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Can we characterize the set $S in GL_n(mathbb C): SAS^-1 in M_n(mathbb R)$
Clash Royale CLAN TAG#URR8PPP
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Let $A in M_n(mathbb R)$ be a fixed real matrix. Let
beginalign*
mathcal E = S in GL_n(mathbb C): S A S^-1 in M_n(mathbb R) .
endalign*
My questions:
- The set should contain $GL_n(mathbb R)$. But is this inclusion proper? In other words, is it true: for any real matrix $A$, there exists an invertible matrix with complex entries, such that $SAS^-1$ has only real entries?
- Is this set $mathcal E$ connected? I guess if the inclusion in question $1$ is always proper, we end up with a set $mathcal E$ that is larger than $GL_n(mathbb R)$ but in general smaller than $GL_n(mathbb C)$. I am wondering whether this set contains "enough elements" to connect the two components of $GL_n(mathbb R)$.
linear-algebra general-topology
asked Jul 28 at 23:47
user9527
923524
add a comment |Â
up vote
1
down vote
favorite
Let $A in M_n(mathbb R)$ be a fixed real matrix. Let
beginalign*
mathcal E = S in GL_n(mathbb C): S A S^-1 in M_n(mathbb R) .
endalign*
My questions:
- The set should contain $GL_n(mathbb R)$. But is this inclusion proper? In other words, is it true: for any real matrix $A$, there exists an invertible matrix with complex entries, such that $SAS^-1$ has only real entries?
- Is this set $mathcal E$ connected? I guess if the inclusion in question $1$ is always proper, we end up with a set $mathcal E$ that is larger than $GL_n(mathbb R)$ but in general smaller than $GL_n(mathbb C)$. I am wondering whether this set contains "enough elements" to connect the two components of $GL_n(mathbb R)$.
linear-algebra general-topology
asked Jul 28 at 23:47
user9527
923524
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A in M_n(mathbb R)$ be a fixed real matrix. Let
beginalign*
mathcal E = S in GL_n(mathbb C): S A S^-1 in M_n(mathbb R) .
endalign*
My questions:
- The set should contain $GL_n(mathbb R)$. But is this inclusion proper? In other words, is it true: for any real matrix $A$, there exists an invertible matrix with complex entries, such that $SAS^-1$ has only real entries?
- Is this set $mathcal E$ connected? I guess if the inclusion in question $1$ is always proper, we end up with a set $mathcal E$ that is larger than $GL_n(mathbb R)$ but in general smaller than $GL_n(mathbb C)$. I am wondering whether this set contains "enough elements" to connect the two components of $GL_n(mathbb R)$.
linear-algebra general-topology
asked Jul 28 at 23:47
user9527
923524
Let $A in M_n(mathbb R)$ be a fixed real matrix. Let
beginalign*
mathcal E = S in GL_n(mathbb C): S A S^-1 in M_n(mathbb R) .
endalign*
My questions:
- The set should contain $GL_n(mathbb R)$. But is this inclusion proper? In other words, is it true: for any real matrix $A$, there exists an invertible matrix with complex entries, such that $SAS^-1$ has only real entries?
- Is this set $mathcal E$ connected? I guess if the inclusion in question $1$ is always proper, we end up with a set $mathcal E$ that is larger than $GL_n(mathbb R)$ but in general smaller than $GL_n(mathbb C)$. I am wondering whether this set contains "enough elements" to connect the two components of $GL_n(mathbb R)$.
linear-algebra general-topology
asked Jul 28 at 23:47
user9527
923524
asked Jul 28 at 23:47
user9527
923524
asked Jul 28 at 23:47
user9527
923524
asked Jul 28 at 23:47
user9527
923524
923524
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Let $ngeq 2,E_A=Sin GL_n(mathbbC);SAS^-1in M_n(mathbbR)$ and $lambdainmathbbR,alphainmathbbR^*$.
Then $(*)$ $E_A+lambda I=E_alpha A=E_A$ and $E_lambda I=GL_n(mathbbC)$.
Let $S=M+iN,S^-1=P+iQ$ where $M,N,P,Q$ are real matrices.
Then the condition $Sin E_A$ is equivalent to $(*)$ $NAP+MAQ=0,NP+MQ=0,MP-NQ=I$.
Thus $E_A$ is a real algebraic set of dimension $dleq 2n^2$; since it contains $GL_n(mathbbR)$, $dgeq n^2$.
$textbfProposition 1.$ $d=2n^2$ iff $A$ is a scalar matrix.
$textbfProof.$ Assume that $A$ is not a scalar matrix. There is $xinmathbbC^n$ s.t. $x,Ax$ is a free system; we consider a basis in the form $x,-iAx,cdots$; then the associated $S_0AS_0^-1$ has $[0,i,0,cdots]^T$ as first column, $(S_0AS_0^-1)_2,1notin mathbbR$ and $S_0notin E_A$. Let $mathcalB$ be the set of bases of $mathbbC^n$.
