$SO(n)$ is connected, alternative form












1












$begingroup$


I have the following exercise:




Show that $SO(n)$ is connected, using the following outline: For the case $n = 1$, there is nothing to show, since a $1times 1$ matrix with determinant one must be $[1]$. Assume, then, that $n geq 2$. Let $e_1$ denote the unit vector with entries $(1,0,ldots, 0)$ in $mathbb R^n$. Given any unit vector $vinmathbb R^n$, show that there exists a continuous path $R(t)$ in $SO(n)$ with $R(O) = I$ and $R(1)v = e_1$. Now, show that any element $R$ of $SO(n)$ can be connected to a block-diagonal matrix of the form
begin{pmatrix}
1 &\&R_1
end{pmatrix}
with $R_1in SO(n- 1)$ and proceed by induction.




I have troubles only with the first part, i.e., Given any unit vector $vinmathbb R^n$, show that there exists a continuous path $R(t)$ in $SO(n)$ with $R(O) = I$ and $R(1)v = e_1$. Please help me, thank you.






linear-algebra lie-groups







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Rotate in the plane spanned by $e_1$ and $v$.
    $endgroup$
    – daw
    Jul 2 '14 at 18:36










  • $begingroup$
    Yes, but how can I express this rotation?
    $endgroup$
    – andrebant
    Jul 2 '14 at 21:03
















1












$begingroup$


I have the following exercise:




Show that $SO(n)$ is connected, using the following outline: For the case $n = 1$, there is nothing to show, since a $1times 1$ matrix with determinant one must be $[1]$. Assume, then, that $n geq 2$. Let $e_1$ denote the unit vector with entries $(1,0,ldots, 0)$ in $mathbb R^n$. Given any unit vector $vinmathbb R^n$, show that there exists a continuous path $R(t)$ in $SO(n)$ with $R(O) = I$ and $R(1)v = e_1$. Now, show that any element $R$ of $SO(n)$ can be connected to a block-diagonal matrix of the form
begin{pmatrix}
1 &\&R_1
end{pmatrix}
with $R_1in SO(n- 1)$ and proceed by induction.




I have troubles only with the first part, i.e., Given any unit vector $vinmathbb R^n$, show that there exists a continuous path $R(t)$ in $SO(n)$ with $R(O) = I$ and $R(1)v = e_1$. Please help me, thank you.






linear-algebra lie-groups







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Rotate in the plane spanned by $e_1$ and $v$.
    $endgroup$
    – daw
    Jul 2 '14 at 18:36










  • $begingroup$
    Yes, but how can I express this rotation?
    $endgroup$
    – andrebant
    Jul 2 '14 at 21:03














1












1








1


0



$begingroup$


I have the following exercise:




Show that $SO(n)$ is connected, using the following outline: For the case $n = 1$, there is nothing to show, since a $1times 1$ matrix with determinant one must be $[1]$. Assume, then, that $n geq 2$. Let $e_1$ denote the unit vector with entries $(1,0,ldots, 0)$ in $mathbb R^n$. Given any unit vector $vinmathbb R^n$, show that there exists a continuous path $R(t)$ in $SO(n)$ with $R(O) = I$ and $R(1)v = e_1$. Now, show that any element $R$ of $SO(n)$ can be connected to a block-diagonal matrix of the form
begin{pmatrix}
1 &\&R_1
end{pmatrix}
with $R_1in SO(n- 1)$ and proceed by induction.




I have troubles only with the first part, i.e., Given any unit vector $vinmathbb R^n$, show that there exists a continuous path $R(t)$ in $SO(n)$ with $R(O) = I$ and $R(1)v = e_1$. Please help me, thank you.






linear-algebra lie-groups







share|cite|improve this question









$endgroup$




I have the following exercise:




Show that $SO(n)$ is connected, using the following outline: For the case $n = 1$, there is nothing to show, since a $1times 1$ matrix with determinant one must be $[1]$. Assume, then, that $n geq 2$. Let $e_1$ denote the unit vector with entries $(1,0,ldots, 0)$ in $mathbb R^n$. Given any unit vector $vinmathbb R^n$, show that there exists a continuous path $R(t)$ in $SO(n)$ with $R(O) = I$ and $R(1)v = e_1$. Now, show that any element $R$ of $SO(n)$ can be connected to a block-diagonal matrix of the form
begin{pmatrix}
1 &\&R_1
end{pmatrix}
with $R_1in SO(n- 1)$ and proceed by induction.




