Set of real polynomials, organized as a vectorspace over $mathbb{R}$.












1












$begingroup$


I'm currently studying for a Linear Algebra exam in january, thus I'm going through some older exam questions, and I'm at the following question.



Let $mathbb{R}[x]$ denote the set of all real polynomials, which is organized as a vectorspace over $mathbb{R}$.




  • A. Assume that $U={pinmathbb{R}_{1}[x]|p(2)=0}$. Determine a basis for $U$ and its dimension.

  • B. Add $p(x)=x-2$ to a basis $B$ for $mathbb{R}_3[x]$, and determine the coordinate vector $[q]_B$ for $q(x)=3+2x+x^2$

  • C. Determine a subspace $Wsubset mathbb{R}_3[x]$ such that $Uoplus W=mathbb{R}_3[x]$. Determine $dim W$.

  • D. Find the matrix $M(T)$ with respect to $B$ when $T=4D+3I$, where $D$ denote differentiation and $I$ denotes the identical map. Determine whether $T$ is injective, surjective or bijective.

  • E. Let $L$ denote the set of all real numbers $lambda$ such that $S=4D-lambda I$ has $span(1,x^2,x^3)$ as a subspace which is invariant under $S$. Determine $L$.


My answers are,



A. $q(x)=x-2$ is a basis for $U$ and its dimension is 1.



B. I add $p(x)=x-2$ to the standard basis of $mathbb{R}_3[x]$, such that $B=(x-2,1,x,x^2,x^3)$, this is not a basis of $mathbb{R}_3[x]$ as $x$ can be written as a linear comb. of the previous elements (vectors?), thus I reduce it to a basis by removing $x$ from the list so $B=(x-2,1,x^2,x^3)$. Now I write the matrix with respect to this basis and add $begin{bmatrix}3\2\1\0end{bmatrix}$ to the matrix, and then I compute the coordinate vector $[q]_B$ by gauss elimination.



$begin{align}
begin{bmatrix}
-2 & 1 & 0 & 0 & 3\
1 & 0 & 0 & 0 & 2\
0 & 0 & 1 & 0& 1\
0 & 0 & 0 & 1 & 0
end{bmatrix}
sim
begin{bmatrix}
1 & 0 & 0 & 0 & 2\
0 & 1 & 0 & 0 & 7\
0 & 0 & 1 & 0 & 1\
0 & 0 & 0 & 1 & 0
end{bmatrix}
end{align}$



Thus
$[q]_B = begin{bmatrix}2 \7\1\0end{bmatrix}$



C. From B. $mathbb{R}_3[x]$ had the basis $(x-2,1,x^2,x^3)$, and the basis of $U=(x-2)$, so $W=(1,x^2,x^3)$, thus $Uoplus W$ will span $mathbb{R}_3[x]$. $dim W=3$



I'm stuck at D. and E. I assume that I start by differentiating the basis of $B$, then representing $B'$ as a matrix which I substitute into $D$ and then add the $3I$? I'm not sure that I'm capable of solving E. without knowing the methods from D.



I hope that some of it is right, and I hope to get some tips, correction and some help with the last two questions.



Best regards Jens.










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I'm currently studying for a Linear Algebra exam in january, thus I'm going through some older exam questions, and I'm at the following question.



    Let $mathbb{R}[x]$ denote the set of all real polynomials, which is organized as a vectorspace over $mathbb{R}$.




    • A. Assume that $U={pinmathbb{R}_{1}[x]|p(2)=0}$. Determine a basis for $U$ and its dimension.

    • B. Add $p(x)=x-2$ to a basis $B$ for $mathbb{R}_3[x]$, and determine the coordinate vector $[q]_B$ for $q(x)=3+2x+x^2$

    • C. Determine a subspace $Wsubset mathbb{R}_3[x]$ such that $Uoplus W=mathbb{R}_3[x]$. Determine $dim W$.

    • D. Find the matrix $M(T)$ with respect to $B$ when $T=4D+3I$, where $D$ denote differentiation and $I$ denotes the identical map. Determine whether $T$ is injective, surjective or bijective.

