Finding the polar decomposition of a $2times2$ matrix.












1












$begingroup$


Specifically, the question is as follows:




Let $B$ be the standard basis of $mathbb{R}^2$. Let $Tinmathcal{L}(mathbb{R}^2)$ be such that
$$mathcal{M}(T,B)=begin{bmatrix}2&3\0&2end{bmatrix}$$
Find the polar decomposition $T=Ssqrt{T^*T}$ of $T$. (Hint: The eigenvectors of $T^*T$ are $(1,2)$ and $(-2,1)$.)




It seems like this should be very straightforward. $T$ and $T^*$ are both invertible, so it should follow that $S=sqrt{T(T^*)^{-1}}$, but this is false, as is easily checked (https://www.wolframalpha.com/input/?i=sqrt(%7B%7B2,3%7D,%7B0,2%7D%7D%7B%7B2,0%7D,%7B3,2%7D%7D%5E-1)sqrt(%7B%7B2,0%7D,%7B3,2%7D%7D%7B%7B2,3%7D,%7B0,2%7D%7D)).



I'm not quite sure why the eigenvectors have been given considering the eigenvalues haven't been given, but the eigenvalues of $T^*T$ are $1$ and $16$. My other thought was to use the singular value decomposition to find the polar decomposition, but I run into similar computational issues. These attempts make me think I'm missing something theoretically. This is a homework question, so I request assistance in finding my error/hints to get to the true solution, but not an actual solution.






linear-algebra matrices proof-verification matrix-decomposition adjoint-operators







share|cite|improve this question









$endgroup$












  • $begingroup$
    $T (sqrt {T^* T})^{-1}$ seems not equal $sqrt {T(T^*)^{-1}}$. This is not the arithmetic operation of real numbers. They are matrices, lots of things are different.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:51










  • $begingroup$
    @xbh Sure, but $T=Ssqrt{T^*T}Rightarrow T^2=S^2T^*TRightarrow T^2T^{-1}=S^2T^*Rightarrow T=S^2T^*Rightarrow T(T^*)^{-1}=S^2$, so in fact they should be equal, for invertible $T$ and $T^*$.
    $endgroup$
    – Atsina
    Dec 3 '18 at 4:54






  • 1




    $begingroup$
    No, $(AB)^2 neq A^2B^2$ generally, and if $AB = BA$ the the equation holds. So the first $implies$ is questionable.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:56












  • $begingroup$
    Also $sqrt T$ makes sense only when $T$ is a symmetric positive definite matrix.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:58










  • $begingroup$
    @xbh Ah, right! Rookie mistake.
    $endgroup$
    – Atsina
    Dec 3 '18 at 4:59
















1












$begingroup$


Specifically, the question is as follows:




Let $B$ be the standard basis of $mathbb{R}^2$. Let $Tinmathcal{L}(mathbb{R}^2)$ be such that
$$mathcal{M}(T,B)=begin{bmatrix}2&3\0&2end{bmatrix}$$
Find the polar decomposition $T=Ssqrt{T^*T}$ of $T$. (Hint: The eigenvectors of $T^*T$ are $(1,2)$ and $(-2,1)$.)




It seems like this should be very straightforward. $T$ and $T^*$ are both invertible, so it should follow that $S=sqrt{T(T^*)^{-1}}$, but this is false, as is easily checked (https://www.wolframalpha.com/input/?i=sqrt(%7B%7B2,3%7D,%7B0,2%7D%7D%7B%7B2,0%7D,%7B3,2%7D%7D%5E-1)sqrt(%7B%7B2,0%7D,%7B3,2%7D%7D%7B%7B2,3%7D,%7B0,2%7D%7D)).



I'm not quite sure why the eigenvectors have been given considering the eigenvalues haven't been given, but the eigenvalues of $T^*T$ are $1$ and $16$. My other thought was to use the singular value decomposition to find the polar decomposition, but I run into similar computational issues. These attempts make me think I'm missing something theoretically. This is a homework question, so I request assistance in finding my error/hints to get to the true solution, but not an actual solution.






linear-algebra matrices proof-verification matrix-decomposition adjoint-operators







share|cite|improve this question









$endgroup$












  • $begingroup$
    $T (sqrt {T^* T})^{-1}$ seems not equal $sqrt {T(T^*)^{-1}}$. This is not the arithmetic operation of real numbers. They are matrices, lots of things are different.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:51










  • $begingroup$
    @xbh Sure, but $T=Ssqrt{T^*T}Rightarrow T^2=S^2T^*TRightarrow T^2T^{-1}=S^2T^*Rightarrow T=S^2T^*Rightarrow T(T^*)^{-1}=S^2$, so in fact they should be equal, for invertible $T$ and $T^*$.
    $endgroup$
    – Atsina
    Dec 3 '18 at 4:54