Let $Z=(e_1,cdots,e_n)inmathcalB;Ae_1=sum_ja_je_j text with a_2inmathbbR$; note that $Znot= mathcalB$ (cf. above).
If $T=[t_i,j]=[e_1,cdots,e_n]$, then $(e_1,cdots,e_n)in Z$ iff $Im((T^-1AT)_2,1)=0$. Since $Im((T^-1AT)_2,1)$ is a rational fraction in the $(Re(t_i,j),Im(t_i,j))$, $Z$ is a real algebric set and, moreover, it is a Zariski closed strict subset of $mathcalB$. We deduce that $d<2n^2$. $square$
To find $d$ in the general case seems difficult (at least for me...). We consider the case $n=2$ and we will see that there are matrices $A$ s.t. $n^2<dim(E_A)<2n^2$.
$textbfProposition 2.$ Assume that $n=2$. If $A$ is a scalar matrix, then $d=8$, otherwise $d=6$.
$textbfProof$. We may assume that $A$ is in the form of its Jordan real form. According to $(*)$, it suffices to consider the three cases
i) $A=beginpmatrix0&0\0&1endpmatrix$, ii) $A=beginpmatrix0&1\0&0endpmatrix$, iii) $A=beginpmatrix0&1\-1&0endpmatrix$.
For i). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&gamma bendpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abnot= 0,betanot= gamma$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For ii). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&beta(b+gamma a)endpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abetagammanot= 0$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For iii). We obtain $S=beginpmatrixa&b\c&dendpmatrix$ where $p=dfracac+bdad-bcinmathbbR,q=dfraca^2+b^2ad-bcinmathbbR^*,r=dfracc^2+d^2ad-bcinmathbbR$ and $ad-bcnot= 0$ (or $pinmathbbR,qinmathbbR,rinmathbbR^*$)
We show that $dim(E_A)=6$. Note that $qnot=0$ and $r=dfrac1+p^2q$ is real if $p,q$ are. Assume that $a,b,p,q$ are independently given (6 degrees of freedom). Then $ad-bc$ and $ac+bd$ are known; since $a^2+b^2not= 0$ we deduce a unique couple $c,d$. $square$
$textbfProposition 3.$ For every $A$, $dim(E_A)geq n^2+1$.
$textbfProof$. If $Sin E_A$, then, for every $thetainmathbbR$, $e^ithetaSin E_A$.
EDIT. I do not know why you are passionate about connected sets... Here, the answer is no in general.
$textbfProposition 4.$ There are matrices $A$ s.t. $E_A$ is not connected.
$textbfProof.$ Let $Ain .M_2(mathbbR)$ s.t. $tr(A)=1,det(A)=4$ and $phi:Sin GL_n(mathbbC)rightarrow SAS^-1in M_n(mathbbC)$. Clearly $phi(E_A)=phi(GL_n(mathbbR))approx (p,q,r,s)inmathbbR^4;p+s=1,ps-rq=4$. In particular, $phi(E_A)$ has one or two connected components.
Let $Z=(p,q,r)inmathbbR^3;p(1-p)-rq=4$. $Z$ has exactly $2$ connected components (draw with Maple if you are in hurry). Suppose $phi(E_A)$ is connected; since $(p,q,r,s)in E_Arightarrow (p,q,r)in Z$ is a continuous function, this would cause $Z$ to be connected, a contradiction. Finally $phi(E_A)$ is not connected and $E_A$ too.
$textbfRemark.$ We can generalize the above proposition to matrices in $ M_n(mathbbR)$ that have conjugate eigenvalues.
answered Jul 30 at 10:05
loup blanc
20.3k21549
Thanks for answering my question. Could you comment on the connectedness of $E_A$? Is it possible the set is connected?
– user9527
Jul 30 at 16:01
Also in defining the variety, why the relations $PN+QM = 0$ and $PM-QN=I$ are not needed? Thanks.
– user9527
Jul 30 at 17:50
add a comment |Â
up vote
1
down vote
Assume that $SA=BS$ for some real matrix $B$. Write $S=M+iN$ where $M$ and $N$ are real. Let us plug in, expand, and then separate the real and imaginary parts of the equation.
$$MA+iNA=BM+iBN; quad textor quad MA=BM, NA=BN.$$
Thus, for fixed $B$, complex solutions to the equation correspond to pairs of real solutions.
answered Jul 29 at 0:03
Aaron
15.1k22552
I guess $M$ and $N$ are not necessarily invertible. So we can conclude the inclusion is proper. What about the second part?
– user9527
Jul 29 at 0:10
I am confused. Doesn't this mean the set contains general linear maps over $mathbb R$ as a proper subset?