I have troubles only with the first part, i.e., Given any unit vector $vinmathbb R^n$, show that there exists a continuous path $R(t)$ in $SO(n)$ with $R(O) = I$ and $R(1)v = e_1$. Please help me, thank you.






linear-algebra lie-groups




linear-algebra lie-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 2 '14 at 15:16









andrebantandrebant

12611




12611








  • 1




    $begingroup$
    Rotate in the plane spanned by $e_1$ and $v$.
    $endgroup$
    – daw
    Jul 2 '14 at 18:36










  • $begingroup$
    Yes, but how can I express this rotation?
    $endgroup$
    – andrebant
    Jul 2 '14 at 21:03














  • 1




    $begingroup$
    Rotate in the plane spanned by $e_1$ and $v$.
    $endgroup$
    – daw
    Jul 2 '14 at 18:36










  • $begingroup$
    Yes, but how can I express this rotation?
    $endgroup$
    – andrebant
    Jul 2 '14 at 21:03








1




1




$begingroup$
Rotate in the plane spanned by $e_1$ and $v$.
$endgroup$
– daw
Jul 2 '14 at 18:36




$begingroup$
Rotate in the plane spanned by $e_1$ and $v$.
$endgroup$
– daw
Jul 2 '14 at 18:36












$begingroup$
Yes, but how can I express this rotation?
$endgroup$
– andrebant
Jul 2 '14 at 21:03




$begingroup$
Yes, but how can I express this rotation?
$endgroup$
– andrebant
Jul 2 '14 at 21:03










1 Answer
1






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oldest

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0












$begingroup$

The matrix
$$
begin{pmatrix}
costheta & -sintheta \
sintheta & costheta \
end{pmatrix}
$$
is in SO($2$) and means rotation.



For $3$-dimension, you first choose $theta_1$ to rotate $(x,y,z)'$ to $(x,sqrt {y^2+z^2},0)'$ continously:
$$
begin{pmatrix}
1 & 0 & 0 \
0 & costheta_1 & -sintheta_1 \
0 & sintheta_1 & costheta_1 \
end{pmatrix}
begin{pmatrix}
x\
y\
z\
end{pmatrix}
=
begin{pmatrix}
x\
sqrt{y^2+z^2}\
0
end{pmatrix}
$$
This matrix A belongs to SO($3$).



And then similarly (called matrix B):
$$
begin{pmatrix}
costheta_2 & -sintheta_2 & 0\
sintheta_2 & costheta_2 & 0\
0 & 0 & 1\
end{pmatrix}
begin{pmatrix}
x\
sqrt{y^2+z^2}\
0\
end{pmatrix}
=
begin{pmatrix}
sqrt{x^2+y^2+z^2}\
0\
0
end{pmatrix}
$$



So, $R(0)=I$, $R(1/2)=A$, $R(1)=BA$






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    1 Answer
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    0












    $begingroup$

    The matrix
    $$
    begin{pmatrix}
    costheta & -sintheta \
    sintheta & costheta \
    end{pmatrix}
    $$
    is in SO($2$) and means rotation.



    For $3$-dimension, you first choose $theta_1$ to rotate $(x,y,z)'$ to $(x,sqrt {y^2+z^2},0)'$ continously:
    $$
    begin{pmatrix}
    1 & 0 & 0 \
    0 & costheta_1 & -sintheta_1 \
    0 & sintheta_1 & costheta_1 \
    end{pmatrix}
    begin{pmatrix}
    x\
    y\
    z\
    end{pmatrix}
    =
    begin{pmatrix}
    x\
    sqrt{y^2+z^2}\
    0
    end{pmatrix}
    $$
    This matrix A belongs to SO($3$).



    And then similarly (called matrix B):
    $$
    begin{pmatrix}
    costheta_2 & -sintheta_2 & 0\
    sintheta_2 & costheta_2 & 0\
    0 & 0 & 1\
    end{pmatrix}
    begin{pmatrix}
    x\
    sqrt{y^2+z^2}\
    0\
    end{pmatrix}
    =
    begin{pmatrix}
    sqrt{x^2+y^2+z^2}\
    0\
    0
    end{pmatrix}
    $$



    So, $R(0)=I$, $R(1/2)=A$, $R(1)=BA$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      The matrix
      $$
      begin{pmatrix}
      costheta & -sintheta \
      sintheta & costheta \
      end{pmatrix}
      $$
      is in SO($2$) and means rotation.



