prove $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$












1












$begingroup$


I'm a student and I'm studying linear algebra. in Polar Decomposition we have:




for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =Ssqrt{T^*T}$$




so if $S$ is a linear transformation then it must be $dim(operatorname{range}(T)) leq dim(operatorname{range}(sqrt{T^*T}))$.



But why?



edit: I think it must be equal, I mean:



$$dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$$



I know that $dim(operatorname{range}(T)) = dim(operatorname{range}(T^*))$ but because of the square root I cannot prove that $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$.






linear-algebra operator-theory adjoint-operators isometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
    $endgroup$
    – egreg
    Jan 6 at 17:24
















1












$begingroup$


I'm a student and I'm studying linear algebra. in Polar Decomposition we have:




for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =Ssqrt{T^*T}$$




so if $S$ is a linear transformation then it must be $dim(operatorname{range}(T)) leq dim(operatorname{range}(sqrt{T^*T}))$.



But why?



edit: I think it must be equal, I mean:



$$dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$$



I know that $dim(operatorname{range}(T)) = dim(operatorname{range}(T^*))$ but because of the square root I cannot prove that $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$.






linear-algebra operator-theory adjoint-operators isometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
    $endgroup$
    – egreg
    Jan 6 at 17:24














1












1








1





$begingroup$


I'm a student and I'm studying linear algebra. in Polar Decomposition we have:




for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =Ssqrt{T^*T}$$




so if $S$ is a linear transformation then it must be $dim(operatorname{range}(T)) leq dim(operatorname{range}(sqrt{T^*T}))$.



But why?



edit: I think it must be equal, I mean:



$$dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$$



I know that $dim(operatorname{range}(T)) = dim(operatorname{range}(T^*))$ but because of the square root I cannot prove that $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$.






linear-algebra operator-theory adjoint-operators isometry







share|cite|improve this question











$endgroup$




I'm a student and I'm studying linear algebra. in Polar Decomposition we have:




for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =Ssqrt{T^*T}$$




so if $S$ is a linear transformation then it must be $dim(operatorname{range}(T)) leq dim(operatorname{range}(sqrt{T^*T}))$.



But why?



edit: I think it must be equal, I mean:



$$dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$$



I know that $dim(operatorname{range}(T)) = dim(operatorname{range}(T^*))$ but because of the square root I cannot prove that $dim(operatorname{range}(T)) = dim(operatorname{range}(sqrt{T^*T}))$.






linear-algebra operator-theory adjoint-operators isometry




linear-algebra operator-theory adjoint-operators isometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 23:05









egreg

181k1485203




181k1485203










asked Jan 6 at 17:17









Peyman mohseni kiasariPeyman mohseni kiasari

1089




1089












  • $begingroup$
    If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
    $endgroup$
    – egreg
    Jan 6 at 17:24


















  • $begingroup$
    If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
    $endgroup$
    – egreg
    Jan 6 at 17:24
















$begingroup$
If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
$endgroup$
– egreg
Jan 6 at 17:24




$begingroup$
If $fcolon Uto V$ and $gcolon Vto W$ are linear maps, then the rank of $gcirc f$ cannot be greater than the ranks of $g$ and $f$.
$endgroup$
– egreg
Jan 6 at 17:24










2 Answers
2






active

oldest

votes


















1












$begingroup$

As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
$$
Sbig|_{text{ran} T} :text{ran} Tto V,
$$ we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)



To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
$$
dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
$$ (Or we can use the fact that $S$ is an isometry.)






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
    $$DeclareMathOperator{range}{range}
    dimrange{ST}ledimrange(T)
    $$
    This follows from the rank-nullity theorem:
    begin{align}
    dim U&=dimrange(T)+dimker(T) \
    dim U&=dimrange(ST)+dimker(ST)
    end{align}
    Therefore
    $$
    dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
    $$
    because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.



    In your case you can conclude that
    $$
    dimrange(T)ledimrange(sqrt{T^*T})
    $$
    On the other hand, $S$ is an isometry, so it is invertible and
    $$
    sqrt{T^*T}=S^{-1}T
    $$
    The same argument as before implies
    $$
    dimrange(sqrt{T^*T})ledimrange(T)
    $$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      1












      $begingroup$

      As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
      $$
      Sbig|_{text{ran} T} :text{ran} Tto V,
      $$       we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)



      To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
      $$
      dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
      $$       (Or we can use the fact that $S$ is an isometry.)






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
        $$
        Sbig|_{text{ran} T} :text{ran} Tto V,
        $$         we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)



        To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
        $$
        dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
        $$         (Or we can use the fact that $S$ is an isometry.)






