n times n$ real diagonal matrix (where $D_11=-1$ and $D_ii=1$ for $i>1$).

Let $R,Q$ be special orthogonal matrices, satisfying $RDQ=D$. Is it true that $R=Q^-1$?

I know that $Q^TDR^T=D$, so $Q^TDR^T=RDQ Rightarrow UDU=D$, where $U=R^TQ^T$.

Is there a slick way to continue from here (to shows that $U=textId$? $UDU=D$ is equivalent to $UD=DU^T$, i.e. $U_ijD_j=D_iU_ji$.

matrix-equations matrix-calculus orthogonality orthogonal-matrices

share|cite|improve this question

edited Mar 26 at 10:01

Asaf Shachar

asked Mar 26 at 9:43

Asaf ShacharAsaf Shachar

5,79931145

$endgroup$

share|cite|improve this question

edited Mar 26 at 10:01

Asaf Shachar

asked Mar 26 at 9:43

Asaf ShacharAsaf Shachar

5,79931145

share|cite|improve this question

edited Mar 26 at 10:01

Asaf Shachar

asked Mar 26 at 9:43

Asaf ShacharAsaf Shachar

5,79931145

asked Mar 26 at 9:43

Asaf ShacharAsaf Shachar

5,79931145

asked Mar 26 at 9:43

Asaf ShacharAsaf Shachar

5,79931145

1 Answer
1

active

oldest

votes

1

$begingroup$

If $R=Q=beginpmatrix0&-1\1&0endpmatrix$ then $RDQ=D$ but $R=-Q^-1neq Q^-1.$

$RDQ=D$ implies $Q^-1=D^-1RD.$ So the question is whether $R=D^-1RD.$ This is true if and only if $R$ is of the form
$$
left(beginarrayc
pm 1 & 0_1times (n-1)\
hline
0_(n-1)times 1 & R'
endarray
right) $$
(Here $R'$ must be orthogonal. It must be special orthogonal if and only if the top left entry is $1.$)

share|cite|improve this answer

answered Mar 26 at 18:23

DapDap

20k842

$endgroup$

add a comment | 

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1

$begingroup$

If $R=Q=beginpmatrix0&-1\1&0endpmatrix$ then $RDQ=D$ but $R=-Q^-1neq Q^-1.$

$RDQ=D$ implies $Q^-1=D^-1RD.$ So the question is whether $R=D^-1RD.$ This is true if and only if $R$ is of the form
$$
left(beginarrayc
pm 1 & 0_1times (n-1)\
hline
0_(n-1)times 1 & R'
endarray
right) $$
(Here $R'$ must be orthogonal. It must be special orthogonal if and only if the top left entry is $1.$)

share|cite|improve this answer

answered Mar 26 at 18:23

DapDap

20k842

$endgroup$

add a comment | 

1

$begingroup$

If $R=Q=beginpmatrix0&-1\1&0endpmatrix$ then $RDQ=D$ but $R=-Q^-1neq Q^-1.$

$RDQ=D$ implies $Q^-1=D^-1RD.$ So the question is whether $R=D^-1RD.$ This is true if and only if $R$ is of the form
$$
left(beginarrayc
pm 1 & 0_1times (n-1)\
hline
0_(n-1)times 1 & R'
endarray
right) $$
(Here $R'$ must be orthogonal. It must be special orthogonal if and only if the top left entry is $1.$)

share|cite|improve this answer

answered Mar 26 at 18:23

DapDap

20k842

$endgroup$

add a comment | 

$begingroup$

If $R=Q=beginpmatrix0&-1\1&0endpmatrix$ then $RDQ=D$ but $R=-Q^-1neq Q^-1.$

$RDQ=D$ implies $Q^-1=D^-1RD.$ So the question is whether $R=D^-1RD.$ This is true if and only if $R$ is of the form
$$
left(beginarrayc
pm 1 & 0_1times (n-1)\
hline
0_(n-1)times 1 & R'
endarray
right) $$
(Here $R'$ must be orthogonal. It must be special orthogonal if and only if the top left entry is $1.$)

share|cite|improve this answer

answered Mar 26 at 18:23

DapDap

20k842

$endgroup$

share|cite|improve this answer

answered Mar 26 at 18:23

DapDap

20k842

answered Mar 26 at 18:23

DapDap

20k842

answered Mar 26 at 18:23

DapDap

20k842