A function evaluated at a operator












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
    2 hours ago
















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
    2 hours ago














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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









Jiexiong687691Jiexiong687691

825




825












  • $begingroup$
    A polynomial is not a function. :P
    $endgroup$
    – darij grinberg
    2 hours ago


















  • $begingroup$
    A polynomial is not a function. :P
    $endgroup$
    – darij grinberg
    2 hours ago
















$begingroup$
A polynomial is not a function. :P
$endgroup$
– darij grinberg
2 hours ago




$begingroup$
A polynomial is not a function. :P
$endgroup$
– darij grinberg
2 hours ago










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



$p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag{10}$



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



Now we recall that



$dim_F mathcal L(V) = n^2; tag{11}$



thus the set



${I, tau, tau^2, ldots, tau^{n - 1}, tau^{n^2 - 1}, tau^{n^2} } subset mathcal L(V), tag{12}$



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



$c_i in F, tag{13}$



not all $0$, such that



$displaystyle sum_0^{n^2} c_i tau_i = 0; tag{14}$



$tau$ then satisfies the polynomial



$c(x) = displaystyle sum_0^{n^2} c_i x^i in F[x]. tag{14}$



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: $mathcal{L}(V)$ 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
        3 hours ago



















      0












      $begingroup$

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






      share|cite|improve this answer








      New contributor




      John Talos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$













        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%2f3085322%2fa-function-evaluated-at-a-operator%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        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



        $p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag{10}$



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



        Now we recall that



        $dim_F mathcal L(V) = n^2; tag{11}$



        thus the set



        ${I, tau, tau^2, ldots, tau^{n - 1}, tau^{n^2 - 1}, tau^{n^2} } subset mathcal L(V), tag{12}$



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



        $c_i in F, tag{13}$



        not all $0$, such that



        $displaystyle sum_0^{n^2} c_i tau_i = 0; tag{14}$



        $tau$ then satisfies the polynomial



        $c(x) = displaystyle sum_0^{n^2} c_i x^i in F[x]. tag{14}$



        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



          $p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag{10}$



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



          Now we recall that



          $dim_F mathcal L(V) = n^2; tag{11}$



          thus the set



          ${I, tau, tau^2, ldots, tau^{n - 1}, tau^{n^2 - 1}, tau^{n^2} } subset mathcal L(V), tag{12}$



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



          $c_i in F, tag{13}$



          not all $0$, such that



          $displaystyle sum_0^{n^2} c_i tau_i = 0; tag{14}$



          $tau$ then satisfies the polynomial



          $c(x) = displaystyle sum_0^{n^2} c_i x^i in F[x]. tag{14}$



          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



            $p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag{10}$



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



            Now we recall that



            $dim_F mathcal L(V) = n^2; tag{11}$



            thus the set



            ${I, tau, tau^2, ldots, tau^{n - 1}, tau^{n^2 - 1}, tau^{n^2} } subset mathcal L(V), tag{12}$



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



            $c_i in F, tag{13}$



            not all $0$, such that



            $displaystyle sum_0^{n^2} c_i tau_i = 0; tag{14}$



            $tau$ then satisfies the polynomial



            $c(x) = displaystyle sum_0^{n^2} c_i x^i in F[x]. tag{14}$



            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



            $p(tau) = displaystyle sum_0^m p_i tau^i in mathcal L(V), tag{10}$



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



            Now we recall that



            $dim_F mathcal L(V) = n^2; tag{11}$



            thus the set



            ${I, tau, tau^2, ldots, tau^{n - 1}, tau^{n^2 - 1}, tau^{n^2} } subset mathcal L(V), tag{12}$



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



            $c_i in F, tag{13}$



            not all $0$, such that



            $displaystyle sum_0^{n^2} c_i tau_i = 0; tag{14}$



            $tau$ then satisfies the polynomial



            $c(x) = displaystyle sum_0^{n^2} c_i x^i in F[x]. tag{14}$



            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 3 hours ago









            Robert LewisRobert Lewis

            45k22964




            45k22964























                3












                $begingroup$

                Hint: $mathcal{L}(V)$ 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: $mathcal{L}(V)$ 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: $mathcal{L}(V)$ 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: $mathcal{L}(V)$ 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 3 hours ago









                    dezdichadodezdichado

                    6,3961929




                    6,3961929























                        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
                          3 hours ago
















                        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
                          3 hours ago














                        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 3 hours ago









                        Robert IsraelRobert Israel

                        321k23210462




                        321k23210462












                        • $begingroup$
                          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
                          3 hours ago


















                        • $begingroup$
                          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
                          3 hours ago
















                        $begingroup$
                        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
                        3 hours ago




                        $begingroup$
                        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
                        3 hours ago











                        0












                        $begingroup$

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






                        share|cite|improve this answer








                        New contributor




                        John Talos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.






                        $endgroup$


















                          0












                          $begingroup$

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






                          share|cite|improve this answer








                          New contributor




                          John Talos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

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






                            share|cite|improve this answer








                            New contributor




                            John Talos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            $endgroup$



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







                            share|cite|improve this answer








                            New contributor




                            John Talos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|cite|improve this answer



                            share|cite|improve this answer






                            New contributor




                            John Talos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            answered 3 hours ago









                            John TalosJohn Talos

                            13




                            13




                            New contributor




                            John Talos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.





                            New contributor





                            John Talos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            John Talos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






























                                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.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085322%2fa-function-evaluated-at-a-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

                                Liste der Baudenkmale in Friedland (Mecklenburg)

                                Single-Malt-Whisky

                                Czorneboh