  • 1




    $begingroup$
    No, $(AB)^2 neq A^2B^2$ generally, and if $AB = BA$ the the equation holds. So the first $implies$ is questionable.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:56












  • $begingroup$
    Also $sqrt T$ makes sense only when $T$ is a symmetric positive definite matrix.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:58










  • $begingroup$
    @xbh Ah, right! Rookie mistake.
    $endgroup$
    – Atsina
    Dec 3 '18 at 4:59














1












1








1





$begingroup$


Specifically, the question is as follows:




Let $B$ be the standard basis of $mathbb{R}^2$. Let $Tinmathcal{L}(mathbb{R}^2)$ be such that
$$mathcal{M}(T,B)=begin{bmatrix}2&3\0&2end{bmatrix}$$
Find the polar decomposition $T=Ssqrt{T^*T}$ of $T$. (Hint: The eigenvectors of $T^*T$ are $(1,2)$ and $(-2,1)$.)




It seems like this should be very straightforward. $T$ and $T^*$ are both invertible, so it should follow that $S=sqrt{T(T^*)^{-1}}$, but this is false, as is easily checked (https://www.wolframalpha.com/input/?i=sqrt(%7B%7B2,3%7D,%7B0,2%7D%7D%7B%7B2,0%7D,%7B3,2%7D%7D%5E-1)sqrt(%7B%7B2,0%7D,%7B3,2%7D%7D%7B%7B2,3%7D,%7B0,2%7D%7D)).



I'm not quite sure why the eigenvectors have been given considering the eigenvalues haven't been given, but the eigenvalues of $T^*T$ are $1$ and $16$. My other thought was to use the singular value decomposition to find the polar decomposition, but I run into similar computational issues. These attempts make me think I'm missing something theoretically. This is a homework question, so I request assistance in finding my error/hints to get to the true solution, but not an actual solution.






linear-algebra matrices proof-verification matrix-decomposition adjoint-operators







share|cite|improve this question









$endgroup$




Specifically, the question is as follows:




Let $B$ be the standard basis of $mathbb{R}^2$. Let $Tinmathcal{L}(mathbb{R}^2)$ be such that
$$mathcal{M}(T,B)=begin{bmatrix}2&3\0&2end{bmatrix}$$
Find the polar decomposition $T=Ssqrt{T^*T}$ of $T$. (Hint: The eigenvectors of $T^*T$ are $(1,2)$ and $(-2,1)$.)




It seems like this should be very straightforward. $T$ and $T^*$ are both invertible, so it should follow that $S=sqrt{T(T^*)^{-1}}$, but this is false, as is easily checked (https://www.wolframalpha.com/input/?i=sqrt(%7B%7B2,3%7D,%7B0,2%7D%7D%7B%7B2,0%7D,%7B3,2%7D%7D%5E-1)sqrt(%7B%7B2,0%7D,%7B3,2%7D%7D%7B%7B2,3%7D,%7B0,2%7D%7D)).



I'm not quite sure why the eigenvectors have been given considering the eigenvalues haven't been given, but the eigenvalues of $T^*T$ are $1$ and $16$. My other thought was to use the singular value decomposition to find the polar decomposition, but I run into similar computational issues. These attempts make me think I'm missing something theoretically. This is a homework question, so I request assistance in finding my error/hints to get to the true solution, but not an actual solution.






linear-algebra matrices proof-verification matrix-decomposition adjoint-operators




linear-algebra matrices proof-verification matrix-decomposition adjoint-operators






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 3:11









AtsinaAtsina

798116




798116












  • $begingroup$
    $T (sqrt {T^* T})^{-1}$ seems not equal $sqrt {T(T^*)^{-1}}$. This is not the arithmetic operation of real numbers. They are matrices, lots of things are different.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:51










  • $begingroup$
    @xbh Sure, but $T=Ssqrt{T^*T}Rightarrow T^2=S^2T^*TRightarrow T^2T^{-1}=S^2T^*Rightarrow T=S^2T^*Rightarrow T(T^*)^{-1}=S^2$, so in fact they should be equal, for invertible $T$ and $T^*$.
    $endgroup$
    – Atsina
    Dec 3 '18 at 4:54






  • 1




    $begingroup$
    No, $(AB)^2 neq A^2B^2$ generally, and if $AB = BA$ the the equation holds. So the first $implies$ is questionable.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:56












  • $begingroup$
    Also $sqrt T$ makes sense only when $T$ is a symmetric positive definite matrix.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:58










  • $begingroup$
    @xbh Ah, right! Rookie mistake.
    $endgroup$
    – Atsina
    Dec 3 '18 at 4:59


















