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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027503%2fshow-that-if-h-is-an-hermitian-matrix-then-u-ii-hiih-1-is-unitary%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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 $UU^*=I$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 0:21









        egregegreg

        183k1486205




        183k1486205






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027503%2fshow-that-if-h-is-an-hermitian-matrix-then-u-ii-hiih-1-is-unitary%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more