$sigma_p(f(A))=f(sigma_p(A))$?












1












$begingroup$


Let $A$ be a bounded linear operator on a Hilbert space, $f$ be a complex polynomial and $sigma_p(A)$ be the point spectrum of $A$ (the set of $lambda$ for which $ker(A-lambda I)neq 0$). I am trying to find a counter example for




$sigma_p(f(A))=f(sigma_p(A))$.




I tried for $A$ the left and right shift operators on $l^2(N)$ with $f=x+1$, and it seems that $sigma_p(f(A))$ and $sigma_p(A)$ are both empty...



Is this correct? If so, how do I more cleverly approach this problem of finding a counter example?






functional-analysis operator-theory spectral-theory







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $A$ be a bounded linear operator on a Hilbert space, $f$ be a complex polynomial and $sigma_p(A)$ be the point spectrum of $A$ (the set of $lambda$ for which $ker(A-lambda I)neq 0$). I am trying to find a counter example for




    $sigma_p(f(A))=f(sigma_p(A))$.




    I tried for $A$ the left and right shift operators on $l^2(N)$ with $f=x+1$, and it seems that $sigma_p(f(A))$ and $sigma_p(A)$ are both empty...



    Is this correct? If so, how do I more cleverly approach this problem of finding a counter example?






    functional-analysis operator-theory spectral-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $A$ be a bounded linear operator on a Hilbert space, $f$ be a complex polynomial and $sigma_p(A)$ be the point spectrum of $A$ (the set of $lambda$ for which $ker(A-lambda I)neq 0$). I am trying to find a counter example for




      $sigma_p(f(A))=f(sigma_p(A))$.




      I tried for $A$ the left and right shift operators on $l^2(N)$ with $f=x+1$, and it seems that $sigma_p(f(A))$ and $sigma_p(A)$ are both empty...



      Is this correct? If so, how do I more cleverly approach this problem of finding a counter example?






      functional-analysis operator-theory spectral-theory







      share|cite|improve this question









      $endgroup$




      Let $A$ be a bounded linear operator on a Hilbert space, $f$ be a complex polynomial and $sigma_p(A)$ be the point spectrum of $A$ (the set of $lambda$ for which $ker(A-lambda I)neq 0$). I am trying to find a counter example for




      $sigma_p(f(A))=f(sigma_p(A))$.




      I tried for $A$ the left and right shift operators on $l^2(N)$ with $f=x+1$, and it seems that $sigma_p(f(A))$ and $sigma_p(A)$ are both empty...



      Is this correct? If so, how do I more cleverly approach this problem of finding a counter example?






      functional-analysis operator-theory spectral-theory




      functional-analysis operator-theory spectral-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 20:13









      Meneer-BeerMeneer-Beer

      1449




      1449






















          1 Answer
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          1












          $begingroup$

          Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
          $sigma_p(A)$ might be empty.



          But these are the only counterexamples.



          If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
          $A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.



          On the other hand, suppose $lambda in sigma_p(f(A))$.
          Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
          If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
          $$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
          so that
          $$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
          For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
          $(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
          $f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
            $sigma_p(A)$ might be empty.



            But these are the only counterexamples.



            If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
            $A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.



            On the other hand, suppose $lambda in sigma_p(f(A))$.
            Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
            If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
            $$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
            so that
            $$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
            For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
            $(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
            $f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
              $sigma_p(A)$ might be empty.



              But these are the only counterexamples.



              If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
              $A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.



              On the other hand, suppose $lambda in sigma_p(f(A))$.
              Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
              If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
              $$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
              so that
              $$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
              For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
              $(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
              $f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
                $sigma_p(A)$ might be empty.



                But these are the only counterexamples.



                If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
                $A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.



                On the other hand, suppose $lambda in sigma_p(f(A))$.
                Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
                If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
                $$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
                so that
                $$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
                For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
                $(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
                $f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.






                share|cite|improve this answer











                $endgroup$



                Let $f$ be a constant polynomial. Then $f(A) = c I$ for some scalar $c$, so $sigma_p(f(A)) = {c}$. But
                $sigma_p(A)$ might be empty.



                But these are the only counterexamples.



                If $lambda in sigma_p(A)$, there is nonzero $v in H$ such that
                $A v = lambda v$, and then $f(A) v = f(lambda) v$, so $p(lambda) in sigma_p(f(A))$. Thus $f(sigma_p(A)) subseteq sigma_p(f(A))$.



                On the other hand, suppose $lambda in sigma_p(f(A))$.
                Thus there is nonzero $v in H$ such that $f(A) v = lambda v$.
                If $f$ has degree $d > 0$, the polynomial $f(z) - lambda$ can be factored over the complex numbers as
                $$f(z) - lambda = prod_{j=1}^d (z - alpha_j)$$
                so that
                $$ 0 = (f(A) - lambda) v = prod_{j=1}^d (A - alpha_j I) v $$
                For some $k$, $1 le k le d$, we must have $prod_{j=k}^d (A - alpha_j I)v = 0$ but $prod_{j=k+1}^d (A - alpha_j I) v ne 0$, i.e. with $w = prod_{j=k+1}^d (A - alpha_j I) v$ (which is just $v$ in the case $k=d$) we have $wne 0$ and
                $(A - alpha_k I) w = 0$, so $alpha_k in sigma_p(A)$, and
                $f(alpha_k) = lambda$. So $sigma_p(f(A)) subseteq f(sigma_p(A))$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 10 '18 at 20:40

























                answered Dec 10 '18 at 20:33









                Robert IsraelRobert Israel

                329k23217470




                329k23217470






























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