Show that if $H$ is an Hermitian matrix, then $U=(iI-H)(iI+H)^{-1}$ is unitary












3












$begingroup$


How would you prove, that when $H$ is a hermitian matrix, then:



$$U=(iI-H)(iI+H)^{-1},$$



where $U$ is unitary, assuming that $(iI+H)$ is invertible.



I thought that because $U$ is unitary, $UU^*=U^*U=I$.



So I tried to take the conjugate transpose of $U$ where:



$$U^*=(-iI-H)(H-iI)^{-1}.$$



Essentially this is where I am stuck:



1) How do I take the transpose of $U$, so I don't just have the conjugate -- or does it not matter?



2) How do I get rid of the inverse because I don't know what to do with it and how to eventually prove $UU^*=I$?



Thank you!






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Take a look at the Cayley transform en.wikipedia.org/wiki/Cayley_transform which is a cousin of the transform you are interested in.
    $endgroup$
    – Jean Marie
    Dec 6 '18 at 23:04
















3












$begingroup$


How would you prove, that when $H$ is a hermitian matrix, then:



$$U=(iI-H)(iI+H)^{-1},$$



where $U$ is unitary, assuming that $(iI+H)$ is invertible.



I thought that because $U$ is unitary, $UU^*=U^*U=I$.



So I tried to take the conjugate transpose of $U$ where:



$$U^*=(-iI-H)(H-iI)^{-1}.$$



Essentially this is where I am stuck:



1) How do I take the transpose of $U$, so I don't just have the conjugate -- or does it not matter?



2) How do I get rid of the inverse because I don't know what to do with it and how to eventually prove $UU^*=I$?



Thank you!






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Take a look at the Cayley transform en.wikipedia.org/wiki/Cayley_transform which is a cousin of the transform you are interested in.
    $endgroup$
    – Jean Marie
    Dec 6 '18 at 23:04














3












3








3





$begingroup$


How would you prove, that when $H$ is a hermitian matrix, then:



$$U=(iI-H)(iI+H)^{-1},$$



where $U$ is unitary, assuming that $(iI+H)$ is invertible.



I thought that because $U$ is unitary, $UU^*=U^*U=I$.



So I tried to take the conjugate transpose of $U$ where:



$$U^*=(-iI-H)(H-iI)^{-1}.$$



Essentially this is where I am stuck:



1) How do I take the transpose of $U$, so I don't just have the conjugate -- or does it not matter?



2) How do I get rid of the inverse because I don't know what to do with it and how to eventually prove $UU^*=I$?



Thank you!






linear-algebra matrices







share|cite|improve this question











$endgroup$




How would you prove, that when $H$ is a hermitian matrix, then:



$$U=(iI-H)(iI+H)^{-1},$$



where $U$ is unitary, assuming that $(iI+H)$ is invertible.



I thought that because $U$ is unitary, $UU^*=U^*U=I$.



So I tried to take the conjugate transpose of $U$ where:



$$U^*=(-iI-H)(H-iI)^{-1}.$$



Essentially this is where I am stuck:



1) How do I take the transpose of $U$, so I don't just have the conjugate -- or does it not matter?



2) How do I get rid of the inverse because I don't know what to do with it and how to eventually prove $UU^*=I$?



Thank you!






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 22:57









Jean Marie

30.5k42153




30.5k42153










asked Dec 5 '18 at 19:06









user623276user623276

162




162








  • 1




    $begingroup$
    Take a look at the Cayley transform en.wikipedia.org/wiki/Cayley_transform which is a cousin of the transform you are interested in.
    $endgroup$
    – Jean Marie
    Dec 6 '18 at 23:04














  • 1




    $begingroup$
    Take a look at the Cayley transform en.wikipedia.org/wiki/Cayley_transform which is a cousin of the transform you are interested in.
    $endgroup$
    – Jean Marie
    Dec 6 '18 at 23:04








1




1




$begingroup$
Take a look at the Cayley transform en.wikipedia.org/wiki/Cayley_transform which is a cousin of the transform you are interested in.
$endgroup$
– Jean Marie
Dec 6 '18 at 23:04




$begingroup$
Take a look at the Cayley transform en.wikipedia.org/wiki/Cayley_transform which is a cousin of the transform you are interested in.
$endgroup$
– Jean Marie
Dec 6 '18 at 23:04










