Inequality between Frobenius norm and L2 Norm












1












$begingroup$


$u , w in S ^ { n - 1 }$ and $v , z in S ^ { m - 1 }$ which means u,w are unit vector in $R^n$, v,z are unit vector in $R^m$
Prove
$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 } leq | u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 }$



$left| right| _ { F }$ is Frobenius norm defined for matrix.



$$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 }=sum _ { i , j } left( u _ { j } v _ { i } - w _ { j } z _ { i } right) ^ { 2 }$$
I tried to first get the lower bound of right hand side, so that we can have product unit between two separate vectors:
$$| u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 } geq 2| u - w | _ { 2 } | v - z | _ { 2 }$$

But when I tried to expand it, it seems hard to rearrange.



Any ideas?






probability matrices statistics inequality







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$endgroup$

















    1












    $begingroup$


    $u , w in S ^ { n - 1 }$ and $v , z in S ^ { m - 1 }$ which means u,w are unit vector in $R^n$, v,z are unit vector in $R^m$
    Prove
    $left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 } leq | u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 }$



    $left| right| _ { F }$ is Frobenius norm defined for matrix.



    $$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 }=sum _ { i , j } left( u _ { j } v _ { i } - w _ { j } z _ { i } right) ^ { 2 }$$
    I tried to first get the lower bound of right hand side, so that we can have product unit between two separate vectors:
    $$| u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 } geq 2| u - w | _ { 2 } | v - z | _ { 2 }$$

    But when I tried to expand it, it seems hard to rearrange.



    Any ideas?






    probability matrices statistics inequality







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      $u , w in S ^ { n - 1 }$ and $v , z in S ^ { m - 1 }$ which means u,w are unit vector in $R^n$, v,z are unit vector in $R^m$
      Prove
      $left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 } leq | u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 }$



      $left| right| _ { F }$ is Frobenius norm defined for matrix.



      $$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 }=sum _ { i , j } left( u _ { j } v _ { i } - w _ { j } z _ { i } right) ^ { 2 }$$
      I tried to first get the lower bound of right hand side, so that we can have product unit between two separate vectors:
      $$| u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 } geq 2| u - w | _ { 2 } | v - z | _ { 2 }$$

      But when I tried to expand it, it seems hard to rearrange.



      Any ideas?






      probability matrices statistics inequality







      share|cite|improve this question











      $endgroup$




      $u , w in S ^ { n - 1 }$ and $v , z in S ^ { m - 1 }$ which means u,w are unit vector in $R^n$, v,z are unit vector in $R^m$
      Prove
      $left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 } leq | u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 }$



      $left| right| _ { F }$ is Frobenius norm defined for matrix.



      $$left| u v ^ { mathrm { T } } - w z ^ { top } right| _ { F } ^ { 2 }=sum _ { i , j } left( u _ { j } v _ { i } - w _ { j } z _ { i } right) ^ { 2 }$$
      I tried to first get the lower bound of right hand side, so that we can have product unit between two separate vectors:
      $$| u - w | _ { 2 } ^ { 2 } + | v - z | _ { 2 } ^ { 2 } geq 2| u - w | _ { 2 } | v - z | _ { 2 }$$

      But when I tried to expand it, it seems hard to rearrange.



      Any ideas?






      probability matrices statistics inequality




      probability matrices statistics inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 11 at 19:43









      Bernard

      122k741116




      122k741116










      asked Jan 11 at 19:25









      DylonDylon

      63




      63






















          1 Answer
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          $begingroup$

          We can expand each of the terms as follows. The Frobenius norm:
          $$
          |uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
          = operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
          = 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
          = 2[1 - (u^Tw)(v^Tz)]
          $$
          The vector norm:
          $$
          |u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
          $$
          To compare these two, we make the following observation:
          $$
          [1 - u^Tw][1 - v^Tz] geq 0 implies\
          1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
          (u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very clear thanks.
            $endgroup$
            – Dylon
            Jan 12 at 6:23











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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

          votes









          0












          $begingroup$

          We can expand each of the terms as follows. The Frobenius norm:
          $$
          |uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
          = operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
          = 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
          = 2[1 - (u^Tw)(v^Tz)]
          $$
          The vector norm:
          $$
          |u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
          $$
          To compare these two, we make the following observation:
          $$
          [1 - u^Tw][1 - v^Tz] geq 0 implies\
          1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
          (u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very clear thanks.
            $endgroup$
            – Dylon
            Jan 12 at 6:23
















          0












          $begingroup$

          We can expand each of the terms as follows. The Frobenius norm:
          $$
          |uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
          = operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
          = 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
          = 2[1 - (u^Tw)(v^Tz)]
          $$
          The vector norm:
          $$
          |u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
          $$
          To compare these two, we make the following observation:
          $$
          [1 - u^Tw][1 - v^Tz] geq 0 implies\
          1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
          (u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very clear thanks.
            $endgroup$
            – Dylon
            Jan 12 at 6:23














          0












          0








          0





          $begingroup$

          We can expand each of the terms as follows. The Frobenius norm:
          $$
          |uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
          = operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
          = 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
          = 2[1 - (u^Tw)(v^Tz)]
          $$
          The vector norm:
          $$
          |u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
          $$
          To compare these two, we make the following observation:
          $$
          [1 - u^Tw][1 - v^Tz] geq 0 implies\
          1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
          (u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
          $$






          share|cite|improve this answer









          $endgroup$



          We can expand each of the terms as follows. The Frobenius norm:
          $$
          |uv^T - wz^T|_F^2 = operatorname{tr}[(uv^T - wz^T)^T(uv^T - wz^T)]\
          = operatorname{tr}[vv^T - (u^Tw)vz^T - (w^Tu)zv^T + zz^T]\
          = 2 - (u^Tw)operatorname{tr}[vz^T + zv^T]\
          = 2[1 - (u^Tw)(v^Tz)]
          $$
          The vector norm:
          $$
          |u-w|^2 + |v-z|^2 = [2 - 2(u^Tw)] + [2 - 2(v^Tz)] = 2[1 - (u^Tw + v^Tz - 1)]
          $$
          To compare these two, we make the following observation:
          $$
          [1 - u^Tw][1 - v^Tz] geq 0 implies\
          1 - u^Tw - v^Tz + (u^Tw)(v^Tz) geq 0 implies\
          (u^Tw)(v^Tz) geq (u^Tw) + (v^Tz) - 1
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 19:46









          OmnomnomnomOmnomnomnom

          128k791186




          128k791186












          • $begingroup$
            Very clear thanks.
            $endgroup$
            – Dylon
            Jan 12 at 6:23


















          • $begingroup$
            Very clear thanks.
            $endgroup$
            – Dylon
            Jan 12 at 6:23
















          $begingroup$
          Very clear thanks.
          $endgroup$
          – Dylon
          Jan 12 at 6:23




          $begingroup$
          Very clear thanks.
          $endgroup$
          – Dylon
          Jan 12 at 6:23


















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