Halmos Finite-Dimensional Vector Spaces: Does $mathcal{P}$ over $mathbb{C}$ with $x(t) = x(1 - t)$ form a...












1












$begingroup$


Paul R. Halmos "Finite-Dimensional Vector Spaces", 2e, chapter I, section 2, exercise 5.d:




Consider the vector space $mathcal{P}$ and the subsets $mathcal{V}$
of $mathcal{P}$ consisting of those vectors (polynomials) $x$ for
which



(d) $x(t) = x(1 - t)$ for all $t$.



In which of these cases is $mathcal{V}$ a vector space?




I would suggest that the subset satisfying (d) does form a vector space, since




  • (d) forms a linear constraint in the polynomial's coefficients $mathbf{a}$ as in


$$
g(mathbf{a}) = x(t) - x(1 - t) = 0,
$$




  • the zero vector is included,

  • every element has an inverse element.


$g$ is a linear constraint, since
$$
g(mathbf{a} + mathbf{b}) = g(mathbf{a}) + g(mathbf{b}) qquad wedge qquad alpha g(mathbf{a}) = g(alpha mathbf{a}),
$$
where $mathbf{a}$, $mathbf{b}$ are coefficients of polynomials in $mathcal{P}$, and $alpha$ is a complex number.



Is that correct?






linear-algebra abstract-algebra vector-spaces







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Paul R. Halmos "Finite-Dimensional Vector Spaces", 2e, chapter I, section 2, exercise 5.d:




    Consider the vector space $mathcal{P}$ and the subsets $mathcal{V}$
    of $mathcal{P}$ consisting of those vectors (polynomials) $x$ for
    which



    (d) $x(t) = x(1 - t)$ for all $t$.



    In which of these cases is $mathcal{V}$ a vector space?




    I would suggest that the subset satisfying (d) does form a vector space, since




    • (d) forms a linear constraint in the polynomial's coefficients $mathbf{a}$ as in


    $$
    g(mathbf{a}) = x(t) - x(1 - t) = 0,
    $$




    • the zero vector is included,

    • every element has an inverse element.


    $g$ is a linear constraint, since
    $$
    g(mathbf{a} + mathbf{b}) = g(mathbf{a}) + g(mathbf{b}) qquad wedge qquad alpha g(mathbf{a}) = g(alpha mathbf{a}),
    $$
    where $mathbf{a}$, $mathbf{b}$ are coefficients of polynomials in $mathcal{P}$, and $alpha$ is a complex number.



    Is that correct?






    linear-algebra abstract-algebra vector-spaces







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Paul R. Halmos "Finite-Dimensional Vector Spaces", 2e, chapter I, section 2, exercise 5.d:




      Consider the vector space $mathcal{P}$ and the subsets $mathcal{V}$
      of $mathcal{P}$ consisting of those vectors (polynomials) $x$ for
      which



      (d) $x(t) = x(1 - t)$ for all $t$.



      In which of these cases is $mathcal{V}$ a vector space?




      I would suggest that the subset satisfying (d) does form a vector space, since




      • (d) forms a linear constraint in the polynomial's coefficients $mathbf{a}$ as in


      $$
      g(mathbf{a}) = x(t) - x(1 - t) = 0,
      $$




      • the zero vector is included,

      • every element has an inverse element.


      $g$ is a linear constraint, since
      $$
      g(mathbf{a} + mathbf{b}) = g(mathbf{a}) + g(mathbf{b}) qquad wedge qquad alpha g(mathbf{a}) = g(alpha mathbf{a}),
      $$
      where $mathbf{a}$, $mathbf{b}$ are coefficients of polynomials in $mathcal{P}$, and $alpha$ is a complex number.



      Is that correct?






      linear-algebra abstract-algebra vector-spaces







      share|cite|improve this question











      $endgroup$




      Paul R. Halmos "Finite-Dimensional Vector Spaces", 2e, chapter I, section 2, exercise 5.d:




      Consider the vector space $mathcal{P}$ and the subsets $mathcal{V}$
      of $mathcal{P}$ consisting of those vectors (polynomials) $x$ for
      which



      (d) $x(t) = x(1 - t)$ for all $t$.



      In which of these cases is $mathcal{V}$ a vector space?




      I would suggest that the subset satisfying (d) does form a vector space, since




      • (d) forms a linear constraint in the polynomial's coefficients $mathbf{a}$ as in


      $$
      g(mathbf{a}) = x(t) - x(1 - t) = 0,
      $$




      • the zero vector is included,

      • every element has an inverse element.