3 Answers
3






active

oldest

votes


















3












$begingroup$

Let $H$ be Hermitian and $U=(iI-H)(iI+H)^{-1}$. Then
$$
U^*=(-iI+H)^{-1}(-iI-H) = (iI-H)^{-1}(iI+H) .
$$



Therefore
$$
begin{split}
UU^*
&= (iI-H)(iI+H)^{-1}(iI-H)^{-1}(iI+H) \
&= (iI-H)(iI+H)^{-1}(iI+H)(iI-H)^{-1} \
&=(iI-H)(iI-H)^{-1} \
&= I.
end{split}
$$




They key fact is that for every $x,yinmathbb C$ and every matrix $A$ you have
$$
(xI+A)(yI+A)=(yI+A)(xI+A).
$$
Multiplying by $(yI+A)^{-1}$ both on the left and the right, you get also
$$
(yI+A)^{-1}(xI+A) = (xI+A)(yI+A)^{-1}.
$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The OP can be said that in general two polynomials in the same matrix commute : $p(A)q(A)=q(A)p(A)$ where $p$ and $q$ are any polynomials.
    $endgroup$
    – Jean Marie
    Dec 6 '18 at 23:00










  • $begingroup$
    Yes, of course, it follows either by a direct computation or from what I stated, once you factorize the polynomials. And then you get also $p(A)q(A)^{-1} = q(A)^{-1}p(A)$, meaning that you can evaluate the rational function $p/q$ however you prefer
    $endgroup$
    – Federico
    Dec 7 '18 at 14:05



















2












$begingroup$

The point is that $(AB)^* = B^* A^*$, which implies $(A^{-1})^* = (A^*)^{-1}$.
Thus if $U = (iI-H)(iI+H)^{-1}$, $$U^* = ((iI + H)^{-1})^* (iI - H)^* =
(-iI+H)^{-1} (-iI-H) = (iI-H)^{-1}(iI+H)$$
and $U U^* = U^* U = I$ (note that $iI+H$ and $iI-H$ and their inverses commute).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think that the point that he missed was that the matrices commute. I wrote that in my answer
    $endgroup$
    – Federico
    Dec 5 '18 at 19:24



















0












$begingroup$

The matrices $iI+H$ and $iI-H$ are invertible, because $H$ only has real eigenvalues, being Hermitian.



Your equality can also be written as $U(iI-H)=(iI+H)$. Taking the Hermitian transpose,
$$
(iI-H)^*U^*=(iI+H)^*
$$
so that
$$
(-iI-H)U^*=-iI+H
$$
which is equivalent to $(iI+H)U^*=iI-H$. Multiply the left-hand side by $U(iI-H)$ and the right hand side by $(iI+H)$:
$$
(iI+H)U^*U(iI-H)=(iI-H)(iI+H)
$$
However, $(iI-H)(iI+H)=(iI+H)(iI-H)$ by direct computation, so
$$
(iI+H)U^*U(iI-H)=(iI+H)(iI-H)
$$
and we can cancel on both sides the invertible matrices, leaving $UU^*=I$.






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Let $H$ be Hermitian and $U=(iI-H)(iI+H)^{-1}$. Then
    $$
    U^*=(-iI+H)^{-1}(-iI-H) = (iI-H)^{-1}(iI+H) .
    $$



    Therefore
    $$
    begin{split}
    UU^*
    &= (iI-H)(iI+H)^{-1}(iI-H)^{-1}(iI+H) \
    &= (iI-H)(iI+H)^{-1}(iI+H)(iI-H)^{-1} \
    &=(iI-H)(iI-H)^{-1} \
    &= I.
    end{split}
    $$




    They key fact is that for every $x,yinmathbb C$ and every matrix $A$ you have
    $$
    (xI+A)(yI+A)=(yI+A)(xI+A).
    $$
    Multiplying by $(yI+A)^{-1}$ both on the left and the right, you get also
    $$
    (yI+A)^{-1}(xI+A) = (xI+A)(yI+A)^{-1}.
    $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      The OP can be said that in general two polynomials in the same matrix commute : $p(A)q(A)=q(A)p(A)$ where $p$ and $q$ are any polynomials.
      $endgroup$
      – Jean Marie
      Dec 6 '18 at 23:00










    • $begingroup$
      Yes, of course, it follows either by a direct computation or from what I stated, once you factorize the polynomials. And then you get also $p(A)q(A)^{-1} = q(A)^{-1}p(A)$, meaning that you can evaluate the rational function $p/q$ however you prefer
      $endgroup$
      – Federico
      Dec 7 '18 at 14:05
















