An inequality of matrix norm












3












$begingroup$


Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.



Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$



I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!






linear-algebra matrices functional-analysis norm







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
    $$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.



    Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
    $$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$



    I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!






    linear-algebra matrices functional-analysis norm







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
      $$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.



      Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
      $$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$



      I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!






      linear-algebra matrices functional-analysis norm







      share|cite|improve this question











      $endgroup$




      Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
      $$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.



      Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
      $$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$



      I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!






      linear-algebra matrices functional-analysis norm




      linear-algebra matrices functional-analysis norm






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      Bernard

      123k741116




      123k741116










      asked 2 hours ago









      bbwbbw

      51739




      51739






















          1 Answer
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          active

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          6












          $begingroup$

          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            1 hour ago











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          1 Answer
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          active

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          6












          $begingroup$

          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            1 hour ago
















          6












          $begingroup$

          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            1 hour ago














          6












          6








          6





          $begingroup$

          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)






          share|cite|improve this answer











          $endgroup$



          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          avsavs

          3,424513




          3,424513












          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            1 hour ago


















          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            1 hour ago
















          $begingroup$
          Thank you so much!
          $endgroup$
          – bbw
          1 hour ago




          $begingroup$
          Thank you so much!
          $endgroup$
          – bbw
          1 hour ago




          1




          1




          $begingroup$
          You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
          $endgroup$
          – avs
          1 hour ago




          $begingroup$
          You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
          $endgroup$
          – avs
          1 hour ago


















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