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
          $$
          Sbig|_{text{ran} T} :text{ran} Tto V,
          $$           we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)



          To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
          $$
          dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
          $$           (Or we can use the fact that $S$ is an isometry.)






          share|cite|improve this answer











          $endgroup$



          As @egreg pointed out, for any linear maps $S,T$, the rank of $ST$ is always less than or equal to that of $T$. To see this, note that by dimension theorem $$dim text{ran} L =dim operatorname{dom} L-dimker Lledim operatorname{dom} L$$ for any linear map $L$. Now, the image of $ST$ can be seen as the image of
          $$
          Sbig|_{text{ran} T} :text{ran} Tto V,
          $$           we can see that $dim text{ran} (ST)le dim text{ran} T$ as wanted. Moreover, equality holds if $dim ker Sbig|_{text{ran}T}=dim [ker Scaptext{ran}T]=0$. (Here, $V$ denotes the vector space where $S,T$ are defined.)



          To see that $dim text{ran}T =dim text{ran}(T^*T)$, note that $Tx = 0$ if and only if $T^*Tx =0$, which is saying that $ker T = ker (T^*T)$. Now, by dimension theorem, we have
          $$
          dim text{ran} T = dim V - dim ker T = dim V - dim ker (T^*T) = dimtext{ran}(T^*T).
          $$           (Or we can use the fact that $S$ is an isometry.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 22:20

























          answered Jan 6 at 22:15









          SongSong

          12.6k631




          12.6k631























              1












              $begingroup$

              If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
              $$DeclareMathOperator{range}{range}
              dimrange{ST}ledimrange(T)
              $$
              This follows from the rank-nullity theorem:
              begin{align}
              dim U&=dimrange(T)+dimker(T) \
              dim U&=dimrange(ST)+dimker(ST)
              end{align}
              Therefore
              $$
              dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
              $$
              because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.



              In your case you can conclude that
              $$
              dimrange(T)ledimrange(sqrt{T^*T})
              $$
              On the other hand, $S$ is an isometry, so it is invertible and
              $$
              sqrt{T^*T}=S^{-1}T
              $$
              The same argument as before implies
              $$
              dimrange(sqrt{T^*T})ledimrange(T)
              $$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
                $$DeclareMathOperator{range}{range}
                dimrange{ST}ledimrange(T)
                $$
                This follows from the rank-nullity theorem:
                begin{align}
                dim U&=dimrange(T)+dimker(T) \
                dim U&=dimrange(ST)+dimker(ST)
                end{align}
                Therefore
                $$
                dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
                $$
                because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.



                In your case you can conclude that
                $$
                dimrange(T)ledimrange(sqrt{T^*T})
                $$
                On the other hand, $S$ is an isometry, so it is invertible and
                $$
                sqrt{T^*T}=S^{-1}T
                $$
                The same argument as before implies
                $$
                dimrange(sqrt{T^*T})ledimrange(T)
                $$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
                  $$DeclareMathOperator{range}{range}
                  dimrange{ST}ledimrange(T)
                  $$
                  This follows from the rank-nullity theorem:
                  begin{align}
                  dim U&=dimrange(T)+dimker(T) \
                  dim U&=dimrange(ST)+dimker(ST)
                  end{align}
                  Therefore
                  $$
                  dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
                  $$
                  because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.



                  In your case you can conclude that
                  $$
                  dimrange(T)ledimrange(sqrt{T^*T})
                  $$
                  On the other hand, $S$ is an isometry, so it is invertible and
                  $$
                  sqrt{T^*T}=S^{-1}T
                  $$
                  The same argument as before implies
                  $$
                  dimrange(sqrt{T^*T})ledimrange(T)
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  If $Tcolon Uto V$ and $Scolon Vto W$ are linear maps between finite dimensional vector spaces, then
                  $$DeclareMathOperator{range}{range}
                  dimrange{ST}ledimrange(T)
                  $$
                  This follows from the rank-nullity theorem:
                  begin{align}
                  dim U&=dimrange(T)+dimker(T) \
                  dim U&=dimrange(ST)+dimker(ST)
                  end{align}
                  Therefore
                  $$
                  dimrange(ST)=dimrange(T)+dimker(T)-dimker(ST)ledimrange(T)
                  $$
                  because from $ker(T)subseteqker(ST)$ we have $dimker(T)ledimker(ST)$.



                  In your case you can conclude that
                  $$
                  dimrange(T)ledimrange(sqrt{T^*T})
                  $$
                  On the other hand, $S$ is an isometry, so it is invertible and
                  $$
                  sqrt{T^*T}=S^{-1}T
                  $$
                  The same argument as before implies
                  $$
                  dimrange(sqrt{T^*T})ledimrange(T)
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 22:34









                  egregegreg

                  181k1485203




                  181k1485203






























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