    3












    $begingroup$

    Let $H$ be Hermitian and $U=(iI-H)(iI+H)^{-1}$. Then
    $$
    U^*=(-iI+H)^{-1}(-iI-H) = (iI-H)^{-1}(iI+H) .
    $$



    Therefore
    $$
    begin{split}
    UU^*
    &= (iI-H)(iI+H)^{-1}(iI-H)^{-1}(iI+H) \
    &= (iI-H)(iI+H)^{-1}(iI+H)(iI-H)^{-1} \
    &=(iI-H)(iI-H)^{-1} \
    &= I.
    end{split}
    $$




    They key fact is that for every $x,yinmathbb C$ and every matrix $A$ you have
    $$
    (xI+A)(yI+A)=(yI+A)(xI+A).
    $$
    Multiplying by $(yI+A)^{-1}$ both on the left and the right, you get also
    $$
    (yI+A)^{-1}(xI+A) = (xI+A)(yI+A)^{-1}.
    $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      The OP can be said that in general two polynomials in the same matrix commute : $p(A)q(A)=q(A)p(A)$ where $p$ and $q$ are any polynomials.
      $endgroup$
      – Jean Marie
      Dec 6 '18 at 23:00










    • $begingroup$
      Yes, of course, it follows either by a direct computation or from what I stated, once you factorize the polynomials. And then you get also $p(A)q(A)^{-1} = q(A)^{-1}p(A)$, meaning that you can evaluate the rational function $p/q$ however you prefer
      $endgroup$
      – Federico
      Dec 7 '18 at 14:05














    3












    3








    3





    $begingroup$

    Let $H$ be Hermitian and $U=(iI-H)(iI+H)^{-1}$. Then
    $$
    U^*=(-iI+H)^{-1}(-iI-H) = (iI-H)^{-1}(iI+H) .
    $$



    Therefore
    $$
    begin{split}
    UU^*
    &= (iI-H)(iI+H)^{-1}(iI-H)^{-1}(iI+H) \
    &= (iI-H)(iI+H)^{-1}(iI+H)(iI-H)^{-1} \
    &=(iI-H)(iI-H)^{-1} \
    &= I.
    end{split}
    $$




    They key fact is that for every $x,yinmathbb C$ and every matrix $A$ you have
    $$
    (xI+A)(yI+A)=(yI+A)(xI+A).
    $$
    Multiplying by $(yI+A)^{-1}$ both on the left and the right, you get also
    $$
    (yI+A)^{-1}(xI+A) = (xI+A)(yI+A)^{-1}.
    $$






    share|cite|improve this answer









    $endgroup$



    Let $H$ be Hermitian and $U=(iI-H)(iI+H)^{-1}$. Then
    $$
    U^*=(-iI+H)^{-1}(-iI-H) = (iI-H)^{-1}(iI+H) .
    $$



    Therefore
    $$
    begin{split}
    UU^*
    &= (iI-H)(iI+H)^{-1}(iI-H)^{-1}(iI+H) \
    &= (iI-H)(iI+H)^{-1}(iI+H)(iI-H)^{-1} \
    &=(iI-H)(iI-H)^{-1} \
    &= I.
    end{split}
    $$




    They key fact is that for every $x,yinmathbb C$ and every matrix $A$ you have
    $$
    (xI+A)(yI+A)=(yI+A)(xI+A).
    $$
    Multiplying by $(yI+A)^{-1}$ both on the left and the right, you get also
    $$
    (yI+A)^{-1}(xI+A) = (xI+A)(yI+A)^{-1}.
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 5 '18 at 19:23









    FedericoFederico

    5,134514




    5,134514








    • 1




      $begingroup$
      The OP can be said that in general two polynomials in the same matrix commute : $p(A)q(A)=q(A)p(A)$ where $p$ and $q$ are any polynomials.
      $endgroup$
      – Jean Marie
      Dec 6 '18 at 23:00










    • $begingroup$
      Yes, of course, it follows either by a direct computation or from what I stated, once you factorize the polynomials. And then you get also $p(A)q(A)^{-1} = q(A)^{-1}p(A)$, meaning that you can evaluate the rational function $p/q$ however you prefer
      $endgroup$
      – Federico
      Dec 7 '18 at 14:05