– user9527
Jul 29 at 0:15
@user9527 Yes, this shows the inclusion is proper. Although, if you want to restrict to where $S$ is invertible, I'm not entirely sure on general conditions. You can always take $N=alpha N$ for any real $alpha$ as long as $N$ is invertible, but I don't see a clean way of finding the intersection of the set with invertible maps. But at a bare minimum, the set contains all complex multiples of real invertible matrices.
– Aaron
Jul 29 at 0:18
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $ngeq 2,E_A=Sin GL_n(mathbbC);SAS^-1in M_n(mathbbR)$ and $lambdainmathbbR,alphainmathbbR^*$.
Then $(*)$ $E_A+lambda I=E_alpha A=E_A$ and $E_lambda I=GL_n(mathbbC)$.
Let $S=M+iN,S^-1=P+iQ$ where $M,N,P,Q$ are real matrices.
Then the condition $Sin E_A$ is equivalent to $(*)$ $NAP+MAQ=0,NP+MQ=0,MP-NQ=I$.
Thus $E_A$ is a real algebraic set of dimension $dleq 2n^2$; since it contains $GL_n(mathbbR)$, $dgeq n^2$.
$textbfProposition 1.$ $d=2n^2$ iff $A$ is a scalar matrix.
$textbfProof.$ Assume that $A$ is not a scalar matrix. There is $xinmathbbC^n$ s.t. $x,Ax$ is a free system; we consider a basis in the form $x,-iAx,cdots$; then the associated $S_0AS_0^-1$ has $[0,i,0,cdots]^T$ as first column, $(S_0AS_0^-1)_2,1notin mathbbR$ and $S_0notin E_A$. Let $mathcalB$ be the set of bases of $mathbbC^n$.
Let $Z=(e_1,cdots,e_n)inmathcalB;Ae_1=sum_ja_je_j text with a_2inmathbbR$; note that $Znot= mathcalB$ (cf. above).
If $T=[t_i,j]=[e_1,cdots,e_n]$, then $(e_1,cdots,e_n)in Z$ iff $Im((T^-1AT)_2,1)=0$. Since $Im((T^-1AT)_2,1)$ is a rational fraction in the $(Re(t_i,j),Im(t_i,j))$, $Z$ is a real algebric set and, moreover, it is a Zariski closed strict subset of $mathcalB$. We deduce that $d<2n^2$. $square$
To find $d$ in the general case seems difficult (at least for me...). We consider the case $n=2$ and we will see that there are matrices $A$ s.t. $n^2<dim(E_A)<2n^2$.
$textbfProposition 2.$ Assume that $n=2$. If $A$ is a scalar matrix, then $d=8$, otherwise $d=6$.
$textbfProof$. We may assume that $A$ is in the form of its Jordan real form. According to $(*)$, it suffices to consider the three cases
i) $A=beginpmatrix0&0\0&1endpmatrix$, ii) $A=beginpmatrix0&1\0&0endpmatrix$, iii) $A=beginpmatrix0&1\-1&0endpmatrix$.
For i). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&gamma bendpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abnot= 0,betanot= gamma$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For ii). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&beta(b+gamma a)endpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abetagammanot= 0$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For iii). We obtain $S=beginpmatrixa&b\c&dendpmatrix$ where $p=dfracac+bdad-bcinmathbbR,q=dfraca^2+b^2ad-bcinmathbbR^*,r=dfracc^2+d^2ad-bcinmathbbR$ and $ad-bcnot= 0$ (or $pinmathbbR,qinmathbbR,rinmathbbR^*$)
We show that $dim(E_A)=6$. Note that $qnot=0$ and $r=dfrac1+p^2q$ is real if $p,q$ are. Assume that $a,b,p,q$ are independently given (6 degrees of freedom). Then $ad-bc$ and $ac+bd$ are known; since $a^2+b^2not= 0$ we deduce a unique couple $c,d$. $square$
$textbfProposition 3.$ For every $A$, $dim(E_A)geq n^2+1$.
$textbfProof$. If $Sin E_A$, then, for every $thetainmathbbR$, $e^ithetaSin E_A$.
EDIT. I do not know why you are passionate about connected sets... Here, the answer is no in general.
$textbfProposition 4.$ There are matrices $A$ s.t. $E_A$ is not connected.
$textbfProof.$ Let $Ain .M_2(mathbbR)$ s.t. $tr(A)=1,det(A)=4$ and $phi:Sin GL_n(mathbbC)rightarrow SAS^-1in M_n(mathbbC)$. Clearly $phi(E_A)=phi(GL_n(mathbbR))approx (p,q,r,s)inmathbbR^4;p+s=1,ps-rq=4$. In particular, $phi(E_A)$ has one or two connected components.
