Can every orthogonal matrix be written as a product of Givens rotations?












3














I'd like to know whether every orthogonal matrix



$$ A in mathcal{O}_n(mathbb{R})$$



can be written as a product of givens-rotations. I know that when we do QR-decomposition of matrix $A$ we get



$$ A = Q R $$



So my idea was to prove that $R$ must be the identity $I_n$, however I'm stuck at that. Can somebody give me a hint on how I could prove this?






matrices matrix-decomposition orthogonal-matrices







share|cite|improve this question




















  • 3




    Orthogonal matrices could be reflections.
    – Doug M
    Nov 28 '18 at 18:01










  • www-old.math.gatech.edu/academic/courses/core/math2601/… page 38
    – Ryan Howe
    Nov 28 '18 at 18:06


















3














I'd like to know whether every orthogonal matrix



$$ A in mathcal{O}_n(mathbb{R})$$



can be written as a product of givens-rotations. I know that when we do QR-decomposition of matrix $A$ we get



$$ A = Q R $$



So my idea was to prove that $R$ must be the identity $I_n$, however I'm stuck at that. Can somebody give me a hint on how I could prove this?






matrices matrix-decomposition orthogonal-matrices







share|cite|improve this question




















  • 3




    Orthogonal matrices could be reflections.
    – Doug M
    Nov 28 '18 at 18:01










  • www-old.math.gatech.edu/academic/courses/core/math2601/… page 38
    – Ryan Howe
    Nov 28 '18 at 18:06
















3












3








3







I'd like to know whether every orthogonal matrix



$$ A in mathcal{O}_n(mathbb{R})$$



can be written as a product of givens-rotations. I know that when we do QR-decomposition of matrix $A$ we get



$$ A = Q R $$



So my idea was to prove that $R$ must be the identity $I_n$, however I'm stuck at that. Can somebody give me a hint on how I could prove this?






matrices matrix-decomposition orthogonal-matrices







share|cite|improve this question















I'd like to know whether every orthogonal matrix



$$ A in mathcal{O}_n(mathbb{R})$$



can be written as a product of givens-rotations. I know that when we do QR-decomposition of matrix $A$ we get



$$ A = Q R $$



So my idea was to prove that $R$ must be the identity $I_n$, however I'm stuck at that. Can somebody give me a hint on how I could prove this?






matrices matrix-decomposition orthogonal-matrices




matrices matrix-decomposition orthogonal-matrices






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share|cite|improve this question













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edited Nov 28 '18 at 18:20









Jean-Claude Arbaut

14.7k63464




14.7k63464










asked Nov 28 '18 at 17:59









msrd0msrd0

1184




1184








  • 3




    Orthogonal matrices could be reflections.
    – Doug M
    Nov 28 '18 at 18:01










  • www-old.math.gatech.edu/academic/courses/core/math2601/… page 38
    – Ryan Howe
    Nov 28 '18 at 18:06
















  • 3




    Orthogonal matrices could be reflections.
    – Doug M
    Nov 28 '18 at 18:01










  • www-old.math.gatech.edu/academic/courses/core/math2601/… page 38
    – Ryan Howe
    Nov 28 '18 at 18:06










3




3




Orthogonal matrices could be reflections.
– Doug M
Nov 28 '18 at 18:01




Orthogonal matrices could be reflections.
– Doug M
Nov 28 '18 at 18:01












www-old.math.gatech.edu/academic/courses/core/math2601/… page 38
– Ryan Howe
Nov 28 '18 at 18:06






www-old.math.gatech.edu/academic/courses/core/math2601/… page 38
– Ryan Howe
Nov 28 '18 at 18:06












2 Answers
2






active

oldest

votes


















2














Givens rotations are... rotations, they preserve orientation ($det(M)=+1$), however $mathcal{O}_n(mathbb{R})$ has two components, one component is rotations ($det(M)=+1$), the other is reflections ($det(M)=-1$).



Example of $Minmathcal{O}_2(mathbb{R})$ that can not be written as a Givens rotation (its determinant is $-1$)
$$
M=left(begin{array}{cc}
-1 & 0 \
0 & 1
end{array}right)
$$
You can not find Givens rotation $G_theta=left(begin{array}{cc}
costheta & -sintheta \
sintheta & costheta
end{array}right)$ such that $M=G_theta$



If you want to have a Givens rotation decomposition you must restrict yourself to $SO_n(mathbb{R})$, the special orthogonal group, which is defined by $M^tM=I_d$ and $det{M}=+1$.






share|cite|improve this answer

















  • 1




    Thanks, I was trying to construct a matrix where it wouldn't work before attempting to prove, I must have missed that one
    – msrd0
    Nov 28 '18 at 22:13



