      $g$ is a linear constraint, since
      $$
      g(mathbf{a} + mathbf{b}) = g(mathbf{a}) + g(mathbf{b}) qquad wedge qquad alpha g(mathbf{a}) = g(alpha mathbf{a}),
      $$
      where $mathbf{a}$, $mathbf{b}$ are coefficients of polynomials in $mathcal{P}$, and $alpha$ is a complex number.



      Is that correct?






      linear-algebra abstract-algebra vector-spaces




      linear-algebra abstract-algebra vector-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 10:18







      Max Herrmann

















      asked Jan 5 at 8:24









      Max HerrmannMax Herrmann

      724419




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          2 Answers
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          active

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          $begingroup$


          I would suggest that the subset satisfying (d) does form a vector space, since




          • (d) forms a linear constraint in the polynomial's coefficients,

          • the zero vector is included,

          • every element has an inverse element.


          Is that correct?




          This is the right idea, but when you say that (d) forms a linear constraint in the coefficients, what do you mean?



          I suppose what you mean is that if the polynomial is $a_0 + a_1 x + a_2 x^2 + cdots ...$ then it is some linear equation in $a_0, a_1, ldots$. But why does this imply that (d) is a vector space?



          However you can formalize (d). I would approach this by defining a map
          $$
          A : mathcal{P} to mathcal{P}
          $$



          where $A(boldsymbol{x}) = boldsymbol{x}(t) - boldsymbol{x}(1-t)$.



          Then, show that $A$ is a linear map -- that is, it preserves addition and scalar multiplication.



          Finally, $mathcal{V}$ is the set of polynomials $boldsymbol{x}$ such that $A(boldsymbol{x}) = boldsymbol{0}$. Since $A$ is linear this is a vector subspace. So this completes your proof.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Your first bullet point is not useful. Just verify the definition. Surely 0 is in there. Every element has an additive inverse. If $c$ is a constant, then $cx(t)=cx(1-t)$, so constant multiples of polynomials are also in there. So you're done.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What if I doubt that every element has an additive inverse?
              $endgroup$
              – Max Herrmann
              Jan 5 at 8:54










            • $begingroup$
              It seems you are misunderstanding the idea of a proof. You need to show this is the case. The constant case is clearly shows the inverse. Just take $c=-1$. However, I did leave out that two functions may not add to satisfy $d$, but this is clear since you just add the contraint equation; i.e. $x(t)+y(t)=x(1-t)+y(1-t)$. You should note that notationally, you should write $(x+y)(t)$ for the aforementioned equation. As well as $(x+y)(1-t)$. This is, by definition, the addition of two polynomials.
              $endgroup$
              – user495490
              Jan 5 at 9:34










            • $begingroup$
              What does $-1 cdot x(t) = -1 cdot x(1-t)$ tell me?
              $endgroup$
              – Max Herrmann
              Jan 5 at 9:45










            • $begingroup$
              That inverse exist. -x(t) also satisfies (d). It is clearly an additive inverse to x(t).
              $endgroup$
              – user495490
              Jan 5 at 12:44










            • $begingroup$
              I find it impressive that it is obvious to you. I need some more hints and smaller steps in argumentation, I'm afraid. E.g. proof by contradiction: Assume there is an element $x'$ in the subset which does not have an additive inverse. Then $g(-mathbf{a}') = -x'(t) + x'(1-t) neq 0$. But $g$ is homogeneous in $mathbf{a}$. Hence, every element has an additive inverse.
              $endgroup$
              – Max Herrmann
              Jan 5 at 14:30













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            2 Answers
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            2 Answers
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            active

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            active

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            active

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            1












            $begingroup$


            I would suggest that the subset satisfying (d) does form a vector space, since




            • (d) forms a linear constraint in the polynomial's coefficients,

            • the zero vector is included,

            • every element has an inverse element.


            Is that correct?




            This is the right idea, but when you say that (d) forms a linear constraint in the coefficients, what do you mean?



            I suppose what you mean is that if the polynomial is $a_0 + a_1 x + a_2 x^2 + cdots ...$ then it is some linear equation in $a_0, a_1, ldots$. But why does this imply that (d) is a vector space?



            However you can formalize (d). I would approach this by defining a map
            $$
            A : mathcal{P} to mathcal{P}
            $$



            where $A(boldsymbol{x}) = boldsymbol{x}(t) - boldsymbol{x}(1-t)$.



            Then, show that $A$ is a linear map -- that is, it preserves addition and scalar multiplication.



