Invertibility of the square root of an operator












2














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





























    2














    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



























      2












      2








      2







      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















      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 yesterday









      Bernard

      118k639112




      118k639112










      asked yesterday









      Schüler

      1,5441421




      1,5441421






















          2 Answers
          2






          active

          oldest

          votes


















          3














          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





















          • Thank you for the answer. Is your equivalence well known?
            – Schüler
            yesterday










          • 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.
            – Song
            yesterday












          • 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.
            – Schüler
            yesterday










          • Of course, I agree with you!
            – Song
            yesterday



















          4














          $N(x)=N(y)$ 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





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            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





















            • Thank you for the answer. Is your equivalence well known?
              – Schüler
              yesterday










            • 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.
              – Song
              yesterday












            • 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.
              – Schüler
              yesterday










            • Of course, I agree with you!
              – Song
              yesterday
















            3














            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





















            • Thank you for the answer. Is your equivalence well known?
              – Schüler
              yesterday










            • 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.
              – Song
              yesterday












            • 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.
              – Schüler
              yesterday










            • Of course, I agree with you!
              – Song
              yesterday














            3












            3








            3






            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












            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 yesterday









            Song

            5,765318




            5,765318












            • Thank you for the answer. Is your equivalence well known?
              – Schüler
              yesterday










            • 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.
              – Song
              yesterday












            • 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.
              – Schüler
              yesterday










            • Of course, I agree with you!
              – Song
              yesterday


















            • Thank you for the answer. Is your equivalence well known?
              – Schüler
              yesterday










            • 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.
              – Song
              yesterday












            • 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.
              – Schüler
              yesterday










            • Of course, I agree with you!
              – Song
              yesterday
















            Thank you for the answer. Is your equivalence well known?
            – Schüler
            yesterday




            Thank you for the answer. Is your equivalence well known?
            – Schüler
            yesterday












            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.
            – Song
            yesterday






            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.
            – Song
            yesterday














            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.
            – Schüler
            yesterday




            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.
            – Schüler
            yesterday












            Of course, I agree with you!
            – Song
            yesterday




            Of course, I agree with you!
            – Song
            yesterday











            4














            $N(x)=N(y)$ 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


























              4














              $N(x)=N(y)$ 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
























                4












                4








                4






                $N(x)=N(y)$ 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












                $N(x)=N(y)$ 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 yesterday









                Tsemo Aristide

                56.3k11444




                56.3k11444






























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