    • E. Let $L$ denote the set of all real numbers $lambda$ such that $S=4D-lambda I$ has $span(1,x^2,x^3)$ as a subspace which is invariant under $S$. Determine $L$.


    My answers are,



    A. $q(x)=x-2$ is a basis for $U$ and its dimension is 1.



    B. I add $p(x)=x-2$ to the standard basis of $mathbb{R}_3[x]$, such that $B=(x-2,1,x,x^2,x^3)$, this is not a basis of $mathbb{R}_3[x]$ as $x$ can be written as a linear comb. of the previous elements (vectors?), thus I reduce it to a basis by removing $x$ from the list so $B=(x-2,1,x^2,x^3)$. Now I write the matrix with respect to this basis and add $begin{bmatrix}3\2\1\0end{bmatrix}$ to the matrix, and then I compute the coordinate vector $[q]_B$ by gauss elimination.



    $begin{align}
    begin{bmatrix}
    -2 & 1 & 0 & 0 & 3\
    1 & 0 & 0 & 0 & 2\
    0 & 0 & 1 & 0& 1\
    0 & 0 & 0 & 1 & 0
    end{bmatrix}
    sim
    begin{bmatrix}
    1 & 0 & 0 & 0 & 2\
    0 & 1 & 0 & 0 & 7\
    0 & 0 & 1 & 0 & 1\
    0 & 0 & 0 & 1 & 0
    end{bmatrix}
    end{align}$



    Thus
    $[q]_B = begin{bmatrix}2 \7\1\0end{bmatrix}$



    C. From B. $mathbb{R}_3[x]$ had the basis $(x-2,1,x^2,x^3)$, and the basis of $U=(x-2)$, so $W=(1,x^2,x^3)$, thus $Uoplus W$ will span $mathbb{R}_3[x]$. $dim W=3$



    I'm stuck at D. and E. I assume that I start by differentiating the basis of $B$, then representing $B'$ as a matrix which I substitute into $D$ and then add the $3I$? I'm not sure that I'm capable of solving E. without knowing the methods from D.



    I hope that some of it is right, and I hope to get some tips, correction and some help with the last two questions.



    Best regards Jens.










    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I'm currently studying for a Linear Algebra exam in january, thus I'm going through some older exam questions, and I'm at the following question.



      Let $mathbb{R}[x]$ denote the set of all real polynomials, which is organized as a vectorspace over $mathbb{R}$.




      • A. Assume that $U={pinmathbb{R}_{1}[x]|p(2)=0}$. Determine a basis for $U$ and its dimension.

      • B. Add $p(x)=x-2$ to a basis $B$ for $mathbb{R}_3[x]$, and determine the coordinate vector $[q]_B$ for $q(x)=3+2x+x^2$

      • C. Determine a subspace $Wsubset mathbb{R}_3[x]$ such that $Uoplus W=mathbb{R}_3[x]$. Determine $dim W$.

      • D. Find the matrix $M(T)$ with respect to $B$ when $T=4D+3I$, where $D$ denote differentiation and $I$ denotes the identical map. Determine whether $T$ is injective, surjective or bijective.

      • E. Let $L$ denote the set of all real numbers $lambda$ such that $S=4D-lambda I$ has $span(1,x^2,x^3)$ as a subspace which is invariant under $S$. Determine $L$.


      My answers are,



      A. $q(x)=x-2$ is a basis for $U$ and its dimension is 1.



      B. I add $p(x)=x-2$ to the standard basis of $mathbb{R}_3[x]$, such that $B=(x-2,1,x,x^2,x^3)$, this is not a basis of $mathbb{R}_3[x]$ as $x$ can be written as a linear comb. of the previous elements (vectors?), thus I reduce it to a basis by removing $x$ from the list so $B=(x-2,1,x^2,x^3)$. Now I write the matrix with respect to this basis and add $begin{bmatrix}3\2\1\0end{bmatrix}$ to the matrix, and then I compute the coordinate vector $[q]_B$ by gauss elimination.