            Finally, $mathcal{V}$ is the set of polynomials $boldsymbol{x}$ such that $A(boldsymbol{x}) = boldsymbol{0}$. Since $A$ is linear this is a vector subspace. So this completes your proof.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$


              I would suggest that the subset satisfying (d) does form a vector space, since




              • (d) forms a linear constraint in the polynomial's coefficients,

              • the zero vector is included,

              • every element has an inverse element.


              Is that correct?




              This is the right idea, but when you say that (d) forms a linear constraint in the coefficients, what do you mean?



              I suppose what you mean is that if the polynomial is $a_0 + a_1 x + a_2 x^2 + cdots ...$ then it is some linear equation in $a_0, a_1, ldots$. But why does this imply that (d) is a vector space?



              However you can formalize (d). I would approach this by defining a map
              $$
              A : mathcal{P} to mathcal{P}
              $$



              where $A(boldsymbol{x}) = boldsymbol{x}(t) - boldsymbol{x}(1-t)$.



              Then, show that $A$ is a linear map -- that is, it preserves addition and scalar multiplication.



              Finally, $mathcal{V}$ is the set of polynomials $boldsymbol{x}$ such that $A(boldsymbol{x}) = boldsymbol{0}$. Since $A$ is linear this is a vector subspace. So this completes your proof.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$


                I would suggest that the subset satisfying (d) does form a vector space, since




                • (d) forms a linear constraint in the polynomial's coefficients,

                • the zero vector is included,

                • every element has an inverse element.


                Is that correct?




                This is the right idea, but when you say that (d) forms a linear constraint in the coefficients, what do you mean?



                I suppose what you mean is that if the polynomial is $a_0 + a_1 x + a_2 x^2 + cdots ...$ then it is some linear equation in $a_0, a_1, ldots$. But why does this imply that (d) is a vector space?



                However you can formalize (d). I would approach this by defining a map
                $$
                A : mathcal{P} to mathcal{P}
                $$



                where $A(boldsymbol{x}) = boldsymbol{x}(t) - boldsymbol{x}(1-t)$.



                Then, show that $A$ is a linear map -- that is, it preserves addition and scalar multiplication.



                Finally, $mathcal{V}$ is the set of polynomials $boldsymbol{x}$ such that $A(boldsymbol{x}) = boldsymbol{0}$. Since $A$ is linear this is a vector subspace. So this completes your proof.






                share|cite|improve this answer









                $endgroup$




                I would suggest that the subset satisfying (d) does form a vector space, since




                • (d) forms a linear constraint in the polynomial's coefficients,

                • the zero vector is included,

                • every element has an inverse element.


                Is that correct?




                This is the right idea, but when you say that (d) forms a linear constraint in the coefficients, what do you mean?



                I suppose what you mean is that if the polynomial is $a_0 + a_1 x + a_2 x^2 + cdots ...$ then it is some linear equation in $a_0, a_1, ldots$. But why does this imply that (d) is a vector space?



                However you can formalize (d). I would approach this by defining a map
                $$
                A : mathcal{P} to mathcal{P}
                $$



                where $A(boldsymbol{x}) = boldsymbol{x}(t) - boldsymbol{x}(1-t)$.



                Then, show that $A$ is a linear map -- that is, it preserves addition and scalar multiplication.



                Finally, $mathcal{V}$ is the set of polynomials $boldsymbol{x}$ such that $A(boldsymbol{x}) = boldsymbol{0}$. Since $A$ is linear this is a vector subspace. So this completes your proof.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 9:57









                60056005

                37k751127




                37k751127























                    0












                    $begingroup$

                    Your first bullet point is not useful. Just verify the definition. Surely 0 is in there. Every element has an additive inverse. If $c$ is a constant, then $cx(t)=cx(1-t)$, so constant multiples of polynomials are also in there. So you're done.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      What if I doubt that every element has an additive inverse?
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 8:54










                    • $begingroup$
                      It seems you are misunderstanding the idea of a proof. You need to show this is the case. The constant case is clearly shows the inverse. Just take $c=-1$. However, I did leave out that two functions may not add to satisfy $d$, but this is clear since you just add the contraint equation; i.e. $x(t)+y(t)=x(1-t)+y(1-t)$. You should note that notationally, you should write $(x+y)(t)$ for the aforementioned equation. As well as $(x+y)(1-t)$. This is, by definition, the addition of two polynomials.
                      $endgroup$
                      – user495490
                      Jan 5 at 9:34










                    • $begingroup$
                      What does $-1 cdot x(t) = -1 cdot x(1-t)$ tell me?
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 9:45










