Invertibility of the square root of an operator












2












$begingroup$


Let $mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $Min mathcal{B}(F)^+$ (i.e. $langle Mx;, ;xranglegeq 0$ for all $xin F$).




  • The square root of $M$ is defined to be the unique operator $N$ such that
    $$N^2=M.$$


  • In this case we write $N=M^{1/2}$




If $M$ is an invertible operators, is $M^{1/2}$ invertible?







functional-analysis operator-theory







share|cite|improve this question











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    2












    $begingroup$


    Let $mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $Min mathcal{B}(F)^+$ (i.e. $langle Mx;, ;xranglegeq 0$ for all $xin F$).




    • The square root of $M$ is defined to be the unique operator $N$ such that
      $$N^2=M.$$


    • In this case we write $N=M^{1/2}$




    If $M$ is an invertible operators, is $M^{1/2}$ invertible?







    functional-analysis operator-theory







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $Min mathcal{B}(F)^+$ (i.e. $langle Mx;, ;xranglegeq 0$ for all $xin F$).




      • The square root of $M$ is defined to be the unique operator $N$ such that
        $$N^2=M.$$


      • In this case we write $N=M^{1/2}$




      If $M$ is an invertible operators, is $M^{1/2}$ invertible?







      functional-analysis operator-theory







      share|cite|improve this question











      $endgroup$




      Let $mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $Min mathcal{B}(F)^+$ (i.e. $langle Mx;, ;xranglegeq 0$ for all $xin F$).




      • The square root of $M$ is defined to be the unique operator $N$ such that
        $$N^2=M.$$


      • In this case we write $N=M^{1/2}$




      If $M$ is an invertible operators, is $M^{1/2}$ invertible?







      functional-analysis operator-theory




      functional-analysis operator-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 14:05









      Bernard

      119k639113




      119k639113










      asked Jan 5 at 14:01









      SchülerSchüler

      1,4461421




      1,4461421






















          2 Answers
          2






          active

          oldest

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          3












          $begingroup$

          Note that $M$ is invertible if and only if there is $epsilon>0$ such that
          $$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the answer. Is your equivalence well known?
            $endgroup$
            – Schüler
            Jan 5 at 14:33










          • $begingroup$
            I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
            $endgroup$
            – Song
            Jan 5 at 14:36












          • $begingroup$
            According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
            $endgroup$
            – Schüler
            Jan 5 at 17:16










          • $begingroup$
            Of course, I agree with you!
            $endgroup$
            – Song
            Jan 5 at 17:20



















          4












          $begingroup$

          $N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.



          For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)



          This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

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            votes









            3












            $begingroup$

            Note that $M$ is invertible if and only if there is $epsilon>0$ such that
            $$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for the answer. Is your equivalence well known?
              $endgroup$
              – Schüler
              Jan 5 at 14:33










            • $begingroup$
              I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
              $endgroup$
              – Song
              Jan 5 at 14:36












            • $begingroup$
              According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
              $endgroup$
              – Schüler
              Jan 5 at 17:16










            • $begingroup$
              Of course, I agree with you!
              $endgroup$
              – Song
              Jan 5 at 17:20
















            3












            $begingroup$

            Note that $M$ is invertible if and only if there is $epsilon>0$ such that
            $$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for the answer. Is your equivalence well known?
              $endgroup$
              – Schüler
              Jan 5 at 14:33










            • $begingroup$
              I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
              $endgroup$
              – Song
              Jan 5 at 14:36












            • $begingroup$
              According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
              $endgroup$
              – Schüler
              Jan 5 at 17:16










            • $begingroup$
              Of course, I agree with you!
              $endgroup$
              – Song
              Jan 5 at 17:20














            3












            3








            3





            $begingroup$

            Note that $M$ is invertible if and only if there is $epsilon>0$ such that
            $$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.






            share|cite|improve this answer









            $endgroup$



            Note that $M$ is invertible if and only if there is $epsilon>0$ such that
            $$sigma(M)subset [epsilon, infty).$$ Thus $sigma(M^{1/2})subset [epsilon^{1/2},infty)$ and $M^{1/2}$ is invertible.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 5 at 14:12









            SongSong

            8,416625




            8,416625












            • $begingroup$
              Thank you for the answer. Is your equivalence well known?
              $endgroup$
              – Schüler
              Jan 5 at 14:33










            • $begingroup$
              I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
              $endgroup$
              – Song
              Jan 5 at 14:36












            • $begingroup$
              According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
              $endgroup$
              – Schüler
              Jan 5 at 17:16










            • $begingroup$
              Of course, I agree with you!
              $endgroup$
              – Song
              Jan 5 at 17:20


















            • $begingroup$
              Thank you for the answer. Is your equivalence well known?
              $endgroup$
              – Schüler
              Jan 5 at 14:33










            • $begingroup$
              I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
              $endgroup$
              – Song
              Jan 5 at 14:36












            • $begingroup$
              According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
              $endgroup$
              – Schüler
              Jan 5 at 17:16










            • $begingroup$
              Of course, I agree with you!
              $endgroup$
              – Song
              Jan 5 at 17:20
















            $begingroup$
            Thank you for the answer. Is your equivalence well known?
            $endgroup$
            – Schüler
            Jan 5 at 14:33




            $begingroup$
            Thank you for the answer. Is your equivalence well known?
            $endgroup$
            – Schüler
            Jan 5 at 14:33












            $begingroup$
            I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
            $endgroup$
            – Song
            Jan 5 at 14:36






            $begingroup$
            I think it is. It is in fact easier than it seems. Since $Mge 0$, we know $sigma(M)subset [0,infty)$. But if $M$ is invertible, we must have $epsilon =min sigma(M) >0$. The other direction is true since $0notin sigma(M)$ means $M$ is invertible.
            $endgroup$
            – Song
            Jan 5 at 14:36














            $begingroup$
            According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
            $endgroup$
            – Schüler
            Jan 5 at 17:16




            $begingroup$
            According to our answer, the converse is also true. That is if $A^{1/2}$ is invertible, then $A$ is also invertible. Do you agree with me? Thanks.
            $endgroup$
            – Schüler
            Jan 5 at 17:16












            $begingroup$
            Of course, I agree with you!
            $endgroup$
            – Song
            Jan 5 at 17:20




            $begingroup$
            Of course, I agree with you!
            $endgroup$
            – Song
            Jan 5 at 17:20











            4












            $begingroup$

            $N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.



            For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)



            This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              $N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.



              For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)



              This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                $N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.



                For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)



                This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.






                share|cite|improve this answer









                $endgroup$



                $N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.



                For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)



                This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 14:06









                Tsemo AristideTsemo Aristide

                56.9k11444




                56.9k11444






























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