A function evaluated at a operator

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$begingroup$


If dim$(V)=n$ and $tau in mathcal L(V)$, prove that there is a nonzero polynomial $p(x)in F[x]$ for which $p(tau)=0$. I don't understand how come a function can be evaluated at a linear operator $tau$.










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  • $begingroup$
    A polynomial is not a function. :P
    $endgroup$
    – darij grinberg
    Jan 24 at 3:03















2












$begingroup$


If dim$(V)=n$ and $tau in mathcal L(V)$, prove that there is a nonzero polynomial $p(x)in F[x]$ for which $p(tau)=0$. I don't understand how come a function can be evaluated at a linear operator $tau$.










share|cite|improve this question









$endgroup$











  • $begingroup$
    A polynomial is not a function. :P
    $endgroup$
    – darij grinberg
    Jan 24 at 3:03













2












2








2


1



$begingroup$


If dim$(V)=n$ and $tau in mathcal L(V)$, prove that there is a nonzero polynomial $p(x)in F[x]$ for which $p(tau)=0$. I don't understand how come a function can be evaluated at a linear operator $tau$.










share|cite|improve this question









$endgroup$




If dim$(V)=n$ and $tau in mathcal L(V)$, prove that there is a nonzero polynomial $p(x)in F[x]$ for which $p(tau)=0$. I don't understand how come a function can be evaluated at a linear operator $tau$.







linear-algebra






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asked Jan 24 at 1:41









Jiexiong687691Jiexiong687691

825




825











  • $begingroup$
    A polynomial is not a function. :P
    $endgroup$
    – darij grinberg
    Jan 24 at 3:03
















  • $begingroup$
    A polynomial is not a function. :P
    $endgroup$
    – darij grinberg
    Jan 24 at 3:03















$begingroup$
A polynomial is not a function. :P
$endgroup$
– darij grinberg
Jan 24 at 3:03




$begingroup$
A polynomial is not a function. :P
$endgroup$
– darij grinberg
Jan 24 at 3:03










4 Answers
4






active

oldest

votes


















2












$begingroup$

First, how to evaluate a polynomial $p(x) in F[x]$ at a linear operator $tau in mathcal L(V)$? The first thing to realize is that powers of $tau$ make sense as elements of $mathcal L(V)$; e.g., for



$vec v in V, tag 1$ we have



$tau vec v in V, tag 2$



so it makes sense to take



$tau^2 vec v = tau(tau vec v); tag 3$



indeed, accepting (3), we may inductively define



$tau^k + 1 vec v = tau (tau^kvec v); tag 4$



thus we define $tau^m$ for any $m in Bbb N$; if $tau$ is represented by a matrix



$[tau_ij], tag 5$



then it is easy to see that $tau^m$ is represented by the matrix



$[(tau^m)_ij] = [tau_ij]^m. tag 6$



In any event, given $tau^m$, we see that



$(alpha tau^m)(vec v) = alpha (tau^m vec v); tag 7$



and we may also construct and evaluate expressions such as



$(alpha tau^l + beta tau^m)vec v = alpha tau^l vec v + beta tau^m vec v tag 8$



using the ordinary linearity properties of $mathcal L(V)$; continuing in this vein, we can for



$p(x) in F[x], ; p(x) = displaystyle sum_0^m p_i x^i, tag 9$



meaningfully define



$p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag10$



which is the general formula for $p(tau)$.



Now we recall that



$dim_F mathcal L(V) = n^2; tag11$



thus the set



$I, tau, tau^2, ldots, tau^n - 1, tau^n^2 - 1, tau^n^2 subset mathcal L(V), tag12$



having as it does $n^2 + 1$ elements, must be linearly dependent over $F$; hence we may find $n^2 +1$ members



$c_i in F, tag13$



not all $0$, such that



$displaystyle sum_0^n^2 c_i tau_i = 0; tag14$



$tau$ then satisfies the polynomial



$c(x) = displaystyle sum_0^n^2 c_i x^i in F[x]. tag14$



We have shown that every $tau in F[x]$ satisfies a polynomial of degree at most $n^2$, but this is far from optimal. For example, the well-known Cayley-Hamilton theorem shows that $tau$ obeys a degree-$n$ polynomial; but the proof of that result is somewhat more involved than what we have done here, and hence will be saved for yet another post.