      $begin{align}
      begin{bmatrix}
      -2 & 1 & 0 & 0 & 3\
      1 & 0 & 0 & 0 & 2\
      0 & 0 & 1 & 0& 1\
      0 & 0 & 0 & 1 & 0
      end{bmatrix}
      sim
      begin{bmatrix}
      1 & 0 & 0 & 0 & 2\
      0 & 1 & 0 & 0 & 7\
      0 & 0 & 1 & 0 & 1\
      0 & 0 & 0 & 1 & 0
      end{bmatrix}
      end{align}$



      Thus
      $[q]_B = begin{bmatrix}2 \7\1\0end{bmatrix}$



      C. From B. $mathbb{R}_3[x]$ had the basis $(x-2,1,x^2,x^3)$, and the basis of $U=(x-2)$, so $W=(1,x^2,x^3)$, thus $Uoplus W$ will span $mathbb{R}_3[x]$. $dim W=3$



      I'm stuck at D. and E. I assume that I start by differentiating the basis of $B$, then representing $B'$ as a matrix which I substitute into $D$ and then add the $3I$? I'm not sure that I'm capable of solving E. without knowing the methods from D.



      I hope that some of it is right, and I hope to get some tips, correction and some help with the last two questions.



      Best regards Jens.










      share|cite|improve this question









      $endgroup$




      I'm currently studying for a Linear Algebra exam in january, thus I'm going through some older exam questions, and I'm at the following question.



      Let $mathbb{R}[x]$ denote the set of all real polynomials, which is organized as a vectorspace over $mathbb{R}$.




      • A. Assume that $U={pinmathbb{R}_{1}[x]|p(2)=0}$. Determine a basis for $U$ and its dimension.

      • B. Add $p(x)=x-2$ to a basis $B$ for $mathbb{R}_3[x]$, and determine the coordinate vector $[q]_B$ for $q(x)=3+2x+x^2$

      • C. Determine a subspace $Wsubset mathbb{R}_3[x]$ such that $Uoplus W=mathbb{R}_3[x]$. Determine $dim W$.

      • D. Find the matrix $M(T)$ with respect to $B$ when $T=4D+3I$, where $D$ denote differentiation and $I$ denotes the identical map. Determine whether $T$ is injective, surjective or bijective.

      • E. Let $L$ denote the set of all real numbers $lambda$ such that $S=4D-lambda I$ has $span(1,x^2,x^3)$ as a subspace which is invariant under $S$. Determine $L$.


      My answers are,



      A. $q(x)=x-2$ is a basis for $U$ and its dimension is 1.



      B. I add $p(x)=x-2$ to the standard basis of $mathbb{R}_3[x]$, such that $B=(x-2,1,x,x^2,x^3)$, this is not a basis of $mathbb{R}_3[x]$ as $x$ can be written as a linear comb. of the previous elements (vectors?), thus I reduce it to a basis by removing $x$ from the list so $B=(x-2,1,x^2,x^3)$. Now I write the matrix with respect to this basis and add $begin{bmatrix}3\2\1\0end{bmatrix}$ to the matrix, and then I compute the coordinate vector $[q]_B$ by gauss elimination.



      $begin{align}
      begin{bmatrix}
      -2 & 1 & 0 & 0 & 3\
      1 & 0 & 0 & 0 & 2\
      0 & 0 & 1 & 0& 1\
      0 & 0 & 0 & 1 & 0
      end{bmatrix}
      sim
      begin{bmatrix}
      1 & 0 & 0 & 0 & 2\
      0 & 1 & 0 & 0 & 7\
      0 & 0 & 1 & 0 & 1\
      0 & 0 & 0 & 1 & 0
      end{bmatrix}
      end{align}$



      Thus
      $[q]_B = begin{bmatrix}2 \7\1\0end{bmatrix}$



      C. From B. $mathbb{R}_3[x]$ had the basis $(x-2,1,x^2,x^3)$, and the basis of $U=(x-2)$, so $W=(1,x^2,x^3)$, thus $Uoplus W$ will span $mathbb{R}_3[x]$. $dim W=3$



      I'm stuck at D. and E. I assume that I start by differentiating the basis of $B$, then representing $B'$ as a matrix which I substitute into $D$ and then add the $3I$? I'm not sure that I'm capable of solving E. without knowing the methods from D.