                    • $begingroup$
                      That inverse exist. -x(t) also satisfies (d). It is clearly an additive inverse to x(t).
                      $endgroup$
                      – user495490
                      Jan 5 at 12:44










                    • $begingroup$
                      I find it impressive that it is obvious to you. I need some more hints and smaller steps in argumentation, I'm afraid. E.g. proof by contradiction: Assume there is an element $x'$ in the subset which does not have an additive inverse. Then $g(-mathbf{a}') = -x'(t) + x'(1-t) neq 0$. But $g$ is homogeneous in $mathbf{a}$. Hence, every element has an additive inverse.
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 14:30


















                    0












                    $begingroup$

                    Your first bullet point is not useful. Just verify the definition. Surely 0 is in there. Every element has an additive inverse. If $c$ is a constant, then $cx(t)=cx(1-t)$, so constant multiples of polynomials are also in there. So you're done.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      What if I doubt that every element has an additive inverse?
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 8:54










                    • $begingroup$
                      It seems you are misunderstanding the idea of a proof. You need to show this is the case. The constant case is clearly shows the inverse. Just take $c=-1$. However, I did leave out that two functions may not add to satisfy $d$, but this is clear since you just add the contraint equation; i.e. $x(t)+y(t)=x(1-t)+y(1-t)$. You should note that notationally, you should write $(x+y)(t)$ for the aforementioned equation. As well as $(x+y)(1-t)$. This is, by definition, the addition of two polynomials.
                      $endgroup$
                      – user495490
                      Jan 5 at 9:34










                    • $begingroup$
                      What does $-1 cdot x(t) = -1 cdot x(1-t)$ tell me?
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 9:45










                    • $begingroup$
                      That inverse exist. -x(t) also satisfies (d). It is clearly an additive inverse to x(t).
                      $endgroup$
                      – user495490
                      Jan 5 at 12:44










                    • $begingroup$
                      I find it impressive that it is obvious to you. I need some more hints and smaller steps in argumentation, I'm afraid. E.g. proof by contradiction: Assume there is an element $x'$ in the subset which does not have an additive inverse. Then $g(-mathbf{a}') = -x'(t) + x'(1-t) neq 0$. But $g$ is homogeneous in $mathbf{a}$. Hence, every element has an additive inverse.
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 14:30
















                    0












                    0








                    0





                    $begingroup$

                    Your first bullet point is not useful. Just verify the definition. Surely 0 is in there. Every element has an additive inverse. If $c$ is a constant, then $cx(t)=cx(1-t)$, so constant multiples of polynomials are also in there. So you're done.






                    share|cite|improve this answer









                    $endgroup$



                    Your first bullet point is not useful. Just verify the definition. Surely 0 is in there. Every element has an additive inverse. If $c$ is a constant, then $cx(t)=cx(1-t)$, so constant multiples of polynomials are also in there. So you're done.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 5 at 8:43









                    user495490user495490

                    285




                    285












                    • $begingroup$
                      What if I doubt that every element has an additive inverse?
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 8:54










                    • $begingroup$
                      It seems you are misunderstanding the idea of a proof. You need to show this is the case. The constant case is clearly shows the inverse. Just take $c=-1$. However, I did leave out that two functions may not add to satisfy $d$, but this is clear since you just add the contraint equation; i.e. $x(t)+y(t)=x(1-t)+y(1-t)$. You should note that notationally, you should write $(x+y)(t)$ for the aforementioned equation. As well as $(x+y)(1-t)$. This is, by definition, the addition of two polynomials.
                      $endgroup$
                      – user495490
                      Jan 5 at 9:34










                    • $begingroup$
                      What does $-1 cdot x(t) = -1 cdot x(1-t)$ tell me?
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 9:45










                    • $begingroup$
                      That inverse exist. -x(t) also satisfies (d). It is clearly an additive inverse to x(t).
                      $endgroup$
                      – user495490
                      Jan 5 at 12:44










                    • $begingroup$
                      I find it impressive that it is obvious to you. I need some more hints and smaller steps in argumentation, I'm afraid. E.g. proof by contradiction: Assume there is an element $x'$ in the subset which does not have an additive inverse. Then $g(-mathbf{a}') = -x'(t) + x'(1-t) neq 0$. But $g$ is homogeneous in $mathbf{a}$. Hence, every element has an additive inverse.
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 14:30




















                    • $begingroup$
                      What if I doubt that every element has an additive inverse?
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 8:54