0














If you write the matrix $A=QR$ then there are two ways you can perform this.



$$ A underbrace{R_{1} R_{2} cdots R_{n}}_{hat{R}^{-1}} = hat{Q} tag{1} $$



or the following, which is what you want



$$ underbrace{Q_{n} Q_{n-1} cdots Q_{2}Q_{1}}_{{Q}^{*}}A = R tag{2} $$



In $1$ we see that Gram-Schmidt applies a sequence of elementary triangular matrices $R_{k}$ on the right of $A$. For $2$ we see a sequence of elementary unitary matrices $Q_{k}$ on the left of $A$. The matrices $Q_{k}$ can be givens matrices.



$$ underbrace{G_{n} G_{n-1} cdots G_{2}G_{1}}_{{G}^{*}}A = R tag{3} $$



All you need to prove is that $G_{k}$ is a unitary matrix. It is. There are notes provided here.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    2














    Givens rotations are... rotations, they preserve orientation ($det(M)=+1$), however $mathcal{O}_n(mathbb{R})$ has two components, one component is rotations ($det(M)=+1$), the other is reflections ($det(M)=-1$).



    Example of $Minmathcal{O}_2(mathbb{R})$ that can not be written as a Givens rotation (its determinant is $-1$)
    $$
    M=left(begin{array}{cc}
    -1 & 0 \
    0 & 1
    end{array}right)
    $$
    You can not find Givens rotation $G_theta=left(begin{array}{cc}
    costheta & -sintheta \
    sintheta & costheta
    end{array}right)$     such that $M=G_theta$



    If you want to have a Givens rotation decomposition you must restrict yourself to $SO_n(mathbb{R})$, the special orthogonal group, which is defined by $M^tM=I_d$ and $det{M}=+1$.






    share|cite|improve this answer

















    • 1




      Thanks, I was trying to construct a matrix where it wouldn't work before attempting to prove, I must have missed that one
      – msrd0
      Nov 28 '18 at 22:13
















    2














    Givens rotations are... rotations, they preserve orientation ($det(M)=+1$), however $mathcal{O}_n(mathbb{R})$ has two components, one component is rotations ($det(M)=+1$), the other is reflections ($det(M)=-1$).



    Example of $Minmathcal{O}_2(mathbb{R})$ that can not be written as a Givens rotation (its determinant is $-1$)
    $$
    M=left(begin{array}{cc}
    -1 & 0 \
    0 & 1
    end{array}right)
    $$
    You can not find Givens rotation $G_theta=left(begin{array}{cc}
    costheta & -sintheta \
    sintheta & costheta
    end{array}right)$     such that $M=G_theta$



    If you want to have a Givens rotation decomposition you must restrict yourself to $SO_n(mathbb{R})$, the special orthogonal group, which is defined by $M^tM=I_d$ and $det{M}=+1$.






    share|cite|improve this answer

















    • 1




      Thanks, I was trying to construct a matrix where it wouldn't work before attempting to prove, I must have missed that one
      – msrd0
      Nov 28 '18 at 22:13














    2












    2








    2






    Givens rotations are... rotations, they preserve orientation ($det(M)=+1$), however $mathcal{O}_n(mathbb{R})$ has two components, one component is rotations ($det(M)=+1$), the other is reflections ($det(M)=-1$).



    Example of $Minmathcal{O}_2(mathbb{R})$ that can not be written as a Givens rotation (its determinant is $-1$)
    $$
    M=left(begin{array}{cc}
    -1 & 0 \
    0 & 1
    end{array}right)
    $$
    You can not find Givens rotation $G_theta=left(begin{array}{cc}
    costheta & -sintheta \
    sintheta & costheta
    end{array}right)$     such that $M=G_theta$



    If you want to have a Givens rotation decomposition you must restrict yourself to $SO_n(mathbb{R})$, the special orthogonal group, which is defined by $M^tM=I_d$ and $det{M}=+1$.






    share|cite|improve this answer












    Givens rotations are... rotations, they preserve orientation ($det(M)=+1$), however $mathcal{O}_n(mathbb{R})$ has two components, one component is rotations ($det(M)=+1$), the other is reflections ($det(M)=-1$).