      For $3$-dimension, you first choose $theta_1$ to rotate $(x,y,z)'$ to $(x,sqrt {y^2+z^2},0)'$ continously:
      $$
      begin{pmatrix}
      1 & 0 & 0 \
      0 & costheta_1 & -sintheta_1 \
      0 & sintheta_1 & costheta_1 \
      end{pmatrix}
      begin{pmatrix}
      x\
      y\
      z\
      end{pmatrix}
      =
      begin{pmatrix}
      x\
      sqrt{y^2+z^2}\
      0
      end{pmatrix}
      $$
      This matrix A belongs to SO($3$).



      And then similarly (called matrix B):
      $$
      begin{pmatrix}
      costheta_2 & -sintheta_2 & 0\
      sintheta_2 & costheta_2 & 0\
      0 & 0 & 1\
      end{pmatrix}
      begin{pmatrix}
      x\
      sqrt{y^2+z^2}\
      0\
      end{pmatrix}
      =
      begin{pmatrix}
      sqrt{x^2+y^2+z^2}\
      0\
      0
      end{pmatrix}
      $$



      So, $R(0)=I$, $R(1/2)=A$, $R(1)=BA$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        The matrix
        $$
        begin{pmatrix}
        costheta & -sintheta \
        sintheta & costheta \
        end{pmatrix}
        $$
        is in SO($2$) and means rotation.



        For $3$-dimension, you first choose $theta_1$ to rotate $(x,y,z)'$ to $(x,sqrt {y^2+z^2},0)'$ continously:
        $$
        begin{pmatrix}
        1 & 0 & 0 \
        0 & costheta_1 & -sintheta_1 \
        0 & sintheta_1 & costheta_1 \
        end{pmatrix}
        begin{pmatrix}
        x\
        y\
        z\
        end{pmatrix}
        =
        begin{pmatrix}
        x\
        sqrt{y^2+z^2}\
        0
        end{pmatrix}
        $$
        This matrix A belongs to SO($3$).



        And then similarly (called matrix B):
        $$
        begin{pmatrix}
        costheta_2 & -sintheta_2 & 0\
        sintheta_2 & costheta_2 & 0\
        0 & 0 & 1\
        end{pmatrix}
        begin{pmatrix}
        x\
        sqrt{y^2+z^2}\
        0\
        end{pmatrix}
        =
        begin{pmatrix}
        sqrt{x^2+y^2+z^2}\
        0\
        0
        end{pmatrix}
        $$



        So, $R(0)=I$, $R(1/2)=A$, $R(1)=BA$






        share|cite|improve this answer











        $endgroup$



        The matrix
        $$
        begin{pmatrix}
        costheta & -sintheta \
        sintheta & costheta \
        end{pmatrix}
        $$
        is in SO($2$) and means rotation.



        For $3$-dimension, you first choose $theta_1$ to rotate $(x,y,z)'$ to $(x,sqrt {y^2+z^2},0)'$ continously:
        $$
        begin{pmatrix}
        1 & 0 & 0 \
        0 & costheta_1 & -sintheta_1 \
        0 & sintheta_1 & costheta_1 \
        end{pmatrix}
        begin{pmatrix}
        x\
        y\
        z\
        end{pmatrix}
        =
        begin{pmatrix}
        x\
        sqrt{y^2+z^2}\
        0
        end{pmatrix}
        $$
        This matrix A belongs to SO($3$).



        And then similarly (called matrix B):
        $$
        begin{pmatrix}
        costheta_2 & -sintheta_2 & 0\
        sintheta_2 & costheta_2 & 0\
        0 & 0 & 1\
        end{pmatrix}
        begin{pmatrix}
        x\
        sqrt{y^2+z^2}\
        0\
        end{pmatrix}
        =
        begin{pmatrix}
        sqrt{x^2+y^2+z^2}\
        0\
        0
        end{pmatrix}
        $$



        So, $R(0)=I$, $R(1/2)=A$, $R(1)=BA$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 23 at 12:33

























        answered Jan 23 at 11:59









        ZWJZWJ

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