      I hope that some of it is right, and I hope to get some tips, correction and some help with the last two questions.



      Best regards Jens.







      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 30 '18 at 23:51









      Jens KramerJens Kramer

      557




      557






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Up till now all correct :)
          For exercise D) you'll need The differentiation Matrix $$D=
          begin{bmatrix}
          0 & 1 & 0 & 0 \
          0 & 0 & 2 & 0 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}$$

          $$T:=4D+3I=begin{bmatrix}
          3 & 4 & 0 & 0 \
          0 & 3 & 8 & 0 \
          0 & 0& 3 & 12 \
          0 & 0 & 0 & 3
          end{bmatrix} $$

          T is bijective because $det(T)=3^4neq 0$ If you have not learned about determinants jet you can simply use the Gauss-algorithm to Calculate $A^{-1}$.
          Then have to write T with as a Matrix with Respect to the Basis B. For this you will need the Transformation Matricies $[id]^B_S$ and $[id]^S_B$ (where S denotes the standart-Basis) then $[T]^B_B=[id]^B_S T [id]^S_B$



          hint:
          $[id]^B_S$ the transformation matrix from S to B is:
          $$[id]^B_S=begin{bmatrix}
          -2 & 1 & 0 & 0 \
          1 & 0 & 0 & 0 \
          0 & 0 & 1& 0 \
          0 & 0 & 0 & 1
          end{bmatrix} $$

          to understand this look at $[id]^B_S e_j $.



          Also note $[id]^B_S=([id]^S_B)^{-1}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think I'm doing something wrong when trying to calculate the $[id]^S_B$ matrix. I get a matrix with a zero in the diagonal such that it isn't invertible.
            $endgroup$
            – Jens Kramer
            Jan 2 at 15:54










          • $begingroup$
            I have misunderstood some things, but correct me if I'm wrong. I just calculated the matrix in the hint, but i thought that it was $[id]^S_B$ not $[id]^B_S$. When you write $e_j$, do you refer to the j'th orthonormal basisvector. I'm not really sure what that should clarify. Last I calculate the inverse of $[id]^B_S$ to be $begin{bmatrix}0 & 1 & 0 & 0\ 1 &2 &0&0\0&0&1&0\0&0&0&1end{bmatrix}$. My previous comment said that I didn't think this was invertible because one of the diagonals had a zero, which seems to be wrong. Is it right to say it isn't diagonizable then?
            $endgroup$
            – Jens Kramer
            Jan 2 at 16:58












          • $begingroup$
            the matrix seems right, it is the inverse of the matrix I've written (i'm not quite sure about my notation i might have done it the wrong way round) , $e_j$ is the j-basisvektor $e_1=(1,0,0,0),e_2=(0,1,0,0),e_3=(0,0,1,0),e_4=(0,0,0,1)$
            $endgroup$
            – A. P
            Jan 2 at 17:07












          • $begingroup$
            You might be right about it, its just I don't use the same notation, it took a minute to figure the id meant the identity :p So multiplying $[id]^B_Se_j$ for each $e_j$ would yield a basis for $B$, right? After finding $[id]^B_S$ and its inverse I determine $[T]^B_B=begin{bmatrix}3&4&0&0\0&3&0&0\8&16&3&0\0&0&12&3end{bmatrix}$ To summarize to find the matrix of $T$, $M(T)=[T]^B_B$, I first compute $T$ from the assumptions, then find an invertible matrix $[id]^B_S$ wrt the basis B, finding its inverse, $[id]^S_B$ then I compute $M(T)=[id]^B_ST_S[id]^S_B$?
            $endgroup$
            – Jens Kramer
            Jan 2 at 17:24








          • 1




            $begingroup$
            👍Wolfram alpha gives me the same answer,
            $endgroup$
            – A. P
            Jan 2 at 18:11