    • 1




      $begingroup$
      The OP can be said that in general two polynomials in the same matrix commute : $p(A)q(A)=q(A)p(A)$ where $p$ and $q$ are any polynomials.
      $endgroup$
      – Jean Marie
      Dec 6 '18 at 23:00










    • $begingroup$
      Yes, of course, it follows either by a direct computation or from what I stated, once you factorize the polynomials. And then you get also $p(A)q(A)^{-1} = q(A)^{-1}p(A)$, meaning that you can evaluate the rational function $p/q$ however you prefer
      $endgroup$
      – Federico
      Dec 7 '18 at 14:05








    1




    1




    $begingroup$
    The OP can be said that in general two polynomials in the same matrix commute : $p(A)q(A)=q(A)p(A)$ where $p$ and $q$ are any polynomials.
    $endgroup$
    – Jean Marie
    Dec 6 '18 at 23:00




    $begingroup$
    The OP can be said that in general two polynomials in the same matrix commute : $p(A)q(A)=q(A)p(A)$ where $p$ and $q$ are any polynomials.
    $endgroup$
    – Jean Marie
    Dec 6 '18 at 23:00












    $begingroup$
    Yes, of course, it follows either by a direct computation or from what I stated, once you factorize the polynomials. And then you get also $p(A)q(A)^{-1} = q(A)^{-1}p(A)$, meaning that you can evaluate the rational function $p/q$ however you prefer
    $endgroup$
    – Federico
    Dec 7 '18 at 14:05




    $begingroup$
    Yes, of course, it follows either by a direct computation or from what I stated, once you factorize the polynomials. And then you get also $p(A)q(A)^{-1} = q(A)^{-1}p(A)$, meaning that you can evaluate the rational function $p/q$ however you prefer
    $endgroup$
    – Federico
    Dec 7 '18 at 14:05











    2












    $begingroup$

    The point is that $(AB)^* = B^* A^*$, which implies $(A^{-1})^* = (A^*)^{-1}$.
    Thus if $U = (iI-H)(iI+H)^{-1}$, $$U^* = ((iI + H)^{-1})^* (iI - H)^* =
    (-iI+H)^{-1} (-iI-H) = (iI-H)^{-1}(iI+H)$$
    and $U U^* = U^* U = I$ (note that $iI+H$ and $iI-H$ and their inverses commute).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think that the point that he missed was that the matrices commute. I wrote that in my answer
      $endgroup$
      – Federico
      Dec 5 '18 at 19:24
















    2












    $begingroup$

    The point is that $(AB)^* = B^* A^*$, which implies $(A^{-1})^* = (A^*)^{-1}$.
    Thus if $U = (iI-H)(iI+H)^{-1}$, $$U^* = ((iI + H)^{-1})^* (iI - H)^* =
    (-iI+H)^{-1} (-iI-H) = (iI-H)^{-1}(iI+H)$$
    and $U U^* = U^* U = I$ (note that $iI+H$ and $iI-H$ and their inverses commute).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think that the point that he missed was that the matrices commute. I wrote that in my answer
      $endgroup$
      – Federico
      Dec 5 '18 at 19:24














    2












    2








    2





    $begingroup$

    The point is that $(AB)^* = B^* A^*$, which implies $(A^{-1})^* = (A^*)^{-1}$.
    Thus if $U = (iI-H)(iI+H)^{-1}$, $$U^* = ((iI + H)^{-1})^* (iI - H)^* =
    (-iI+H)^{-1} (-iI-H) = (iI-H)^{-1}(iI+H)$$
    and $U U^* = U^* U = I$ (note that $iI+H$ and $iI-H$ and their inverses commute).






    share|cite|improve this answer









    $endgroup$



    The point is that $(AB)^* = B^* A^*$, which implies $(A^{-1})^* = (A^*)^{-1}$.
    Thus if $U = (iI-H)(iI+H)^{-1}$, $$U^* = ((iI + H)^{-1})^* (iI - H)^* =
    (-iI+H)^{-1} (-iI-H) = (iI-H)^{-1}(iI+H)$$
    and $U U^* = U^* U = I$ (note that $iI+H$ and $iI-H$ and their inverses commute).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 5 '18 at 19:23









    Robert IsraelRobert Israel

    326k23215469




    326k23215469












    • $begingroup$
      I think that the point that he missed was that the matrices commute. I wrote that in my answer
      $endgroup$
      – Federico
      Dec 5 '18 at 19:24


