    Example of $Minmathcal{O}_2(mathbb{R})$ that can not be written as a Givens rotation (its determinant is $-1$)
    $$
    M=left(begin{array}{cc}
    -1 & 0 \
    0 & 1
    end{array}right)
    $$
    You can not find Givens rotation $G_theta=left(begin{array}{cc}
    costheta & -sintheta \
    sintheta & costheta
    end{array}right)$     such that $M=G_theta$



    If you want to have a Givens rotation decomposition you must restrict yourself to $SO_n(mathbb{R})$, the special orthogonal group, which is defined by $M^tM=I_d$ and $det{M}=+1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 '18 at 18:14









    Picaud VincentPicaud Vincent

    1,29539




    1,29539








    • 1




      Thanks, I was trying to construct a matrix where it wouldn't work before attempting to prove, I must have missed that one
      – msrd0
      Nov 28 '18 at 22:13














    • 1




      Thanks, I was trying to construct a matrix where it wouldn't work before attempting to prove, I must have missed that one
      – msrd0
      Nov 28 '18 at 22:13








    1




    1




    Thanks, I was trying to construct a matrix where it wouldn't work before attempting to prove, I must have missed that one
    – msrd0
    Nov 28 '18 at 22:13




    Thanks, I was trying to construct a matrix where it wouldn't work before attempting to prove, I must have missed that one
    – msrd0
    Nov 28 '18 at 22:13











    0














    If you write the matrix $A=QR$ then there are two ways you can perform this.



    $$ A underbrace{R_{1} R_{2} cdots R_{n}}_{hat{R}^{-1}} = hat{Q} tag{1} $$



    or the following, which is what you want



    $$ underbrace{Q_{n} Q_{n-1} cdots Q_{2}Q_{1}}_{{Q}^{*}}A = R tag{2} $$



    In $1$ we see that Gram-Schmidt applies a sequence of elementary triangular matrices $R_{k}$ on the right of $A$. For $2$ we see a sequence of elementary unitary matrices $Q_{k}$ on the left of $A$. The matrices $Q_{k}$ can be givens matrices.



    $$ underbrace{G_{n} G_{n-1} cdots G_{2}G_{1}}_{{G}^{*}}A = R tag{3} $$



    All you need to prove is that $G_{k}$ is a unitary matrix. It is. There are notes provided here.






    share|cite|improve this answer


























      0














      If you write the matrix $A=QR$ then there are two ways you can perform this.



      $$ A underbrace{R_{1} R_{2} cdots R_{n}}_{hat{R}^{-1}} = hat{Q} tag{1} $$



      or the following, which is what you want



      $$ underbrace{Q_{n} Q_{n-1} cdots Q_{2}Q_{1}}_{{Q}^{*}}A = R tag{2} $$



      In $1$ we see that Gram-Schmidt applies a sequence of elementary triangular matrices $R_{k}$ on the right of $A$. For $2$ we see a sequence of elementary unitary matrices $Q_{k}$ on the left of $A$. The matrices $Q_{k}$ can be givens matrices.



      $$ underbrace{G_{n} G_{n-1} cdots G_{2}G_{1}}_{{G}^{*}}A = R tag{3} $$



      All you need to prove is that $G_{k}$ is a unitary matrix. It is. There are notes provided here.






      share|cite|improve this answer
























        0












        0








        0






        If you write the matrix $A=QR$ then there are two ways you can perform this.



        $$ A underbrace{R_{1} R_{2} cdots R_{n}}_{hat{R}^{-1}} = hat{Q} tag{1} $$



        or the following, which is what you want



        $$ underbrace{Q_{n} Q_{n-1} cdots Q_{2}Q_{1}}_{{Q}^{*}}A = R tag{2} $$



        In $1$ we see that Gram-Schmidt applies a sequence of elementary triangular matrices $R_{k}$ on the right of $A$. For $2$ we see a sequence of elementary unitary matrices $Q_{k}$ on the left of $A$. The matrices $Q_{k}$ can be givens matrices.



        $$ underbrace{G_{n} G_{n-1} cdots G_{2}G_{1}}_{{G}^{*}}A = R tag{3} $$



        All you need to prove is that $G_{k}$ is a unitary matrix. It is. There are notes provided here.






        share|cite|improve this answer












        If you write the matrix $A=QR$ then there are two ways you can perform this.



        $$ A underbrace{R_{1} R_{2} cdots R_{n}}_{hat{R}^{-1}} = hat{Q} tag{1} $$



        or the following, which is what you want



        $$ underbrace{Q_{n} Q_{n-1} cdots Q_{2}Q_{1}}_{{Q}^{*}}A = R tag{2} $$



        In $1$ we see that Gram-Schmidt applies a sequence of elementary triangular matrices $R_{k}$ on the right of $A$. For $2$ we see a sequence of elementary unitary matrices $Q_{k}$ on the left of $A$. The matrices $Q_{k}$ can be givens matrices.



        $$ underbrace{G_{n} G_{n-1} cdots G_{2}G_{1}}_{{G}^{*}}A = R tag{3} $$



        All you need to prove is that $G_{k}$ is a unitary matrix. It is. There are notes provided here.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 18:15









        Ryan HoweRyan Howe

        2,41911323




        2,41911323






























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