  • $begingroup$
    $T (sqrt {T^* T})^{-1}$ seems not equal $sqrt {T(T^*)^{-1}}$. This is not the arithmetic operation of real numbers. They are matrices, lots of things are different.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:51










  • $begingroup$
    @xbh Sure, but $T=Ssqrt{T^*T}Rightarrow T^2=S^2T^*TRightarrow T^2T^{-1}=S^2T^*Rightarrow T=S^2T^*Rightarrow T(T^*)^{-1}=S^2$, so in fact they should be equal, for invertible $T$ and $T^*$.
    $endgroup$
    – Atsina
    Dec 3 '18 at 4:54






  • 1




    $begingroup$
    No, $(AB)^2 neq A^2B^2$ generally, and if $AB = BA$ the the equation holds. So the first $implies$ is questionable.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:56












  • $begingroup$
    Also $sqrt T$ makes sense only when $T$ is a symmetric positive definite matrix.
    $endgroup$
    – xbh
    Dec 3 '18 at 4:58










  • $begingroup$
    @xbh Ah, right! Rookie mistake.
    $endgroup$
    – Atsina
    Dec 3 '18 at 4:59
















$begingroup$
$T (sqrt {T^* T})^{-1}$ seems not equal $sqrt {T(T^*)^{-1}}$. This is not the arithmetic operation of real numbers. They are matrices, lots of things are different.
$endgroup$
– xbh
Dec 3 '18 at 4:51




$begingroup$
$T (sqrt {T^* T})^{-1}$ seems not equal $sqrt {T(T^*)^{-1}}$. This is not the arithmetic operation of real numbers. They are matrices, lots of things are different.
$endgroup$
– xbh
Dec 3 '18 at 4:51












$begingroup$
@xbh Sure, but $T=Ssqrt{T^*T}Rightarrow T^2=S^2T^*TRightarrow T^2T^{-1}=S^2T^*Rightarrow T=S^2T^*Rightarrow T(T^*)^{-1}=S^2$, so in fact they should be equal, for invertible $T$ and $T^*$.
$endgroup$
– Atsina
Dec 3 '18 at 4:54




$begingroup$
@xbh Sure, but $T=Ssqrt{T^*T}Rightarrow T^2=S^2T^*TRightarrow T^2T^{-1}=S^2T^*Rightarrow T=S^2T^*Rightarrow T(T^*)^{-1}=S^2$, so in fact they should be equal, for invertible $T$ and $T^*$.
$endgroup$
– Atsina
Dec 3 '18 at 4:54




1




1




$begingroup$
No, $(AB)^2 neq A^2B^2$ generally, and if $AB = BA$ the the equation holds. So the first $implies$ is questionable.
$endgroup$
– xbh
Dec 3 '18 at 4:56






$begingroup$
No, $(AB)^2 neq A^2B^2$ generally, and if $AB = BA$ the the equation holds. So the first $implies$ is questionable.
$endgroup$
– xbh
Dec 3 '18 at 4:56














$begingroup$
Also $sqrt T$ makes sense only when $T$ is a symmetric positive definite matrix.
$endgroup$
– xbh
Dec 3 '18 at 4:58




$begingroup$
Also $sqrt T$ makes sense only when $T$ is a symmetric positive definite matrix.
$endgroup$
– xbh
Dec 3 '18 at 4:58












$begingroup$
@xbh Ah, right! Rookie mistake.
$endgroup$
– Atsina
Dec 3 '18 at 4:59




$begingroup$
@xbh Ah, right! Rookie mistake.
$endgroup$
– Atsina
Dec 3 '18 at 4:59










1 Answer
1






active

oldest

votes


















0












$begingroup$

I believe there's a more elegant solution, but this works:



Observe that $T^*T=begin{bmatrix}4&6\6&16end{bmatrix}$ and $sqrt{T^*T}=begin{bmatrix}2&sqrt{6}\sqrt{6}&sqrt{13}end{bmatrix}$. Also, as is easily verified with a computational algebra engine, $$sqrt{T^*T}^{-1}=begin{bmatrix}frac{sqrt{13}}{-6+2sqrt{13}}&-frac{sqrt{6}}{-6+2sqrt{13}}\-frac{sqrt{6}}{-6+2sqrt{13}}&frac{2}{-6+2sqrt{13}}end{bmatrix}.$$
Similarly,
$$S=Tsqrt{T^*T}^{-1}=begin{bmatrix}-frac{3sqrt{6}-2sqrt{13}}{2(-3+sqrt{13})}&-frac{-3+sqrt{6}}{-3+sqrt{13}}\
-frac{sqrt{6}}{-3+sqrt{13}}&frac{2}{-3+sqrt{13}}end{bmatrix}.$$
It is easily verified that this is the desired $S$.






