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





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062747%2finvertibility-of-the-square-root-of-an-operator%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062747%2finvertibility-of-the-square-root-of-an-operator%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    How to make a Squid Proxy server?

                    Image
                    1 I want to set up squid proxy server for monitoring user activity in a network and Internet Web URL filtering for user. I also want to configure the user login, user log, user access group, and block unwanted web sites in my network. Please help me and guide me the full process to do this. server proxy squid share | improve this question edited Jul 31 '12 at 14:45 Jorge Castro 37.1k 107 422 617 asked Jul 31 '12 at 5:16 Goivind Tiwari Goivind Tiwari ...
                    Read more

                    Is this a new Fibonacci Identity?

                    Image
                    6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w...
                    Read more

                    19世紀

                    Image
                    この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料...
                    Read more