Let $Z=(p,q,r)inmathbbR^3;p(1-p)-rq=4$. $Z$ has exactly $2$ connected components (draw with Maple if you are in hurry). Suppose $phi(E_A)$ is connected; since $(p,q,r,s)in E_Arightarrow (p,q,r)in Z$ is a continuous function, this would cause $Z$ to be connected, a contradiction. Finally $phi(E_A)$ is not connected and $E_A$ too.
$textbfRemark.$ We can generalize the above proposition to matrices in $ M_n(mathbbR)$ that have conjugate eigenvalues.
answered Jul 30 at 10:05
loup blanc
20.3k21549
Thanks for answering my question. Could you comment on the connectedness of $E_A$? Is it possible the set is connected?
– user9527
Jul 30 at 16:01
Also in defining the variety, why the relations $PN+QM = 0$ and $PM-QN=I$ are not needed? Thanks.
– user9527
Jul 30 at 17:50
add a comment |Â
up vote
1
down vote
accepted
Let $ngeq 2,E_A=Sin GL_n(mathbbC);SAS^-1in M_n(mathbbR)$ and $lambdainmathbbR,alphainmathbbR^*$.
Then $(*)$ $E_A+lambda I=E_alpha A=E_A$ and $E_lambda I=GL_n(mathbbC)$.
Let $S=M+iN,S^-1=P+iQ$ where $M,N,P,Q$ are real matrices.
Then the condition $Sin E_A$ is equivalent to $(*)$ $NAP+MAQ=0,NP+MQ=0,MP-NQ=I$.
Thus $E_A$ is a real algebraic set of dimension $dleq 2n^2$; since it contains $GL_n(mathbbR)$, $dgeq n^2$.
$textbfProposition 1.$ $d=2n^2$ iff $A$ is a scalar matrix.
$textbfProof.$ Assume that $A$ is not a scalar matrix. There is $xinmathbbC^n$ s.t. $x,Ax$ is a free system; we consider a basis in the form $x,-iAx,cdots$; then the associated $S_0AS_0^-1$ has $[0,i,0,cdots]^T$ as first column, $(S_0AS_0^-1)_2,1notin mathbbR$ and $S_0notin E_A$. Let $mathcalB$ be the set of bases of $mathbbC^n$.
Let $Z=(e_1,cdots,e_n)inmathcalB;Ae_1=sum_ja_je_j text with a_2inmathbbR$; note that $Znot= mathcalB$ (cf. above).
If $T=[t_i,j]=[e_1,cdots,e_n]$, then $(e_1,cdots,e_n)in Z$ iff $Im((T^-1AT)_2,1)=0$. Since $Im((T^-1AT)_2,1)$ is a rational fraction in the $(Re(t_i,j),Im(t_i,j))$, $Z$ is a real algebric set and, moreover, it is a Zariski closed strict subset of $mathcalB$. We deduce that $d<2n^2$. $square$
To find $d$ in the general case seems difficult (at least for me...). We consider the case $n=2$ and we will see that there are matrices $A$ s.t. $n^2<dim(E_A)<2n^2$.
$textbfProposition 2.$ Assume that $n=2$. If $A$ is a scalar matrix, then $d=8$, otherwise $d=6$.
$textbfProof$. We may assume that $A$ is in the form of its Jordan real form. According to $(*)$, it suffices to consider the three cases
i) $A=beginpmatrix0&0\0&1endpmatrix$, ii) $A=beginpmatrix0&1\0&0endpmatrix$, iii) $A=beginpmatrix0&1\-1&0endpmatrix$.
For i). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&gamma bendpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abnot= 0,betanot= gamma$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For ii). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&beta(b+gamma a)endpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abetagammanot= 0$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For iii). We obtain $S=beginpmatrixa&b\c&dendpmatrix$ where $p=dfracac+bdad-bcinmathbbR,q=dfraca^2+b^2ad-bcinmathbbR^*,r=dfracc^2+d^2ad-bcinmathbbR$ and $ad-bcnot= 0$ (or $pinmathbbR,qinmathbbR,rinmathbbR^*$)
We show that $dim(E_A)=6$. Note that $qnot=0$ and $r=dfrac1+p^2q$ is real if $p,q$ are. Assume that $a,b,p,q$ are independently given (6 degrees of freedom). Then $ad-bc$ and $ac+bd$ are known; since $a^2+b^2not= 0$ we deduce a unique couple $c,d$. $square$
$textbfProposition 3.$ For every $A$, $dim(E_A)geq n^2+1$.
$textbfProof$. If $Sin E_A$, then, for every $thetainmathbbR$, $e^ithetaSin E_A$.
EDIT. I do not know why you are passionate about connected sets... Here, the answer is no in general.
$textbfProposition 4.$ There are matrices $A$ s.t. $E_A$ is not connected.
