There does not exist two continuous functions that the values constitute the spectrum of the matrix $A(z) =...











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I read an example from Rajendra Bhatia's Matrix Analysis.



Let $A(z) = pmatrix{0 & z \ 1 & 0}$ with $z in mathbb C$. The eigenvalues are clearly $pm sqrt{z}$. Then he says these cannot be represented by two continuous functions over any domain $G$ contains $0$. I guess this should violate some basic property of complex-valued functions, which unfortunately I am not familiar with. Could someone point this out for me? Thanks.






linear-algebra complex-analysis







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    up vote
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    down vote

    favorite












    I read an example from Rajendra Bhatia's Matrix Analysis.



    Let $A(z) = pmatrix{0 & z \ 1 & 0}$ with $z in mathbb C$. The eigenvalues are clearly $pm sqrt{z}$. Then he says these cannot be represented by two continuous functions over any domain $G$ contains $0$. I guess this should violate some basic property of complex-valued functions, which unfortunately I am not familiar with. Could someone point this out for me? Thanks.






    linear-algebra complex-analysis







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I read an example from Rajendra Bhatia's Matrix Analysis.



      Let $A(z) = pmatrix{0 & z \ 1 & 0}$ with $z in mathbb C$. The eigenvalues are clearly $pm sqrt{z}$. Then he says these cannot be represented by two continuous functions over any domain $G$ contains $0$. I guess this should violate some basic property of complex-valued functions, which unfortunately I am not familiar with. Could someone point this out for me? Thanks.






      linear-algebra complex-analysis







      share|cite|improve this question













      I read an example from Rajendra Bhatia's Matrix Analysis.



      Let $A(z) = pmatrix{0 & z \ 1 & 0}$ with $z in mathbb C$. The eigenvalues are clearly $pm sqrt{z}$. Then he says these cannot be represented by two continuous functions over any domain $G$ contains $0$. I guess this should violate some basic property of complex-valued functions, which unfortunately I am not familiar with. Could someone point this out for me? Thanks.






      linear-algebra complex-analysis




      linear-algebra complex-analysis






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      asked Nov 27 at 17:38









      user1101010

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          If such continuous functions $f$ and $g$ existed, then the would satisfy
          $$
          f=-g quad & quad f^2(z)=g^2(z)=z.
          $$
          Thus for every $zne 0$, we have that $f(z)ne 0$, and hence
          $$
          frac{f(z+h)-f(z)}{h}=frac{1}{f(z+h)+f(z)}cdotfrac{f^2(z+h)-f^2(z)}{h}=
          frac{1}{f(z+h)+f(z)}cdotfrac{z+h-z}{h}\=frac{1}{f(z+h)+f(z)}to frac{1}{2f(z)}.
          $$
          Therefore, $f$ (and similarly $g$) is analytic in a region
          $$
          U={zinmathbb C: 0<|z|<R}
          $$
          for some $R>0$, and as $|,f(z)|=|z|^{1/2}le R^{1/2}$, then $f$ is bounded in $U$ and therefore extends analytically in $0$, and as $lim_{zto 0}f^2(z)=0$, then
          $lim_{zto 0}f(z)=0$. And thus,
          $$
          hat f(z)=left{
          begin{array}{ccc}
          f(z) & if & zne 0, \
          0 & if & z=0,
          end{array}
          right.
          $$
          is analytic in $Ucup{0}$, and $z=0$ is a zero of $hat f$. Hence, there is a $kinmathbb N$, such that
          $$
          hat f(z)=z^kh(z),
          $$
          where $h(z)$ is analytic and $h(0)ne 0$. Thus
          $$
          z=hat f^2(z)=z^{2k}h^2(z),
          $$
          and finally
          $$
          1=hat f^2(z)=z^{2k-1}h^2(z),
          $$
          for all $zin U$. This is impossible, since the right hand side tends to zero, as $zto 0$.






          share|cite|improve this answer





















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            If such continuous functions $f$ and $g$ existed, then the would satisfy
            $$
            f=-g quad & quad f^2(z)=g^2(z)=z.
            $$
            Thus for every $zne 0$, we have that $f(z)ne 0$, and hence
            $$
            frac{f(z+h)-f(z)}{h}=frac{1}{f(z+h)+f(z)}cdotfrac{f^2(z+h)-f^2(z)}{h}=
            frac{1}{f(z+h)+f(z)}cdotfrac{z+h-z}{h}\=frac{1}{f(z+h)+f(z)}to frac{1}{2f(z)}.
            $$
            Therefore, $f$ (and similarly $g$) is analytic in a region
            $$
            U={zinmathbb C: 0<|z|<R}
            $$
            for some $R>0$, and as $|,f(z)|=|z|^{1/2}le R^{1/2}$, then $f$ is bounded in $U$ and therefore extends analytically in $0$, and as $lim_{zto 0}f^2(z)=0$, then
            $lim_{zto 0}f(z)=0$. And thus,
            $$
            hat f(z)=left{
            begin{array}{ccc}
            f(z) & if & zne 0, \
            0 & if & z=0,
            end{array}
            right.
            $$
            is analytic in $Ucup{0}$, and $z=0$ is a zero of $hat f$. Hence, there is a $kinmathbb N$, such that
            $$
            hat f(z)=z^kh(z),
            $$
            where $h(z)$ is analytic and $h(0)ne 0$. Thus
            $$
            z=hat f^2(z)=z^{2k}h^2(z),
            $$
            and finally
            $$
            1=hat f^2(z)=z^{2k-1}h^2(z),
            $$
            for all $zin U$. This is impossible, since the right hand side tends to zero, as $zto 0$.