    • $begingroup$
      I think that the point that he missed was that the matrices commute. I wrote that in my answer
      $endgroup$
      – Federico
      Dec 5 '18 at 19:24
















    $begingroup$
    I think that the point that he missed was that the matrices commute. I wrote that in my answer
    $endgroup$
    – Federico
    Dec 5 '18 at 19:24




    $begingroup$
    I think that the point that he missed was that the matrices commute. I wrote that in my answer
    $endgroup$
    – Federico
    Dec 5 '18 at 19:24











    0












    $begingroup$

    The matrices $iI+H$ and $iI-H$ are invertible, because $H$ only has real eigenvalues, being Hermitian.



    Your equality can also be written as $U(iI-H)=(iI+H)$. Taking the Hermitian transpose,
    $$
    (iI-H)^*U^*=(iI+H)^*
    $$
    so that
    $$
    (-iI-H)U^*=-iI+H
    $$
    which is equivalent to $(iI+H)U^*=iI-H$. Multiply the left-hand side by $U(iI-H)$ and the right hand side by $(iI+H)$:
    $$
    (iI+H)U^*U(iI-H)=(iI-H)(iI+H)
    $$
    However, $(iI-H)(iI+H)=(iI+H)(iI-H)$ by direct computation, so
    $$
    (iI+H)U^*U(iI-H)=(iI+H)(iI-H)
    $$
    and we can cancel on both sides the invertible matrices, leaving $UU^*=I$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The matrices $iI+H$ and $iI-H$ are invertible, because $H$ only has real eigenvalues, being Hermitian.



      Your equality can also be written as $U(iI-H)=(iI+H)$. Taking the Hermitian transpose,
      $$
      (iI-H)^*U^*=(iI+H)^*
      $$
      so that
      $$
      (-iI-H)U^*=-iI+H
      $$
      which is equivalent to $(iI+H)U^*=iI-H$. Multiply the left-hand side by $U(iI-H)$ and the right hand side by $(iI+H)$:
      $$
      (iI+H)U^*U(iI-H)=(iI-H)(iI+H)
      $$
      However, $(iI-H)(iI+H)=(iI+H)(iI-H)$ by direct computation, so
      $$
      (iI+H)U^*U(iI-H)=(iI+H)(iI-H)
      $$
      and we can cancel on both sides the invertible matrices, leaving $UU^*=I$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The matrices $iI+H$ and $iI-H$ are invertible, because $H$ only has real eigenvalues, being Hermitian.



        Your equality can also be written as $U(iI-H)=(iI+H)$. Taking the Hermitian transpose,
        $$
        (iI-H)^*U^*=(iI+H)^*
        $$
        so that
        $$
        (-iI-H)U^*=-iI+H
        $$
        which is equivalent to $(iI+H)U^*=iI-H$. Multiply the left-hand side by $U(iI-H)$ and the right hand side by $(iI+H)$:
        $$
        (iI+H)U^*U(iI-H)=(iI-H)(iI+H)
        $$
        However, $(iI-H)(iI+H)=(iI+H)(iI-H)$ by direct computation, so
        $$
        (iI+H)U^*U(iI-H)=(iI+H)(iI-H)
        $$
        and we can cancel on both sides the invertible matrices, leaving $UU^*=I$.






        share|cite|improve this answer









        $endgroup$



        The matrices $iI+H$ and $iI-H$ are invertible, because $H$ only has real eigenvalues, being Hermitian.



        Your equality can also be written as $U(iI-H)=(iI+H)$. Taking the Hermitian transpose,
        $$
        (iI-H)^*U^*=(iI+H)^*
        $$
        so that
        $$
        (-iI-H)U^*=-iI+H
        $$
        which is equivalent to $(iI+H)U^*=iI-H$. Multiply the left-hand side by $U(iI-H)$ and the right hand side by $(iI+H)$:
        $$
        (iI+H)U^*U(iI-H)=(iI-H)(iI+H)
        $$
        However, $(iI-H)(iI+H)=(iI+H)(iI-H)$ by direct computation, so
        $$
        (iI+H)U^*U(iI-H)=(iI+H)(iI-H)
        $$
        and we can cancel on both sides the invertible matrices, leaving $UU^*=I$.







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        answered Dec 7 '18 at 0:21









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