$textbfProof.$ Let $Ain .M_2(mathbbR)$ s.t. $tr(A)=1,det(A)=4$ and $phi:Sin GL_n(mathbbC)rightarrow SAS^-1in M_n(mathbbC)$. Clearly $phi(E_A)=phi(GL_n(mathbbR))approx (p,q,r,s)inmathbbR^4;p+s=1,ps-rq=4$. In particular, $phi(E_A)$ has one or two connected components.
Let $Z=(p,q,r)inmathbbR^3;p(1-p)-rq=4$. $Z$ has exactly $2$ connected components (draw with Maple if you are in hurry). Suppose $phi(E_A)$ is connected; since $(p,q,r,s)in E_Arightarrow (p,q,r)in Z$ is a continuous function, this would cause $Z$ to be connected, a contradiction. Finally $phi(E_A)$ is not connected and $E_A$ too.
$textbfRemark.$ We can generalize the above proposition to matrices in $ M_n(mathbbR)$ that have conjugate eigenvalues.
answered Jul 30 at 10:05
loup blanc
20.3k21549
Thanks for answering my question. Could you comment on the connectedness of $E_A$? Is it possible the set is connected?
– user9527
Jul 30 at 16:01
Also in defining the variety, why the relations $PN+QM = 0$ and $PM-QN=I$ are not needed? Thanks.
– user9527
Jul 30 at 17:50
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $ngeq 2,E_A=Sin GL_n(mathbbC);SAS^-1in M_n(mathbbR)$ and $lambdainmathbbR,alphainmathbbR^*$.
Then $(*)$ $E_A+lambda I=E_alpha A=E_A$ and $E_lambda I=GL_n(mathbbC)$.
Let $S=M+iN,S^-1=P+iQ$ where $M,N,P,Q$ are real matrices.
Then the condition $Sin E_A$ is equivalent to $(*)$ $NAP+MAQ=0,NP+MQ=0,MP-NQ=I$.
Thus $E_A$ is a real algebraic set of dimension $dleq 2n^2$; since it contains $GL_n(mathbbR)$, $dgeq n^2$.
$textbfProposition 1.$ $d=2n^2$ iff $A$ is a scalar matrix.
$textbfProof.$ Assume that $A$ is not a scalar matrix. There is $xinmathbbC^n$ s.t. $x,Ax$ is a free system; we consider a basis in the form $x,-iAx,cdots$; then the associated $S_0AS_0^-1$ has $[0,i,0,cdots]^T$ as first column, $(S_0AS_0^-1)_2,1notin mathbbR$ and $S_0notin E_A$. Let $mathcalB$ be the set of bases of $mathbbC^n$.
Let $Z=(e_1,cdots,e_n)inmathcalB;Ae_1=sum_ja_je_j text with a_2inmathbbR$; note that $Znot= mathcalB$ (cf. above).
If $T=[t_i,j]=[e_1,cdots,e_n]$, then $(e_1,cdots,e_n)in Z$ iff $Im((T^-1AT)_2,1)=0$. Since $Im((T^-1AT)_2,1)$ is a rational fraction in the $(Re(t_i,j),Im(t_i,j))$, $Z$ is a real algebric set and, moreover, it is a Zariski closed strict subset of $mathcalB$. We deduce that $d<2n^2$. $square$
To find $d$ in the general case seems difficult (at least for me...). We consider the case $n=2$ and we will see that there are matrices $A$ s.t. $n^2<dim(E_A)<2n^2$.
$textbfProposition 2.$ Assume that $n=2$. If $A$ is a scalar matrix, then $d=8$, otherwise $d=6$.
$textbfProof$. We may assume that $A$ is in the form of its Jordan real form. According to $(*)$, it suffices to consider the three cases
i) $A=beginpmatrix0&0\0&1endpmatrix$, ii) $A=beginpmatrix0&1\0&0endpmatrix$, iii) $A=beginpmatrix0&1\-1&0endpmatrix$.
For i). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&gamma bendpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abnot= 0,betanot= gamma$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For ii). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&beta(b+gamma a)endpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abetagammanot= 0$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For iii). We obtain $S=beginpmatrixa&b\c&dendpmatrix$ where $p=dfracac+bdad-bcinmathbbR,q=dfraca^2+b^2ad-bcinmathbbR^*,r=dfracc^2+d^2ad-bcinmathbbR$ and $ad-bcnot= 0$ (or $pinmathbbR,qinmathbbR,rinmathbbR^*$)
We show that $dim(E_A)=6$. Note that $qnot=0$ and $r=dfrac1+p^2q$ is real if $p,q$ are. Assume that $a,b,p,q$ are independently given (6 degrees of freedom). Then $ad-bc$ and $ac+bd$ are known; since $a^2+b^2not= 0$ we deduce a unique couple $c,d$. $square$
$textbfProposition 3.$ For every $A$, $dim(E_A)geq n^2+1$.