            share|cite|improve this answer

























              up vote
              4
              down vote



              accepted










              If such continuous functions $f$ and $g$ existed, then the would satisfy
              $$
              f=-g quad & quad f^2(z)=g^2(z)=z.
              $$
              Thus for every $zne 0$, we have that $f(z)ne 0$, and hence
              $$
              frac{f(z+h)-f(z)}{h}=frac{1}{f(z+h)+f(z)}cdotfrac{f^2(z+h)-f^2(z)}{h}=
              frac{1}{f(z+h)+f(z)}cdotfrac{z+h-z}{h}\=frac{1}{f(z+h)+f(z)}to frac{1}{2f(z)}.
              $$
              Therefore, $f$ (and similarly $g$) is analytic in a region
              $$
              U={zinmathbb C: 0<|z|<R}
              $$
              for some $R>0$, and as $|,f(z)|=|z|^{1/2}le R^{1/2}$, then $f$ is bounded in $U$ and therefore extends analytically in $0$, and as $lim_{zto 0}f^2(z)=0$, then
              $lim_{zto 0}f(z)=0$. And thus,
              $$
              hat f(z)=left{
              begin{array}{ccc}
              f(z) & if & zne 0, \
              0 & if & z=0,
              end{array}
              right.
              $$
              is analytic in $Ucup{0}$, and $z=0$ is a zero of $hat f$. Hence, there is a $kinmathbb N$, such that
              $$
              hat f(z)=z^kh(z),
              $$
              where $h(z)$ is analytic and $h(0)ne 0$. Thus
              $$
              z=hat f^2(z)=z^{2k}h^2(z),
              $$
              and finally
              $$
              1=hat f^2(z)=z^{2k-1}h^2(z),
              $$
              for all $zin U$. This is impossible, since the right hand side tends to zero, as $zto 0$.






              share|cite|improve this answer























                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                If such continuous functions $f$ and $g$ existed, then the would satisfy
                $$
                f=-g quad & quad f^2(z)=g^2(z)=z.
                $$
                Thus for every $zne 0$, we have that $f(z)ne 0$, and hence
                $$
                frac{f(z+h)-f(z)}{h}=frac{1}{f(z+h)+f(z)}cdotfrac{f^2(z+h)-f^2(z)}{h}=
                frac{1}{f(z+h)+f(z)}cdotfrac{z+h-z}{h}\=frac{1}{f(z+h)+f(z)}to frac{1}{2f(z)}.
                $$
                Therefore, $f$ (and similarly $g$) is analytic in a region
                $$
                U={zinmathbb C: 0<|z|<R}
                $$
                for some $R>0$, and as $|,f(z)|=|z|^{1/2}le R^{1/2}$, then $f$ is bounded in $U$ and therefore extends analytically in $0$, and as $lim_{zto 0}f^2(z)=0$, then
                $lim_{zto 0}f(z)=0$. And thus,
                $$
                hat f(z)=left{
                begin{array}{ccc}
                f(z) & if & zne 0, \
                0 & if & z=0,
                end{array}
                right.
                $$
                is analytic in $Ucup{0}$, and $z=0$ is a zero of $hat f$. Hence, there is a $kinmathbb N$, such that
                $$
                hat f(z)=z^kh(z),
                $$
                where $h(z)$ is analytic and $h(0)ne 0$. Thus
                $$
                z=hat f^2(z)=z^{2k}h^2(z),
                $$
                and finally
                $$
                1=hat f^2(z)=z^{2k-1}h^2(z),
                $$
                for all $zin U$. This is impossible, since the right hand side tends to zero, as $zto 0$.






                share|cite|improve this answer












                If such continuous functions $f$ and $g$ existed, then the would satisfy
                $$
                f=-g quad & quad f^2(z)=g^2(z)=z.
                $$
                Thus for every $zne 0$, we have that $f(z)ne 0$, and hence
                $$
                frac{f(z+h)-f(z)}{h}=frac{1}{f(z+h)+f(z)}cdotfrac{f^2(z+h)-f^2(z)}{h}=
                frac{1}{f(z+h)+f(z)}cdotfrac{z+h-z}{h}\=frac{1}{f(z+h)+f(z)}to frac{1}{2f(z)}.
                $$
                Therefore, $f$ (and similarly $g$) is analytic in a region
                $$
                U={zinmathbb C: 0<|z|<R}
                $$
                for some $R>0$, and as $|,f(z)|=|z|^{1/2}le R^{1/2}$, then $f$ is bounded in $U$ and therefore extends analytically in $0$, and as $lim_{zto 0}f^2(z)=0$, then
                $lim_{zto 0}f(z)=0$. And thus,
                $$
                hat f(z)=left{
                begin{array}{ccc}
                f(z) & if & zne 0, \
                0 & if & z=0,
                end{array}
                right.
                $$
                is analytic in $Ucup{0}$, and $z=0$ is a zero of $hat f$. Hence, there is a $kinmathbb N$, such that
                $$
                hat f(z)=z^kh(z),
                $$
                where $h(z)$ is analytic and $h(0)ne 0$. Thus
                $$
                z=hat f^2(z)=z^{2k}h^2(z),
                $$
                and finally
                $$
                1=hat f^2(z)=z^{2k-1}h^2(z),
                $$
                for all $zin U$. This is impossible, since the right hand side tends to zero, as $zto 0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 at 18:14









                Yiorgos S. Smyrlis

                62.2k1383162




                62.2k1383162






























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