Inverse of matrix product












0












$begingroup$


Let A be an $ntimes n$ matrix and B be an $ntimes m$ matrix with $m<n$ can I use this identity?



$(AB)^{+}=B^+A^{-1}$



I am not sure that this is the right inverse of the product $AB$ if this is correct and works as the inverse of square matrices product can you give me a reference?
Otherwise I need ideas. Thanks






matrices inverse







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $AB$ is not a square matrix. But you can write $(AB)^+ = B^+ A^+ = B^+ A^{-1}$ if $A$ has an inverse
    $endgroup$
    – Damien
    Jan 4 at 9:05


















0












$begingroup$


Let A be an $ntimes n$ matrix and B be an $ntimes m$ matrix with $m<n$ can I use this identity?



$(AB)^{+}=B^+A^{-1}$



I am not sure that this is the right inverse of the product $AB$ if this is correct and works as the inverse of square matrices product can you give me a reference?
Otherwise I need ideas. Thanks






matrices inverse







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $AB$ is not a square matrix. But you can write $(AB)^+ = B^+ A^+ = B^+ A^{-1}$ if $A$ has an inverse
    $endgroup$
    – Damien
    Jan 4 at 9:05
















0












0








0





$begingroup$


Let A be an $ntimes n$ matrix and B be an $ntimes m$ matrix with $m<n$ can I use this identity?



$(AB)^{+}=B^+A^{-1}$



I am not sure that this is the right inverse of the product $AB$ if this is correct and works as the inverse of square matrices product can you give me a reference?
Otherwise I need ideas. Thanks






matrices inverse







share|cite|improve this question











$endgroup$




Let A be an $ntimes n$ matrix and B be an $ntimes m$ matrix with $m<n$ can I use this identity?



$(AB)^{+}=B^+A^{-1}$



I am not sure that this is the right inverse of the product $AB$ if this is correct and works as the inverse of square matrices product can you give me a reference?
Otherwise I need ideas. Thanks






matrices inverse




matrices inverse






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 14:12







iacopo

















asked Jan 4 at 9:02









iacopoiacopo

1127




1127








  • 1




    $begingroup$
    $AB$ is not a square matrix. But you can write $(AB)^+ = B^+ A^+ = B^+ A^{-1}$ if $A$ has an inverse
    $endgroup$
    – Damien
    Jan 4 at 9:05
















  • 1




    $begingroup$
    $AB$ is not a square matrix. But you can write $(AB)^+ = B^+ A^+ = B^+ A^{-1}$ if $A$ has an inverse
    $endgroup$
    – Damien
    Jan 4 at 9:05










1




1




$begingroup$
$AB$ is not a square matrix. But you can write $(AB)^+ = B^+ A^+ = B^+ A^{-1}$ if $A$ has an inverse
$endgroup$
– Damien
Jan 4 at 9:05






$begingroup$
$AB$ is not a square matrix. But you can write $(AB)^+ = B^+ A^+ = B^+ A^{-1}$ if $A$ has an inverse
$endgroup$
– Damien
Jan 4 at 9:05












2 Answers
2






active

oldest

votes


















1












$begingroup$

I suppose that you are asking whether $(AB)^{color{red}{+}}=B^+A^{-1}$ rather than whether $(AB)^{color{red}{-1}}=B^+A^{-1}$. As other users have pointed out $AB$ is not a square matrix in general and it makes no sense to talk about $(AB)^{-1}$.



Is $(AB)^+=B^+A^{-1}$? Unfortunately, the answer is no. Out of the four defining properties of Moore-Penrose pseudoinverse, $B^+A^{-1}$ only satisfies three. In particular, $(AB)(AB)^+$ should be Hermitian, but that in general does not hold if you put $B^+A^{-1}$ in place of $(AB)^+$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Correct and thank you
    $endgroup$
    – iacopo
    Jan 4 at 14:12



















0












$begingroup$

No, you cannot do that, since $AB$ is not a square matrix and therefore it has no inverse.