$textbfProof$. If $Sin E_A$, then, for every $thetainmathbbR$, $e^ithetaSin E_A$.
EDIT. I do not know why you are passionate about connected sets... Here, the answer is no in general.
$textbfProposition 4.$ There are matrices $A$ s.t. $E_A$ is not connected.
$textbfProof.$ Let $Ain .M_2(mathbbR)$ s.t. $tr(A)=1,det(A)=4$ and $phi:Sin GL_n(mathbbC)rightarrow SAS^-1in M_n(mathbbC)$. Clearly $phi(E_A)=phi(GL_n(mathbbR))approx (p,q,r,s)inmathbbR^4;p+s=1,ps-rq=4$. In particular, $phi(E_A)$ has one or two connected components.
Let $Z=(p,q,r)inmathbbR^3;p(1-p)-rq=4$. $Z$ has exactly $2$ connected components (draw with Maple if you are in hurry). Suppose $phi(E_A)$ is connected; since $(p,q,r,s)in E_Arightarrow (p,q,r)in Z$ is a continuous function, this would cause $Z$ to be connected, a contradiction. Finally $phi(E_A)$ is not connected and $E_A$ too.
$textbfRemark.$ We can generalize the above proposition to matrices in $ M_n(mathbbR)$ that have conjugate eigenvalues.
answered Jul 30 at 10:05
loup blanc
20.3k21549
Let $ngeq 2,E_A=Sin GL_n(mathbbC);SAS^-1in M_n(mathbbR)$ and $lambdainmathbbR,alphainmathbbR^*$.
Then $(*)$ $E_A+lambda I=E_alpha A=E_A$ and $E_lambda I=GL_n(mathbbC)$.
Let $S=M+iN,S^-1=P+iQ$ where $M,N,P,Q$ are real matrices.
Then the condition $Sin E_A$ is equivalent to $(*)$ $NAP+MAQ=0,NP+MQ=0,MP-NQ=I$.
Thus $E_A$ is a real algebraic set of dimension $dleq 2n^2$; since it contains $GL_n(mathbbR)$, $dgeq n^2$.
$textbfProposition 1.$ $d=2n^2$ iff $A$ is a scalar matrix.
$textbfProof.$ Assume that $A$ is not a scalar matrix. There is $xinmathbbC^n$ s.t. $x,Ax$ is a free system; we consider a basis in the form $x,-iAx,cdots$; then the associated $S_0AS_0^-1$ has $[0,i,0,cdots]^T$ as first column, $(S_0AS_0^-1)_2,1notin mathbbR$ and $S_0notin E_A$. Let $mathcalB$ be the set of bases of $mathbbC^n$.
Let $Z=(e_1,cdots,e_n)inmathcalB;Ae_1=sum_ja_je_j text with a_2inmathbbR$; note that $Znot= mathcalB$ (cf. above).
If $T=[t_i,j]=[e_1,cdots,e_n]$, then $(e_1,cdots,e_n)in Z$ iff $Im((T^-1AT)_2,1)=0$. Since $Im((T^-1AT)_2,1)$ is a rational fraction in the $(Re(t_i,j),Im(t_i,j))$, $Z$ is a real algebric set and, moreover, it is a Zariski closed strict subset of $mathcalB$. We deduce that $d<2n^2$. $square$
To find $d$ in the general case seems difficult (at least for me...). We consider the case $n=2$ and we will see that there are matrices $A$ s.t. $n^2<dim(E_A)<2n^2$.
$textbfProposition 2.$ Assume that $n=2$. If $A$ is a scalar matrix, then $d=8$, otherwise $d=6$.
$textbfProof$. We may assume that $A$ is in the form of its Jordan real form. According to $(*)$, it suffices to consider the three cases
i) $A=beginpmatrix0&0\0&1endpmatrix$, ii) $A=beginpmatrix0&1\0&0endpmatrix$, iii) $A=beginpmatrix0&1\-1&0endpmatrix$.
For i). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&gamma bendpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abnot= 0,betanot= gamma$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For ii). We obtain $Sin E_A$ iff $S=beginpmatrixa&b\beta a&beta(b+gamma a)endpmatrix$ where $a,bin mathbbC,beta,gammainmathbbR,abetagammanot= 0$ or $S$ is in another form that has no more degrees of freedom. Thus $dim(E_A)=6$.