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    Hint: $mathcalL(V)$ is also a finite dimensional vector space (over what?) Thus, you can find some integer $m$ (you can even explicitly find $m$ in terms of $n$) such that:
    $$1,tau, tau^2,dots tau^m$$ are linearly dependent. I think you can continue from here.






    share|cite|improve this answer









    $endgroup$




















      3












      $begingroup$

      It's not a function, it's a polynomial. For polynomials, it's easy to understand:
      if the polynomial is $f(x) = a_0 + a_1 x + ldots + a_n x^n$, and the operator is $tau$,
      then $f(tau) = a_0 I + a_1 tau + ldots + a_n tau^n$.
      This works in any algebra over a field $F$, where the polynomial has coefficients in $F$.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Note that in this context, the powers of $x$ are evaluated by repeated composition of $tau$ with itself, while the coefficients of the polynomial are treated as scalar multiples of those powers. Finally the terms of the polynomial are added together as one adds linear mappings to get a linear mapping as a result.
        $endgroup$
        – hardmath
        Jan 24 at 2:37


















      0












      $begingroup$

      Let $p$ be the characterstic Polynomial of $tau$. Then there exist $a_0,a_1,cdots a_n-1inmathbbC$ such that $det(tau-lambda I)=p(t)=a_0+a_1 t+cdots a_n-1t^n-1=0$. Plugging $tau$ yeilds
      $p(tau)=0$






      share|cite|improve this answer









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        4 Answers
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        active

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        4 Answers
        4






        active

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        active

        oldest

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        active

        oldest

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        2












        $begingroup$

        First, how to evaluate a polynomial $p(x) in F[x]$ at a linear operator $tau in mathcal L(V)$? The first thing to realize is that powers of $tau$ make sense as elements of $mathcal L(V)$; e.g., for



        $vec v in V, tag 1$ we have



        $tau vec v in V, tag 2$



        so it makes sense to take



        $tau^2 vec v = tau(tau vec v); tag 3$



        indeed, accepting (3), we may inductively define



        $tau^k + 1 vec v = tau (tau^kvec v); tag 4$



        thus we define $tau^m$ for any $m in Bbb N$; if $tau$ is represented by a matrix



        $[tau_ij], tag 5$



        then it is easy to see that $tau^m$ is represented by the matrix



        $[(tau^m)_ij] = [tau_ij]^m. tag 6$



        In any event, given $tau^m$, we see that



        $(alpha tau^m)(vec v) = alpha (tau^m vec v); tag 7$



        and we may also construct and evaluate expressions such as



        $(alpha tau^l + beta tau^m)vec v = alpha tau^l vec v + beta tau^m vec v tag 8$



        using the ordinary linearity properties of $mathcal L(V)$; continuing in this vein, we can for



        $p(x) in F[x], ; p(x) = displaystyle sum_0^m p_i x^i, tag 9$



        meaningfully define



        $p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag10$



        which is the general formula for $p(tau)$.



        Now we recall that



        $dim_F mathcal L(V) = n^2; tag11$



        thus the set



        $I, tau, tau^2, ldots, tau^n - 1, tau^n^2 - 1, tau^n^2 subset mathcal L(V), tag12$



        having as it does $n^2 + 1$ elements, must be linearly dependent over $F$; hence we may find $n^2 +1$ members



        $c_i in F, tag13$



        not all $0$, such that



        $displaystyle sum_0^n^2 c_i tau_i = 0; tag14$



        $tau$ then satisfies the polynomial



        $c(x) = displaystyle sum_0^n^2 c_i x^i in F[x]. tag14$



        We have shown that every $tau in F[x]$ satisfies a polynomial of degree at most $n^2$, but this is far from optimal. For example, the well-known Cayley-Hamilton theorem shows that $tau$ obeys a degree-$n$ polynomial; but the proof of that result is somewhat more involved than what we have done here, and hence will be saved for yet another post.






        share|cite|improve this answer









        $endgroup$

















          2












          $begingroup$

          First, how to evaluate a polynomial $p(x) in F[x]$ at a linear operator $tau in mathcal L(V)$? The first thing to realize is that powers of $tau$ make sense as elements of $mathcal L(V)$; e.g., for



          $vec v in V, tag 1$ we have



          $tau vec v in V, tag 2$



          so it makes sense to take



          $tau^2 vec v = tau(tau vec v); tag 3$



          indeed, accepting (3), we may inductively define



          $tau^k + 1 vec v = tau (tau^kvec v); tag 4$



          thus we define $tau^m$ for any $m in Bbb N$; if $tau$ is represented by a matrix



          $[tau_ij], tag 5$



          then it is easy to see that $tau^m$ is represented by the matrix



          $[(tau^m)_ij] = [tau_ij]^m. tag 6$



          In any event, given $tau^m$, we see that



          $(alpha tau^m)(vec v) = alpha (tau^m vec v); tag 7$



          and we may also construct and evaluate expressions such as



          $(alpha tau^l + beta tau^m)vec v = alpha tau^l vec v + beta tau^m vec v tag 8$



          using the ordinary linearity properties of $mathcal L(V)$; continuing in this vein, we can for



          $p(x) in F[x], ; p(x) = displaystyle sum_0^m p_i x^i, tag 9$



          meaningfully define



          $p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag10$



          which is the general formula for $p(tau)$.



