The rank of a matrix of Gaussian random vectors












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If I generate a matrix $Rinmathbb{R^{ktimes p}}$, $k<p$, with i.i.d. entries $R_{i,j}simmathcal{N}(0,sigma^2)$, is there a guarantee that this matrix will have rank $k$, i.e. its columns will be independent?






linear-algebra probability random-matrices







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    $begingroup$


    If I generate a matrix $Rinmathbb{R^{ktimes p}}$, $k<p$, with i.i.d. entries $R_{i,j}simmathcal{N}(0,sigma^2)$, is there a guarantee that this matrix will have rank $k$, i.e. its columns will be independent?






    linear-algebra probability random-matrices







    share|cite|improve this question









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      0












      0








      0





      $begingroup$


      If I generate a matrix $Rinmathbb{R^{ktimes p}}$, $k<p$, with i.i.d. entries $R_{i,j}simmathcal{N}(0,sigma^2)$, is there a guarantee that this matrix will have rank $k$, i.e. its columns will be independent?






      linear-algebra probability random-matrices







      share|cite|improve this question









      $endgroup$




      If I generate a matrix $Rinmathbb{R^{ktimes p}}$, $k<p$, with i.i.d. entries $R_{i,j}simmathcal{N}(0,sigma^2)$, is there a guarantee that this matrix will have rank $k$, i.e. its columns will be independent?






      linear-algebra probability random-matrices




      linear-algebra probability random-matrices






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 16 at 8:28









      UndertherainbowUndertherainbow

      312417




      312417






















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          It is not 'surely' true, but we can say that
          $$
          P(;text{rank}(R)=k ;)=1,
          $$           i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
          text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
          $$           Thus, we have
          $$
          P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
          $$           where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
          $
          {p(mathcal{R})=0}
          $           has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
          $$
          P(;text{rank}(R)<k ;)=0
          $$           as desired.






          share|cite|improve this answer









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            active

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            1












            $begingroup$

            It is not 'surely' true, but we can say that
            $$
            P(;text{rank}(R)=k ;)=1,
            $$             i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
            text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
            $$             Thus, we have
            $$
            P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
            $$             where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
            $
            {p(mathcal{R})=0}
            $             has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
            $$
            P(;text{rank}(R)<k ;)=0
            $$             as desired.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              It is not 'surely' true, but we can say that
              $$
              P(;text{rank}(R)=k ;)=1,
              $$               i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
              text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
              $$               Thus, we have
              $$
              P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
              $$               where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
              $
              {p(mathcal{R})=0}
              $               has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
              $$
              P(;text{rank}(R)<k ;)=0
              $$               as desired.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                It is not 'surely' true, but we can say that
                $$
                P(;text{rank}(R)=k ;)=1,
                $$                 i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
                text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
                $$                 Thus, we have
                $$
                P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
                $$                 where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
                $
                {p(mathcal{R})=0}
                $                 has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
                $$
                P(;text{rank}(R)<k ;)=0
                $$                 as desired.






                share|cite|improve this answer









                $endgroup$



                It is not 'surely' true, but we can say that
                $$
                P(;text{rank}(R)=k ;)=1,
                $$                 i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
                text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
                $$                 Thus, we have
                $$
                P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
                $$                 where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
                $
                {p(mathcal{R})=0}
                $                 has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
                $$
                P(;text{rank}(R)<k ;)=0
                $$                 as desired.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 16 at 8:42









                SongSong

                14.2k1633




                14.2k1633






























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