                    • $begingroup$
                      It seems you are misunderstanding the idea of a proof. You need to show this is the case. The constant case is clearly shows the inverse. Just take $c=-1$. However, I did leave out that two functions may not add to satisfy $d$, but this is clear since you just add the contraint equation; i.e. $x(t)+y(t)=x(1-t)+y(1-t)$. You should note that notationally, you should write $(x+y)(t)$ for the aforementioned equation. As well as $(x+y)(1-t)$. This is, by definition, the addition of two polynomials.
                      $endgroup$
                      – user495490
                      Jan 5 at 9:34










                    • $begingroup$
                      What does $-1 cdot x(t) = -1 cdot x(1-t)$ tell me?
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 9:45










                    • $begingroup$
                      That inverse exist. -x(t) also satisfies (d). It is clearly an additive inverse to x(t).
                      $endgroup$
                      – user495490
                      Jan 5 at 12:44










                    • $begingroup$
                      I find it impressive that it is obvious to you. I need some more hints and smaller steps in argumentation, I'm afraid. E.g. proof by contradiction: Assume there is an element $x'$ in the subset which does not have an additive inverse. Then $g(-mathbf{a}') = -x'(t) + x'(1-t) neq 0$. But $g$ is homogeneous in $mathbf{a}$. Hence, every element has an additive inverse.
                      $endgroup$
                      – Max Herrmann
                      Jan 5 at 14:30


















                    $begingroup$
                    What if I doubt that every element has an additive inverse?
                    $endgroup$
                    – Max Herrmann
                    Jan 5 at 8:54




                    $begingroup$
                    What if I doubt that every element has an additive inverse?
                    $endgroup$
                    – Max Herrmann
                    Jan 5 at 8:54












                    $begingroup$
                    It seems you are misunderstanding the idea of a proof. You need to show this is the case. The constant case is clearly shows the inverse. Just take $c=-1$. However, I did leave out that two functions may not add to satisfy $d$, but this is clear since you just add the contraint equation; i.e. $x(t)+y(t)=x(1-t)+y(1-t)$. You should note that notationally, you should write $(x+y)(t)$ for the aforementioned equation. As well as $(x+y)(1-t)$. This is, by definition, the addition of two polynomials.
                    $endgroup$
                    – user495490
                    Jan 5 at 9:34




                    $begingroup$
                    It seems you are misunderstanding the idea of a proof. You need to show this is the case. The constant case is clearly shows the inverse. Just take $c=-1$. However, I did leave out that two functions may not add to satisfy $d$, but this is clear since you just add the contraint equation; i.e. $x(t)+y(t)=x(1-t)+y(1-t)$. You should note that notationally, you should write $(x+y)(t)$ for the aforementioned equation. As well as $(x+y)(1-t)$. This is, by definition, the addition of two polynomials.
                    $endgroup$
                    – user495490
                    Jan 5 at 9:34












                    $begingroup$
                    What does $-1 cdot x(t) = -1 cdot x(1-t)$ tell me?
                    $endgroup$
                    – Max Herrmann
                    Jan 5 at 9:45




                    $begingroup$
                    What does $-1 cdot x(t) = -1 cdot x(1-t)$ tell me?
                    $endgroup$
                    – Max Herrmann
                    Jan 5 at 9:45












                    $begingroup$
                    That inverse exist. -x(t) also satisfies (d). It is clearly an additive inverse to x(t).
                    $endgroup$
                    – user495490
                    Jan 5 at 12:44




                    $begingroup$
                    That inverse exist. -x(t) also satisfies (d). It is clearly an additive inverse to x(t).
                    $endgroup$
                    – user495490
                    Jan 5 at 12:44












                    $begingroup$
                    I find it impressive that it is obvious to you. I need some more hints and smaller steps in argumentation, I'm afraid. E.g. proof by contradiction: Assume there is an element $x'$ in the subset which does not have an additive inverse. Then $g(-mathbf{a}') = -x'(t) + x'(1-t) neq 0$. But $g$ is homogeneous in $mathbf{a}$. Hence, every element has an additive inverse.
                    $endgroup$
                    – Max Herrmann
                    Jan 5 at 14:30






                    $begingroup$
                    I find it impressive that it is obvious to you. I need some more hints and smaller steps in argumentation, I'm afraid. E.g. proof by contradiction: Assume there is an element $x'$ in the subset which does not have an additive inverse. Then $g(-mathbf{a}') = -x'(t) + x'(1-t) neq 0$. But $g$ is homogeneous in $mathbf{a}$. Hence, every element has an additive inverse.
                    $endgroup$
                    – Max Herrmann
                    Jan 5 at 14:30




















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