          Now we recall that



          $dim_F mathcal L(V) = n^2; tag11$



          thus the set



          $I, tau, tau^2, ldots, tau^n - 1, tau^n^2 - 1, tau^n^2 subset mathcal L(V), tag12$



          having as it does $n^2 + 1$ elements, must be linearly dependent over $F$; hence we may find $n^2 +1$ members



          $c_i in F, tag13$



          not all $0$, such that



          $displaystyle sum_0^n^2 c_i tau_i = 0; tag14$



          $tau$ then satisfies the polynomial



          $c(x) = displaystyle sum_0^n^2 c_i x^i in F[x]. tag14$



          We have shown that every $tau in F[x]$ satisfies a polynomial of degree at most $n^2$, but this is far from optimal. For example, the well-known Cayley-Hamilton theorem shows that $tau$ obeys a degree-$n$ polynomial; but the proof of that result is somewhat more involved than what we have done here, and hence will be saved for yet another post.






          share|cite|improve this answer









          $endgroup$















            2












            2








            2





            $begingroup$

            First, how to evaluate a polynomial $p(x) in F[x]$ at a linear operator $tau in mathcal L(V)$? The first thing to realize is that powers of $tau$ make sense as elements of $mathcal L(V)$; e.g., for



            $vec v in V, tag 1$ we have



            $tau vec v in V, tag 2$



            so it makes sense to take



            $tau^2 vec v = tau(tau vec v); tag 3$



            indeed, accepting (3), we may inductively define



            $tau^k + 1 vec v = tau (tau^kvec v); tag 4$



            thus we define $tau^m$ for any $m in Bbb N$; if $tau$ is represented by a matrix



            $[tau_ij], tag 5$



            then it is easy to see that $tau^m$ is represented by the matrix



            $[(tau^m)_ij] = [tau_ij]^m. tag 6$



            In any event, given $tau^m$, we see that



            $(alpha tau^m)(vec v) = alpha (tau^m vec v); tag 7$



            and we may also construct and evaluate expressions such as



            $(alpha tau^l + beta tau^m)vec v = alpha tau^l vec v + beta tau^m vec v tag 8$



            using the ordinary linearity properties of $mathcal L(V)$; continuing in this vein, we can for



            $p(x) in F[x], ; p(x) = displaystyle sum_0^m p_i x^i, tag 9$



            meaningfully define



            $p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag10$



            which is the general formula for $p(tau)$.



            Now we recall that



            $dim_F mathcal L(V) = n^2; tag11$



            thus the set



            $I, tau, tau^2, ldots, tau^n - 1, tau^n^2 - 1, tau^n^2 subset mathcal L(V), tag12$



            having as it does $n^2 + 1$ elements, must be linearly dependent over $F$; hence we may find $n^2 +1$ members



            $c_i in F, tag13$



            not all $0$, such that



            $displaystyle sum_0^n^2 c_i tau_i = 0; tag14$



            $tau$ then satisfies the polynomial



            $c(x) = displaystyle sum_0^n^2 c_i x^i in F[x]. tag14$



            We have shown that every $tau in F[x]$ satisfies a polynomial of degree at most $n^2$, but this is far from optimal. For example, the well-known Cayley-Hamilton theorem shows that $tau$ obeys a degree-$n$ polynomial; but the proof of that result is somewhat more involved than what we have done here, and hence will be saved for yet another post.






            share|cite|improve this answer









            $endgroup$



            First, how to evaluate a polynomial $p(x) in F[x]$ at a linear operator $tau in mathcal L(V)$? The first thing to realize is that powers of $tau$ make sense as elements of $mathcal L(V)$; e.g., for



