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









$endgroup$













    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%2f3023566%2ffinding-the-polar-decomposition-of-a-2-times2-matrix%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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









    $endgroup$


















      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






























            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%2f3023566%2ffinding-the-polar-decomposition-of-a-2-times2-matrix%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

            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

            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