For iii). We obtain $S=beginpmatrixa&b\c&dendpmatrix$ where $p=dfracac+bdad-bcinmathbbR,q=dfraca^2+b^2ad-bcinmathbbR^*,r=dfracc^2+d^2ad-bcinmathbbR$ and $ad-bcnot= 0$ (or $pinmathbbR,qinmathbbR,rinmathbbR^*$)
We show that $dim(E_A)=6$. Note that $qnot=0$ and $r=dfrac1+p^2q$ is real if $p,q$ are. Assume that $a,b,p,q$ are independently given (6 degrees of freedom). Then $ad-bc$ and $ac+bd$ are known; since $a^2+b^2not= 0$ we deduce a unique couple $c,d$. $square$
$textbfProposition 3.$ For every $A$, $dim(E_A)geq n^2+1$.
$textbfProof$. If $Sin E_A$, then, for every $thetainmathbbR$, $e^ithetaSin E_A$.
EDIT. I do not know why you are passionate about connected sets... Here, the answer is no in general.
$textbfProposition 4.$ There are matrices $A$ s.t. $E_A$ is not connected.
$textbfProof.$ Let $Ain .M_2(mathbbR)$ s.t. $tr(A)=1,det(A)=4$ and $phi:Sin GL_n(mathbbC)rightarrow SAS^-1in M_n(mathbbC)$. Clearly $phi(E_A)=phi(GL_n(mathbbR))approx (p,q,r,s)inmathbbR^4;p+s=1,ps-rq=4$. In particular, $phi(E_A)$ has one or two connected components.
Let $Z=(p,q,r)inmathbbR^3;p(1-p)-rq=4$. $Z$ has exactly $2$ connected components (draw with Maple if you are in hurry). Suppose $phi(E_A)$ is connected; since $(p,q,r,s)in E_Arightarrow (p,q,r)in Z$ is a continuous function, this would cause $Z$ to be connected, a contradiction. Finally $phi(E_A)$ is not connected and $E_A$ too.
$textbfRemark.$ We can generalize the above proposition to matrices in $ M_n(mathbbR)$ that have conjugate eigenvalues.
answered Jul 30 at 10:05
loup blanc
20.3k21549
edited Jul 31 at 16:08
answered Jul 30 at 10:05
loup blanc
20.3k21549
answered Jul 30 at 10:05
loup blanc
20.3k21549
answered Jul 30 at 10:05
loup blanc
20.3k21549
20.3k21549
Thanks for answering my question. Could you comment on the connectedness of $E_A$? Is it possible the set is connected?
– user9527
Jul 30 at 16:01
Also in defining the variety, why the relations $PN+QM = 0$ and $PM-QN=I$ are not needed? Thanks.
– user9527
Jul 30 at 17:50
add a comment |Â
Thanks for answering my question. Could you comment on the connectedness of $E_A$? Is it possible the set is connected?
– user9527
Jul 30 at 16:01
Also in defining the variety, why the relations $PN+QM = 0$ and $PM-QN=I$ are not needed? Thanks.
– user9527
Jul 30 at 17:50
Thanks for answering my question. Could you comment on the connectedness of $E_A$? Is it possible the set is connected?
– user9527
Jul 30 at 16:01
Thanks for answering my question. Could you comment on the connectedness of $E_A$? Is it possible the set is connected?
– user9527
Jul 30 at 16:01
Also in defining the variety, why the relations $PN+QM = 0$ and $PM-QN=I$ are not needed? Thanks.
– user9527
Jul 30 at 17:50
Also in defining the variety, why the relations $PN+QM = 0$ and $PM-QN=I$ are not needed? Thanks.
– user9527
Jul 30 at 17:50
add a comment |Â
up vote
1
down vote
Assume that $SA=BS$ for some real matrix $B$. Write $S=M+iN$ where $M$ and $N$ are real. Let us plug in, expand, and then separate the real and imaginary parts of the equation.
$$MA+iNA=BM+iBN; quad textor quad MA=BM, NA=BN.$$
Thus, for fixed $B$, complex solutions to the equation correspond to pairs of real solutions.
answered Jul 29 at 0:03
Aaron
15.1k22552
I guess $M$ and $N$ are not necessarily invertible. So we can conclude the inclusion is proper. What about the second part?
– user9527
Jul 29 at 0:10
I am confused. Doesn't this mean the set contains general linear maps over $mathbb R$ as a proper subset?
– user9527
Jul 29 at 0:15
@user9527 Yes, this shows the inclusion is proper. Although, if you want to restrict to where $S$ is invertible, I'm not entirely sure on general conditions. You can always take $N=alpha N$ for any real $alpha$ as long as $N$ is invertible, but I don't see a clean way of finding the intersection of the set with invertible maps. But at a bare minimum, the set contains all complex multiples of real invertible matrices.
– Aaron
Jul 29 at 0:18
add a comment |Â
up vote
1
down vote
Assume that $SA=BS$ for some real matrix $B$. Write $S=M+iN$ where $M$ and $N$ are real. Let us plug in, expand, and then separate the real and imaginary parts of the equation.
$$MA+iNA=BM+iBN; quad textor quad MA=BM, NA=BN.$$
Thus, for fixed $B$, complex solutions to the equation correspond to pairs of real solutions.
answered Jul 29 at 0:03
Aaron
15.1k22552
I guess $M$ and $N$ are not necessarily invertible. So we can conclude the inclusion is proper. What about the second part?
