Why there are n-r dimensional set of vectors x for rank-deficient least square problem












0














I saw this theorem on the lecture slides (http://www2.aueb.gr/users/douros/docs_master/Least_Square_pr.pdf) with topic Rank-Deficient Least Squares Problem




Given $Ain R^{mtimes n}, m > n, , r = text{rank(A)}<n , text{and} , b in R^m$ and we want to find $x$ such that $min_x Vert Ax-b Vert_2$, then there are n-r dimensional set of vectors x that minimize $Vert Ax-b Vert_2$




What's the reason behind the "$n-r$" theorem?






linear-algebra optimization numerical-linear-algebra least-squares







share|cite|improve this question






















  • $n-r$ is the dimension of Nullity of $A$.
    – Yadati Kiran
    Nov 22 '18 at 3:47












  • @YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
    – WeiShan Ng
    Nov 22 '18 at 4:50










  • Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
    – Yadati Kiran
    Nov 22 '18 at 4:55


















0














I saw this theorem on the lecture slides (http://www2.aueb.gr/users/douros/docs_master/Least_Square_pr.pdf) with topic Rank-Deficient Least Squares Problem




Given $Ain R^{mtimes n}, m > n, , r = text{rank(A)}<n , text{and} , b in R^m$ and we want to find $x$ such that $min_x Vert Ax-b Vert_2$, then there are n-r dimensional set of vectors x that minimize $Vert Ax-b Vert_2$




What's the reason behind the "$n-r$" theorem?






linear-algebra optimization numerical-linear-algebra least-squares







share|cite|improve this question






















  • $n-r$ is the dimension of Nullity of $A$.
    – Yadati Kiran
    Nov 22 '18 at 3:47












  • @YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
    – WeiShan Ng
    Nov 22 '18 at 4:50










  • Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
    – Yadati Kiran
    Nov 22 '18 at 4:55
















0












0








0







I saw this theorem on the lecture slides (http://www2.aueb.gr/users/douros/docs_master/Least_Square_pr.pdf) with topic Rank-Deficient Least Squares Problem




Given $Ain R^{mtimes n}, m > n, , r = text{rank(A)}<n , text{and} , b in R^m$ and we want to find $x$ such that $min_x Vert Ax-b Vert_2$, then there are n-r dimensional set of vectors x that minimize $Vert Ax-b Vert_2$




What's the reason behind the "$n-r$" theorem?






linear-algebra optimization numerical-linear-algebra least-squares







share|cite|improve this question













I saw this theorem on the lecture slides (http://www2.aueb.gr/users/douros/docs_master/Least_Square_pr.pdf) with topic Rank-Deficient Least Squares Problem




Given $Ain R^{mtimes n}, m > n, , r = text{rank(A)}<n , text{and} , b in R^m$ and we want to find $x$ such that $min_x Vert Ax-b Vert_2$, then there are n-r dimensional set of vectors x that minimize $Vert Ax-b Vert_2$




What's the reason behind the "$n-r$" theorem?






linear-algebra optimization numerical-linear-algebra least-squares




linear-algebra optimization numerical-linear-algebra least-squares






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 '18 at 3:04









WeiShan Ng

617




617












  • $n-r$ is the dimension of Nullity of $A$.
    – Yadati Kiran
    Nov 22 '18 at 3:47












  • @YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
    – WeiShan Ng
    Nov 22 '18 at 4:50










  • Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
    – Yadati Kiran
    Nov 22 '18 at 4:55




















  • $n-r$ is the dimension of Nullity of $A$.
    – Yadati Kiran
    Nov 22 '18 at 3:47












  • @YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
    – WeiShan Ng
    Nov 22 '18 at 4:50










  • Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
    – Yadati Kiran
    Nov 22 '18 at 4:55


















$n-r$ is the dimension of Nullity of $A$.
– Yadati Kiran
Nov 22 '18 at 3:47






$n-r$ is the dimension of Nullity of $A$.
– Yadati Kiran
Nov 22 '18 at 3:47














@YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
– WeiShan Ng
Nov 22 '18 at 4:50




@YadatiKiran Why is the least square solution in the nullity of A?? I thought it will be in the nullity of $A^T$ so it will have a dimension of m-r?
– WeiShan Ng
Nov 22 '18 at 4:50












Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 '18 at 4:55






Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 '18 at 4:55












1 Answer
1






active

oldest

votes


















0














I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
$$A^T A vec{x} = A^T vec{b}$$

And we also know that




$R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$




So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$



Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$



And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
$$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$

which is of $n-r$ dimension






share|cite|improve this answer





















    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%2f3008698%2fwhy-there-are-n-r-dimensional-set-of-vectors-x-for-rank-deficient-least-square-p%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














    I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
    $$A^T A vec{x} = A^T vec{b}$$

    And we also know that




    $R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$




    So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$



    Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$



    And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
    $$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$

    which is of $n-r$ dimension






    share|cite|improve this answer


























      0














      I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
      $$A^T A vec{x} = A^T vec{b}$$

      And we also know that




      $R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$




      So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$



      Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$



      And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
      $$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$

      which is of $n-r$ dimension






      share|cite|improve this answer
























        0












        0








        0






        I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
        $$A^T A vec{x} = A^T vec{b}$$

        And we also know that




        $R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$




        So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$



        Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$



        And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
        $$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$

        which is of $n-r$ dimension






        share|cite|improve this answer












        I think I get it now....Given a matrix $A in R^{mtimes n}, , m>n$ and $b in R^m$, if we want to find a least square solution $mathbf{hat{x}}$ for the system $Avec{x} = vec{b}$, we are actually trying to solve the normal equation
        $$A^T A vec{x} = A^T vec{b}$$

        And we also know that




        $R(A^T)=R(A) = R(A^T A)$ and $N(A) = N(A^T A)$




        So if the A is rank deficient, $r < n$ then the $nullity(A^T A)= n-r$, which means the the solution space of $A^T A vec{v}=0$ is non-trivial and of $n-r$ dimension. We will now have infinity solution for the normal equation $A^T A vec{x} = A^T vec{b}$



        Let $vec{v} in N(A^T A)$, then $$vec{v} = sum_{i=1}^{n-r} alpha_i mathbf{n_i} $$ where $N(A^T A) = span {mathbf{n_1, cdots , n_{n-r}}}$



        And let $vec{w} in R(A^T A)$. So the least square solution can now be written as
        $$mathbf{hat{x}} = vec{w} + alpha_1 mathbf{n_1} + cdots + alpha_{n-r} mathbf{n_{n-r}}$$

        which is of $n-r$ dimension







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '18 at 5:13









        WeiShan Ng

        617




        617






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3008698%2fwhy-there-are-n-r-dimensional-set-of-vectors-x-for-rank-deficient-least-square-p%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

            mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

            Image
            0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
            Read more

            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