share|cite|improve this answer









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    0












    $begingroup$

    I believe there's a more elegant solution, but this works:



    Observe that $T^*T=begin{bmatrix}4&6\6&16end{bmatrix}$ and $sqrt{T^*T}=begin{bmatrix}2&sqrt{6}\sqrt{6}&sqrt{13}end{bmatrix}$. Also, as is easily verified with a computational algebra engine, $$sqrt{T^*T}^{-1}=begin{bmatrix}frac{sqrt{13}}{-6+2sqrt{13}}&-frac{sqrt{6}}{-6+2sqrt{13}}\-frac{sqrt{6}}{-6+2sqrt{13}}&frac{2}{-6+2sqrt{13}}end{bmatrix}.$$
    Similarly,
    $$S=Tsqrt{T^*T}^{-1}=begin{bmatrix}-frac{3sqrt{6}-2sqrt{13}}{2(-3+sqrt{13})}&-frac{-3+sqrt{6}}{-3+sqrt{13}}\
    -frac{sqrt{6}}{-3+sqrt{13}}&frac{2}{-3+sqrt{13}}end{bmatrix}.$$
    It is easily verified that this is the desired $S$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I believe there's a more elegant solution, but this works:



      Observe that $T^*T=begin{bmatrix}4&6\6&16end{bmatrix}$ and $sqrt{T^*T}=begin{bmatrix}2&sqrt{6}\sqrt{6}&sqrt{13}end{bmatrix}$. Also, as is easily verified with a computational algebra engine, $$sqrt{T^*T}^{-1}=begin{bmatrix}frac{sqrt{13}}{-6+2sqrt{13}}&-frac{sqrt{6}}{-6+2sqrt{13}}\-frac{sqrt{6}}{-6+2sqrt{13}}&frac{2}{-6+2sqrt{13}}end{bmatrix}.$$
      Similarly,
      $$S=Tsqrt{T^*T}^{-1}=begin{bmatrix}-frac{3sqrt{6}-2sqrt{13}}{2(-3+sqrt{13})}&-frac{-3+sqrt{6}}{-3+sqrt{13}}\
      -frac{sqrt{6}}{-3+sqrt{13}}&frac{2}{-3+sqrt{13}}end{bmatrix}.$$
      It is easily verified that this is the desired $S$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I believe there's a more elegant solution, but this works:



        Observe that $T^*T=begin{bmatrix}4&6\6&16end{bmatrix}$ and $sqrt{T^*T}=begin{bmatrix}2&sqrt{6}\sqrt{6}&sqrt{13}end{bmatrix}$. Also, as is easily verified with a computational algebra engine, $$sqrt{T^*T}^{-1}=begin{bmatrix}frac{sqrt{13}}{-6+2sqrt{13}}&-frac{sqrt{6}}{-6+2sqrt{13}}\-frac{sqrt{6}}{-6+2sqrt{13}}&frac{2}{-6+2sqrt{13}}end{bmatrix}.$$
        Similarly,
        $$S=Tsqrt{T^*T}^{-1}=begin{bmatrix}-frac{3sqrt{6}-2sqrt{13}}{2(-3+sqrt{13})}&-frac{-3+sqrt{6}}{-3+sqrt{13}}\
        -frac{sqrt{6}}{-3+sqrt{13}}&frac{2}{-3+sqrt{13}}end{bmatrix}.$$
        It is easily verified that this is the desired $S$.






        share|cite|improve this answer









        $endgroup$



        I believe there's a more elegant solution, but this works:



        Observe that $T^*T=begin{bmatrix}4&6\6&16end{bmatrix}$ and $sqrt{T^*T}=begin{bmatrix}2&sqrt{6}\sqrt{6}&sqrt{13}end{bmatrix}$. Also, as is easily verified with a computational algebra engine, $$sqrt{T^*T}^{-1}=begin{bmatrix}frac{sqrt{13}}{-6+2sqrt{13}}&-frac{sqrt{6}}{-6+2sqrt{13}}\-frac{sqrt{6}}{-6+2sqrt{13}}&frac{2}{-6+2sqrt{13}}end{bmatrix}.$$
        Similarly,
        $$S=Tsqrt{T^*T}^{-1}=begin{bmatrix}-frac{3sqrt{6}-2sqrt{13}}{2(-3+sqrt{13})}&-frac{-3+sqrt{6}}{-3+sqrt{13}}\
        -frac{sqrt{6}}{-3+sqrt{13}}&frac{2}{-3+sqrt{13}}end{bmatrix}.$$
        It is easily verified that this is the desired $S$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 5:30









        AtsinaAtsina

        798116




        798116






























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