Is every matrix conjugate to its transpose in a continuous way?












19














It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $ntimes n$ matrix $M$ with real entries, there is a matrix $S_Min GL(n,mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:




Is there a continuous map $psicolon M_{n,n}(mathbb{R})longrightarrow GL(n,mathbb{R})$ such that$$bigl(forall Min M_{ntimes n}(mathbb{R})bigr):psi(M)^{-1}.M.psi(M)=M^T?$$




My guess is that the answer is negative even for $n=2$.



Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=begin{bmatrix}x&y\z&tend{bmatrix},$$then you can take$$S_M=begin{bmatrix}az&bz\bz&bt-bx+ayend{bmatrix},$$with $a$ and $b$ chosen such that $det(S_M)neq0$ but, of course, this will only work if $zneq0$. What if $z=0$? Then you can take$$S_M=begin{bmatrix}-at+ax&ay\ay&byend{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $det(S_M)neq0$; the problem now is that, of course, this will only work if $yneq0$. And so on. This looks like the problem of finding a logarithm for each $zinmathbb{C}setminus{0}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.










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    19














    It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $ntimes n$ matrix $M$ with real entries, there is a matrix $S_Min GL(n,mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:




    Is there a continuous map $psicolon M_{n,n}(mathbb{R})longrightarrow GL(n,mathbb{R})$ such that$$bigl(forall Min M_{ntimes n}(mathbb{R})bigr):psi(M)^{-1}.M.psi(M)=M^T?$$




    My guess is that the answer is negative even for $n=2$.



    Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=begin{bmatrix}x&y\z&tend{bmatrix},$$then you can take$$S_M=begin{bmatrix}az&bz\bz&bt-bx+ayend{bmatrix},$$with $a$ and $b$ chosen such that $det(S_M)neq0$ but, of course, this will only work if $zneq0$. What if $z=0$? Then you can take$$S_M=begin{bmatrix}-at+ax&ay\ay&byend{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $det(S_M)neq0$; the problem now is that, of course, this will only work if $yneq0$. And so on. This looks like the problem of finding a logarithm for each $zinmathbb{C}setminus{0}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.










    share|cite|improve this question



























      19












      19








      19


      5





      It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $ntimes n$ matrix $M$ with real entries, there is a matrix $S_Min GL(n,mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:




      Is there a continuous map $psicolon M_{n,n}(mathbb{R})longrightarrow GL(n,mathbb{R})$ such that$$bigl(forall Min M_{ntimes n}(mathbb{R})bigr):psi(M)^{-1}.M.psi(M)=M^T?$$




      My guess is that the answer is negative even for $n=2$.



      Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=begin{bmatrix}x&y\z&tend{bmatrix},$$then you can take$$S_M=begin{bmatrix}az&bz\bz&bt-bx+ayend{bmatrix},$$with $a$ and $b$ chosen such that $det(S_M)neq0$ but, of course, this will only work if $zneq0$. What if $z=0$? Then you can take$$S_M=begin{bmatrix}-at+ax&ay\ay&byend{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $det(S_M)neq0$; the problem now is that, of course, this will only work if $yneq0$. And so on. This looks like the problem of finding a logarithm for each $zinmathbb{C}setminus{0}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.










      share|cite|improve this question















      It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $ntimes n$ matrix $M$ with real entries, there is a matrix $S_Min GL(n,mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:




      Is there a continuous map $psicolon M_{n,n}(mathbb{R})longrightarrow GL(n,mathbb{R})$ such that$$bigl(forall Min M_{ntimes n}(mathbb{R})bigr):psi(M)^{-1}.M.psi(M)=M^T?$$




      My guess is that the answer is negative even for $n=2$.



      Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=begin{bmatrix}x&y\z&tend{bmatrix},$$then you can take$$S_M=begin{bmatrix}az&bz\bz&bt-bx+ayend{bmatrix},$$with $a$ and $b$ chosen such that $det(S_M)neq0$ but, of course, this will only work if $zneq0$. What if $z=0$? Then you can take$$S_M=begin{bmatrix}-at+ax&ay\ay&byend{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $det(S_M)neq0$; the problem now is that, of course, this will only work if $yneq0$. And so on. This looks like the problem of finding a logarithm for each $zinmathbb{C}setminus{0}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.







      linear-algebra matrices continuity






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      edited Jan 3 at 23:23







      José Carlos Santos

















      asked Dec 14 '18 at 10:03









      José Carlos SantosJosé Carlos Santos

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          11














          My first thought is to look at a simple example - $2times 2$ rotation matrices.
          Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.



          Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.



          Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.



          Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.



          Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.



          OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.






          share|cite|improve this answer































            6














            Consider the continuous function
            $$
            M_t=begin{cases}
            tpmatrix{1&1\ 0&2}&text{ when } tge0,\
            tpmatrix{1&0\ -1&2}&text{ when } t<0.
            end{cases}
            $$

            It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
            $$
            S_t=begin{cases}
            pmatrix{a&b\ b&b}&text{ when } t>0,\
            pmatrix{b&b\ b&a}&text{ when } t<0.
            end{cases}
            $$

            It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.






            share|cite|improve this answer























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              11














              My first thought is to look at a simple example - $2times 2$ rotation matrices.
              Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.



              Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.



              Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.



              Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.



              Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.



              OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.






              share|cite|improve this answer




























                11














                My first thought is to look at a simple example - $2times 2$ rotation matrices.
                Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.



                Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.



                Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.



                Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.



                Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.



                OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.






                share|cite|improve this answer


























                  11












                  11








                  11






                  My first thought is to look at a simple example - $2times 2$ rotation matrices.
                  Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.



                  Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.



                  Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.



                  Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.



                  Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.



                  OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.






                  share|cite|improve this answer














                  My first thought is to look at a simple example - $2times 2$ rotation matrices.
                  Oops - rotation by $theta$ and rotation by $-theta$ are conjugate by any (real) reflection. No help there.



                  Second thought - OK, powers of something all work with the same $S$. So then, what happens at the identity? Which one do we choose? Actually, I can make those powers continuous by bringing in the matrix exponential.



                  Now we're ready. Consider a matrix $A$ with some eigenvalue $lambda$ of multiplicity $1$ and associated eigenvector $v$. $A^T$ has $lambda$ as an eigenvalue with multiplicity $1$ and associated eigenvector $w$. While we can't pin down $S(A)$ completely, we do know that $S(A)w=av$ for some nonzero $a$. This will also be true for any nonzero power of $A$, and for $exp(tA)$ for any nonzero $t$. As $tto 0$, we then have $S(I)w = lim_{tto 0}S(exp(tA))w=lim_{tto 0}a(exp(tA))v=cv$ for some $c$, possibly zero.



                  Almost there; all we need now are some concrete examples of what $v$ and $w$ can be. As it turns out, any pair of non-orthogonal nonzero real vectors are possible. Let $A$ be the rank-1 matrix $vw^T$, so $A^T=wv^T$. Then $Av=langle v,wrangle v$ and $A^Tw = langle v,wrangle w$, so these are the lone eigenvectors for the nonzero eigenvalue of $A$.



                  Combining these, $S(I)$ takes an arbitrary nonzero vector $w$ to something that's simultaneously a multiple of almost every nonzero vector $v$, which must be zero. That gives $S(I)=0$, an impossibility. By this contradiction, there is no way to choose $S$ continuously.



                  OK, I actually proved specifically that $S$ can't be continuous at the identity. Continuity elsewhere isn't ruled out yet.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 14 '18 at 11:21

























                  answered Dec 14 '18 at 11:05









                  jmerryjmerry

                  2,521312




                  2,521312























                      6














                      Consider the continuous function
                      $$
                      M_t=begin{cases}
                      tpmatrix{1&1\ 0&2}&text{ when } tge0,\
                      tpmatrix{1&0\ -1&2}&text{ when } t<0.
                      end{cases}
                      $$

                      It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
                      $$
                      S_t=begin{cases}
                      pmatrix{a&b\ b&b}&text{ when } t>0,\
                      pmatrix{b&b\ b&a}&text{ when } t<0.
                      end{cases}
                      $$

                      It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.






                      share|cite|improve this answer




























                        6














                        Consider the continuous function
                        $$
                        M_t=begin{cases}
                        tpmatrix{1&1\ 0&2}&text{ when } tge0,\
                        tpmatrix{1&0\ -1&2}&text{ when } t<0.
                        end{cases}
                        $$

                        It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
                        $$
                        S_t=begin{cases}
                        pmatrix{a&b\ b&b}&text{ when } t>0,\
                        pmatrix{b&b\ b&a}&text{ when } t<0.
                        end{cases}
                        $$

                        It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.






                        share|cite|improve this answer


























                          6












                          6








                          6






                          Consider the continuous function
                          $$
                          M_t=begin{cases}
                          tpmatrix{1&1\ 0&2}&text{ when } tge0,\
                          tpmatrix{1&0\ -1&2}&text{ when } t<0.
                          end{cases}
                          $$

                          It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
                          $$
                          S_t=begin{cases}
                          pmatrix{a&b\ b&b}&text{ when } t>0,\
                          pmatrix{b&b\ b&a}&text{ when } t<0.
                          end{cases}
                          $$

                          It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.






                          share|cite|improve this answer














                          Consider the continuous function
                          $$
                          M_t=begin{cases}
                          tpmatrix{1&1\ 0&2}&text{ when } tge0,\
                          tpmatrix{1&0\ -1&2}&text{ when } t<0.
                          end{cases}
                          $$

                          It can be shown that all solutions to the equation $M_tS_t=S_tM_t^T$ are given by matrices of the form
                          $$
                          S_t=begin{cases}
                          pmatrix{a&b\ b&b}&text{ when } t>0,\
                          pmatrix{b&b\ b&a}&text{ when } t<0.
                          end{cases}
                          $$

                          It follows that if $S_t$ is chosen continuously, it must be in the form of $pmatrix{b&b\ b&b}$ at $t=0$ and hence it cannot remain non-singular.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 14 '18 at 14:52

























                          answered Dec 14 '18 at 11:20









                          user1551user1551

                          71.8k566125




                          71.8k566125






























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