          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%2f3057295%2fset-of-real-polynomials-organized-as-a-vectorspace-over-mathbbr%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Up till now all correct :)
          For exercise D) you'll need The differentiation Matrix $$D=
          begin{bmatrix}
          0 & 1 & 0 & 0 \
          0 & 0 & 2 & 0 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}$$

          $$T:=4D+3I=begin{bmatrix}
          3 & 4 & 0 & 0 \
          0 & 3 & 8 & 0 \
          0 & 0& 3 & 12 \
          0 & 0 & 0 & 3
          end{bmatrix} $$

          T is bijective because $det(T)=3^4neq 0$ If you have not learned about determinants jet you can simply use the Gauss-algorithm to Calculate $A^{-1}$.
          Then have to write T with as a Matrix with Respect to the Basis B. For this you will need the Transformation Matricies $[id]^B_S$ and $[id]^S_B$ (where S denotes the standart-Basis) then $[T]^B_B=[id]^B_S T [id]^S_B$



          hint:
          $[id]^B_S$ the transformation matrix from S to B is:
          $$[id]^B_S=begin{bmatrix}
          -2 & 1 & 0 & 0 \
          1 & 0 & 0 & 0 \
          0 & 0 & 1& 0 \
          0 & 0 & 0 & 1
          end{bmatrix} $$

          to understand this look at $[id]^B_S e_j $.



          Also note $[id]^B_S=([id]^S_B)^{-1}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think I'm doing something wrong when trying to calculate the $[id]^S_B$ matrix. I get a matrix with a zero in the diagonal such that it isn't invertible.
            $endgroup$
            – Jens Kramer
            Jan 2 at 15:54










          • $begingroup$
            I have misunderstood some things, but correct me if I'm wrong. I just calculated the matrix in the hint, but i thought that it was $[id]^S_B$ not $[id]^B_S$. When you write $e_j$, do you refer to the j'th orthonormal basisvector. I'm not really sure what that should clarify. Last I calculate the inverse of $[id]^B_S$ to be $begin{bmatrix}0 & 1 & 0 & 0\ 1 &2 &0&0\0&0&1&0\0&0&0&1end{bmatrix}$. My previous comment said that I didn't think this was invertible because one of the diagonals had a zero, which seems to be wrong. Is it right to say it isn't diagonizable then?
            $endgroup$
            – Jens Kramer
            Jan 2 at 16:58












          • $begingroup$
            the matrix seems right, it is the inverse of the matrix I've written (i'm not quite sure about my notation i might have done it the wrong way round) , $e_j$ is the j-basisvektor $e_1=(1,0,0,0),e_2=(0,1,0,0),e_3=(0,0,1,0),e_4=(0,0,0,1)$
            $endgroup$
            – A. P
            Jan 2 at 17:07












          • $begingroup$
            You might be right about it, its just I don't use the same notation, it took a minute to figure the id meant the identity :p So multiplying $[id]^B_Se_j$ for each $e_j$ would yield a basis for $B$, right? After finding $[id]^B_S$ and its inverse I determine $[T]^B_B=begin{bmatrix}3&4&0&0\0&3&0&0\8&16&3&0\0&0&12&3end{bmatrix}$ To summarize to find the matrix of $T$, $M(T)=[T]^B_B$, I first compute $T$ from the assumptions, then find an invertible matrix $[id]^B_S$ wrt the basis B, finding its inverse, $[id]^S_B$ then I compute $M(T)=[id]^B_ST_S[id]^S_B$?
            $endgroup$
            – Jens Kramer
            Jan 2 at 17:24








          • 1




            $begingroup$
            👍Wolfram alpha gives me the same answer,
            $endgroup$
            – A. P
            Jan 2 at 18:11
















          1












          $begingroup$

          Up till now all correct :)
          For exercise D) you'll need The differentiation Matrix $$D=
          begin{bmatrix}
          0 & 1 & 0 & 0 \
          0 & 0 & 2 & 0 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}$$

          $$T:=4D+3I=begin{bmatrix}
          3 & 4 & 0 & 0 \
          0 & 3 & 8 & 0 \
          0 & 0& 3 & 12 \
          0 & 0 & 0 & 3
          end{bmatrix} $$

