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







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  • $n-r$ is the dimension of Nullity of $A$.
    – Yadati Kiran
    Nov 22 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 at 4:50










  • Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
    – Yadati Kiran
    Nov 22 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 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 at 4:50










  • Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
    – Yadati Kiran
    Nov 22 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 at 3:04









WeiShan Ng

597




597












  • $n-r$ is the dimension of Nullity of $A$.
    – Yadati Kiran
    Nov 22 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 at 4:50










  • Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
    – Yadati Kiran
    Nov 22 at 4:55




















  • $n-r$ is the dimension of Nullity of $A$.
    – Yadati Kiran
    Nov 22 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 at 4:50










  • Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
    – Yadati Kiran
    Nov 22 at 4:55


















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






$n-r$ is the dimension of Nullity of $A$.
– Yadati Kiran
Nov 22 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 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 at 4:50












Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 at 4:55






Rank of $A^{T}$ is less than or equal to $n$ right since $m>n$?
– Yadati Kiran
Nov 22 at 4:55












1 Answer
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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






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    1 Answer
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    1 Answer
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    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 at 5:13









        WeiShan Ng

        597




        597






























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