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%2f3061443%2finverse-of-matrix-product%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    I suppose that you are asking whether $(AB)^{color{red}{+}}=B^+A^{-1}$ rather than whether $(AB)^{color{red}{-1}}=B^+A^{-1}$. As other users have pointed out $AB$ is not a square matrix in general and it makes no sense to talk about $(AB)^{-1}$.



    Is $(AB)^+=B^+A^{-1}$? Unfortunately, the answer is no. Out of the four defining properties of Moore-Penrose pseudoinverse, $B^+A^{-1}$ only satisfies three. In particular, $(AB)(AB)^+$ should be Hermitian, but that in general does not hold if you put $B^+A^{-1}$ in place of $(AB)^+$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Correct and thank you
      $endgroup$
      – iacopo
      Jan 4 at 14:12
















    1












    $begingroup$

    I suppose that you are asking whether $(AB)^{color{red}{+}}=B^+A^{-1}$ rather than whether $(AB)^{color{red}{-1}}=B^+A^{-1}$. As other users have pointed out $AB$ is not a square matrix in general and it makes no sense to talk about $(AB)^{-1}$.



    Is $(AB)^+=B^+A^{-1}$? Unfortunately, the answer is no. Out of the four defining properties of Moore-Penrose pseudoinverse, $B^+A^{-1}$ only satisfies three. In particular, $(AB)(AB)^+$ should be Hermitian, but that in general does not hold if you put $B^+A^{-1}$ in place of $(AB)^+$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Correct and thank you
      $endgroup$
      – iacopo
      Jan 4 at 14:12














    1












    1








    1





    $begingroup$

    I suppose that you are asking whether $(AB)^{color{red}{+}}=B^+A^{-1}$ rather than whether $(AB)^{color{red}{-1}}=B^+A^{-1}$. As other users have pointed out $AB$ is not a square matrix in general and it makes no sense to talk about $(AB)^{-1}$.



    Is $(AB)^+=B^+A^{-1}$? Unfortunately, the answer is no. Out of the four defining properties of Moore-Penrose pseudoinverse, $B^+A^{-1}$ only satisfies three. In particular, $(AB)(AB)^+$ should be Hermitian, but that in general does not hold if you put $B^+A^{-1}$ in place of $(AB)^+$.






    share|cite|improve this answer









    $endgroup$



    I suppose that you are asking whether $(AB)^{color{red}{+}}=B^+A^{-1}$ rather than whether $(AB)^{color{red}{-1}}=B^+A^{-1}$. As other users have pointed out $AB$ is not a square matrix in general and it makes no sense to talk about $(AB)^{-1}$.



    Is $(AB)^+=B^+A^{-1}$? Unfortunately, the answer is no. Out of the four defining properties of Moore-Penrose pseudoinverse, $B^+A^{-1}$ only satisfies three. In particular, $(AB)(AB)^+$ should be Hermitian, but that in general does not hold if you put $B^+A^{-1}$ in place of $(AB)^+$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 4 at 11:09









    user1551user1551

    73.4k566128




    73.4k566128












    • $begingroup$
      Correct and thank you
      $endgroup$
      – iacopo
      Jan 4 at 14:12


















    • $begingroup$
      Correct and thank you
      $endgroup$
      – iacopo
      Jan 4 at 14:12
















    $begingroup$
    Correct and thank you
    $endgroup$
    – iacopo
    Jan 4 at 14:12




    $begingroup$
    Correct and thank you
    $endgroup$
    – iacopo
    Jan 4 at 14:12











    0












    $begingroup$

    No, you cannot do that, since $AB$ is not a square matrix and therefore it has no inverse.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      No, you cannot do that, since $AB$ is not a square matrix and therefore it has no inverse.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        No, you cannot do that, since $AB$ is not a square matrix and therefore it has no inverse.






        share|cite|improve this answer









        $endgroup$



        No, you cannot do that, since $AB$ is not a square matrix and therefore it has no inverse.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 9:05









        José Carlos SantosJosé Carlos Santos

        165k22132235




        165k22132235






























            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%2f3061443%2finverse-of-matrix-product%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

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more