            $vec v in V, tag 1$ we have



            $tau vec v in V, tag 2$



            so it makes sense to take



            $tau^2 vec v = tau(tau vec v); tag 3$



            indeed, accepting (3), we may inductively define



            $tau^k + 1 vec v = tau (tau^kvec v); tag 4$



            thus we define $tau^m$ for any $m in Bbb N$; if $tau$ is represented by a matrix



            $[tau_ij], tag 5$



            then it is easy to see that $tau^m$ is represented by the matrix



            $[(tau^m)_ij] = [tau_ij]^m. tag 6$



            In any event, given $tau^m$, we see that



            $(alpha tau^m)(vec v) = alpha (tau^m vec v); tag 7$



            and we may also construct and evaluate expressions such as



            $(alpha tau^l + beta tau^m)vec v = alpha tau^l vec v + beta tau^m vec v tag 8$



            using the ordinary linearity properties of $mathcal L(V)$; continuing in this vein, we can for



            $p(x) in F[x], ; p(x) = displaystyle sum_0^m p_i x^i, tag 9$



            meaningfully define



            $p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag10$



            which is the general formula for $p(tau)$.



            Now we recall that



            $dim_F mathcal L(V) = n^2; tag11$



            thus the set



            $I, tau, tau^2, ldots, tau^n - 1, tau^n^2 - 1, tau^n^2 subset mathcal L(V), tag12$



            having as it does $n^2 + 1$ elements, must be linearly dependent over $F$; hence we may find $n^2 +1$ members



            $c_i in F, tag13$



            not all $0$, such that



            $displaystyle sum_0^n^2 c_i tau_i = 0; tag14$



            $tau$ then satisfies the polynomial



            $c(x) = displaystyle sum_0^n^2 c_i x^i in F[x]. tag14$



            We have shown that every $tau in F[x]$ satisfies a polynomial of degree at most $n^2$, but this is far from optimal. For example, the well-known Cayley-Hamilton theorem shows that $tau$ obeys a degree-$n$ polynomial; but the proof of that result is somewhat more involved than what we have done here, and hence will be saved for yet another post.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 24 at 2:54









            Robert LewisRobert Lewis

            46.2k23066




            46.2k23066





















                3












                $begingroup$

                Hint: $mathcalL(V)$ is also a finite dimensional vector space (over what?) Thus, you can find some integer $m$ (you can even explicitly find $m$ in terms of $n$) such that:
                $$1,tau, tau^2,dots tau^m$$ are linearly dependent. I think you can continue from here.






                share|cite|improve this answer









                $endgroup$

















                  3












                  $begingroup$

                  Hint: $mathcalL(V)$ is also a finite dimensional vector space (over what?) Thus, you can find some integer $m$ (you can even explicitly find $m$ in terms of $n$) such that:
                  $$1,tau, tau^2,dots tau^m$$ are linearly dependent. I think you can continue from here.






                  share|cite|improve this answer









                  $endgroup$















                    3












                    3








                    3





                    $begingroup$

                    Hint: $mathcalL(V)$ is also a finite dimensional vector space (over what?) Thus, you can find some integer $m$ (you can even explicitly find $m$ in terms of $n$) such that:
                    $$1,tau, tau^2,dots tau^m$$ are linearly dependent. I think you can continue from here.






                    share|cite|improve this answer









                    $endgroup$



                    Hint: $mathcalL(V)$ is also a finite dimensional vector space (over what?) Thus, you can find some integer $m$ (you can even explicitly find $m$ in terms of $n$) such that:
                    $$1,tau, tau^2,dots tau^m$$ are linearly dependent. I think you can continue from here.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 24 at 2:00









                    dezdichadodezdichado

                    6,3711929




                    6,3711929





















                        3












                        $begingroup$

                        It's not a function, it's a polynomial. For polynomials, it's easy to understand:
                        if the polynomial is $f(x) = a_0 + a_1 x + ldots + a_n x^n$, and the operator is $tau$,
                        then $f(tau) = a_0 I + a_1 tau + ldots + a_n tau^n$.
                        This works in any algebra over a field $F$, where the polynomial has coefficients in $F$.






                        share|cite|improve this answer









                        $endgroup$












                        • $begingroup$
                          Note that in this context, the powers of $x$ are evaluated by repeated composition of $tau$ with itself, while the coefficients of the polynomial are treated as scalar multiples of those powers. Finally the terms of the polynomial are added together as one adds linear mappings to get a linear mapping as a result.
                          $endgroup$
                          – hardmath
                          Jan 24 at 2:37















                        3












                        $begingroup$

                        It's not a function, it's a polynomial. For polynomials, it's easy to understand:
                        if the polynomial is $f(x) = a_0 + a_1 x + ldots + a_n x^n$, and the operator is $tau$,
                        then $f(tau) = a_0 I + a_1 tau + ldots + a_n tau^n$.
                        This works in any algebra over a field $F$, where the polynomial has coefficients in $F$.