– user9527
Jul 29 at 0:10
I am confused. Doesn't this mean the set contains general linear maps over $mathbb R$ as a proper subset?
– user9527
Jul 29 at 0:15
@user9527 Yes, this shows the inclusion is proper. Although, if you want to restrict to where $S$ is invertible, I'm not entirely sure on general conditions. You can always take $N=alpha N$ for any real $alpha$ as long as $N$ is invertible, but I don't see a clean way of finding the intersection of the set with invertible maps. But at a bare minimum, the set contains all complex multiples of real invertible matrices.
– Aaron
Jul 29 at 0:18
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assume that $SA=BS$ for some real matrix $B$. Write $S=M+iN$ where $M$ and $N$ are real. Let us plug in, expand, and then separate the real and imaginary parts of the equation.
$$MA+iNA=BM+iBN; quad textor quad MA=BM, NA=BN.$$
Thus, for fixed $B$, complex solutions to the equation correspond to pairs of real solutions.
answered Jul 29 at 0:03
Aaron
15.1k22552
Assume that $SA=BS$ for some real matrix $B$. Write $S=M+iN$ where $M$ and $N$ are real. Let us plug in, expand, and then separate the real and imaginary parts of the equation.
$$MA+iNA=BM+iBN; quad textor quad MA=BM, NA=BN.$$
Thus, for fixed $B$, complex solutions to the equation correspond to pairs of real solutions.
answered Jul 29 at 0:03
Aaron
15.1k22552
answered Jul 29 at 0:03
Aaron
15.1k22552
answered Jul 29 at 0:03
Aaron
15.1k22552
answered Jul 29 at 0:03
Aaron
15.1k22552
15.1k22552
I guess $M$ and $N$ are not necessarily invertible. So we can conclude the inclusion is proper. What about the second part?
– user9527
Jul 29 at 0:10
I am confused. Doesn't this mean the set contains general linear maps over $mathbb R$ as a proper subset?
– user9527
Jul 29 at 0:15
@user9527 Yes, this shows the inclusion is proper. Although, if you want to restrict to where $S$ is invertible, I'm not entirely sure on general conditions. You can always take $N=alpha N$ for any real $alpha$ as long as $N$ is invertible, but I don't see a clean way of finding the intersection of the set with invertible maps. But at a bare minimum, the set contains all complex multiples of real invertible matrices.
– Aaron
Jul 29 at 0:18
add a comment |Â
I guess $M$ and $N$ are not necessarily invertible. So we can conclude the inclusion is proper. What about the second part?
– user9527
Jul 29 at 0:10
I am confused. Doesn't this mean the set contains general linear maps over $mathbb R$ as a proper subset?
– user9527
Jul 29 at 0:15
@user9527 Yes, this shows the inclusion is proper. Although, if you want to restrict to where $S$ is invertible, I'm not entirely sure on general conditions. You can always take $N=alpha N$ for any real $alpha$ as long as $N$ is invertible, but I don't see a clean way of finding the intersection of the set with invertible maps. But at a bare minimum, the set contains all complex multiples of real invertible matrices.
– Aaron
Jul 29 at 0:18
I guess $M$ and $N$ are not necessarily invertible. So we can conclude the inclusion is proper. What about the second part?
– user9527
Jul 29 at 0:10
I guess $M$ and $N$ are not necessarily invertible. So we can conclude the inclusion is proper. What about the second part?
– user9527
Jul 29 at 0:10
I am confused. Doesn't this mean the set contains general linear maps over $mathbb R$ as a proper subset?
– user9527
Jul 29 at 0:15
I am confused. Doesn't this mean the set contains general linear maps over $mathbb R$ as a proper subset?
– user9527
Jul 29 at 0:15
@user9527 Yes, this shows the inclusion is proper. Although, if you want to restrict to where $S$ is invertible, I'm not entirely sure on general conditions. You can always take $N=alpha N$ for any real $alpha$ as long as $N$ is invertible, but I don't see a clean way of finding the intersection of the set with invertible maps. But at a bare minimum, the set contains all complex multiples of real invertible matrices.
– Aaron
Jul 29 at 0:18
@user9527 Yes, this shows the inclusion is proper. Although, if you want to restrict to where $S$ is invertible, I'm not entirely sure on general conditions. You can always take $N=alpha N$ for any real $alpha$ as long as $N$ is invertible, but I don't see a clean way of finding the intersection of the set with invertible maps. But at a bare minimum, the set contains all complex multiples of real invertible matrices.
– Aaron
Jul 29 at 0:18
add a comment |Â
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