                        share|cite|improve this answer









                        $endgroup$












                        • $begingroup$
                          Note that in this context, the powers of $x$ are evaluated by repeated composition of $tau$ with itself, while the coefficients of the polynomial are treated as scalar multiples of those powers. Finally the terms of the polynomial are added together as one adds linear mappings to get a linear mapping as a result.
                          $endgroup$
                          – hardmath
                          Jan 24 at 2:37













                        3












                        3








                        3





                        $begingroup$

                        It's not a function, it's a polynomial. For polynomials, it's easy to understand:
                        if the polynomial is $f(x) = a_0 + a_1 x + ldots + a_n x^n$, and the operator is $tau$,
                        then $f(tau) = a_0 I + a_1 tau + ldots + a_n tau^n$.
                        This works in any algebra over a field $F$, where the polynomial has coefficients in $F$.






                        share|cite|improve this answer









                        $endgroup$



                        It's not a function, it's a polynomial. For polynomials, it's easy to understand:
                        if the polynomial is $f(x) = a_0 + a_1 x + ldots + a_n x^n$, and the operator is $tau$,
                        then $f(tau) = a_0 I + a_1 tau + ldots + a_n tau^n$.
                        This works in any algebra over a field $F$, where the polynomial has coefficients in $F$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 24 at 2:25









                        Robert IsraelRobert Israel

                        323k23212466




                        323k23212466











                        • $begingroup$
                          Note that in this context, the powers of $x$ are evaluated by repeated composition of $tau$ with itself, while the coefficients of the polynomial are treated as scalar multiples of those powers. Finally the terms of the polynomial are added together as one adds linear mappings to get a linear mapping as a result.
                          $endgroup$
                          – hardmath
                          Jan 24 at 2:37
















                        • $begingroup$
                          Note that in this context, the powers of $x$ are evaluated by repeated composition of $tau$ with itself, while the coefficients of the polynomial are treated as scalar multiples of those powers. Finally the terms of the polynomial are added together as one adds linear mappings to get a linear mapping as a result.
                          $endgroup$
                          – hardmath
                          Jan 24 at 2:37















                        $begingroup$
                        Note that in this context, the powers of $x$ are evaluated by repeated composition of $tau$ with itself, while the coefficients of the polynomial are treated as scalar multiples of those powers. Finally the terms of the polynomial are added together as one adds linear mappings to get a linear mapping as a result.
                        $endgroup$
                        – hardmath
                        Jan 24 at 2:37




                        $begingroup$
                        Note that in this context, the powers of $x$ are evaluated by repeated composition of $tau$ with itself, while the coefficients of the polynomial are treated as scalar multiples of those powers. Finally the terms of the polynomial are added together as one adds linear mappings to get a linear mapping as a result.
                        $endgroup$
                        – hardmath
                        Jan 24 at 2:37











                        0












                        $begingroup$

                        Let $p$ be the characterstic Polynomial of $tau$. Then there exist $a_0,a_1,cdots a_n-1inmathbbC$ such that $det(tau-lambda I)=p(t)=a_0+a_1 t+cdots a_n-1t^n-1=0$. Plugging $tau$ yeilds
                        $p(tau)=0$






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          Let $p$ be the characterstic Polynomial of $tau$. Then there exist $a_0,a_1,cdots a_n-1inmathbbC$ such that $det(tau-lambda I)=p(t)=a_0+a_1 t+cdots a_n-1t^n-1=0$. Plugging $tau$ yeilds
                          $p(tau)=0$






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            Let $p$ be the characterstic Polynomial of $tau$. Then there exist $a_0,a_1,cdots a_n-1inmathbbC$ such that $det(tau-lambda I)=p(t)=a_0+a_1 t+cdots a_n-1t^n-1=0$. Plugging $tau$ yeilds
                            $p(tau)=0$






                            share|cite|improve this answer









                            $endgroup$



                            Let $p$ be the characterstic Polynomial of $tau$. Then there exist $a_0,a_1,cdots a_n-1inmathbbC$ such that $det(tau-lambda I)=p(t)=a_0+a_1 t+cdots a_n-1t^n-1=0$. Plugging $tau$ yeilds
                            $p(tau)=0$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 24 at 2:16









                            John TalosJohn Talos

                            14




                            14



























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