          T is bijective because $det(T)=3^4neq 0$ If you have not learned about determinants jet you can simply use the Gauss-algorithm to Calculate $A^{-1}$.
          Then have to write T with as a Matrix with Respect to the Basis B. For this you will need the Transformation Matricies $[id]^B_S$ and $[id]^S_B$ (where S denotes the standart-Basis) then $[T]^B_B=[id]^B_S T [id]^S_B$



          hint:
          $[id]^B_S$ the transformation matrix from S to B is:
          $$[id]^B_S=begin{bmatrix}
          -2 & 1 & 0 & 0 \
          1 & 0 & 0 & 0 \
          0 & 0 & 1& 0 \
          0 & 0 & 0 & 1
          end{bmatrix} $$

          to understand this look at $[id]^B_S e_j $.



          Also note $[id]^B_S=([id]^S_B)^{-1}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think I'm doing something wrong when trying to calculate the $[id]^S_B$ matrix. I get a matrix with a zero in the diagonal such that it isn't invertible.
            $endgroup$
            – Jens Kramer
            Jan 2 at 15:54










          • $begingroup$
            I have misunderstood some things, but correct me if I'm wrong. I just calculated the matrix in the hint, but i thought that it was $[id]^S_B$ not $[id]^B_S$. When you write $e_j$, do you refer to the j'th orthonormal basisvector. I'm not really sure what that should clarify. Last I calculate the inverse of $[id]^B_S$ to be $begin{bmatrix}0 & 1 & 0 & 0\ 1 &2 &0&0\0&0&1&0\0&0&0&1end{bmatrix}$. My previous comment said that I didn't think this was invertible because one of the diagonals had a zero, which seems to be wrong. Is it right to say it isn't diagonizable then?
            $endgroup$
            – Jens Kramer
            Jan 2 at 16:58












          • $begingroup$
            the matrix seems right, it is the inverse of the matrix I've written (i'm not quite sure about my notation i might have done it the wrong way round) , $e_j$ is the j-basisvektor $e_1=(1,0,0,0),e_2=(0,1,0,0),e_3=(0,0,1,0),e_4=(0,0,0,1)$
            $endgroup$
            – A. P
            Jan 2 at 17:07












          • $begingroup$
            You might be right about it, its just I don't use the same notation, it took a minute to figure the id meant the identity :p So multiplying $[id]^B_Se_j$ for each $e_j$ would yield a basis for $B$, right? After finding $[id]^B_S$ and its inverse I determine $[T]^B_B=begin{bmatrix}3&4&0&0\0&3&0&0\8&16&3&0\0&0&12&3end{bmatrix}$ To summarize to find the matrix of $T$, $M(T)=[T]^B_B$, I first compute $T$ from the assumptions, then find an invertible matrix $[id]^B_S$ wrt the basis B, finding its inverse, $[id]^S_B$ then I compute $M(T)=[id]^B_ST_S[id]^S_B$?
            $endgroup$
            – Jens Kramer
            Jan 2 at 17:24








          • 1




            $begingroup$
            👍Wolfram alpha gives me the same answer,
            $endgroup$
            – A. P
            Jan 2 at 18:11














          1












          1








          1





          $begingroup$

          Up till now all correct :)
          For exercise D) you'll need The differentiation Matrix $$D=
          begin{bmatrix}
          0 & 1 & 0 & 0 \
          0 & 0 & 2 & 0 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}$$

          $$T:=4D+3I=begin{bmatrix}
          3 & 4 & 0 & 0 \
          0 & 3 & 8 & 0 \
          0 & 0& 3 & 12 \
          0 & 0 & 0 & 3
          end{bmatrix} $$

          T is bijective because $det(T)=3^4neq 0$ If you have not learned about determinants jet you can simply use the Gauss-algorithm to Calculate $A^{-1}$.
          Then have to write T with as a Matrix with Respect to the Basis B. For this you will need the Transformation Matricies $[id]^B_S$ and $[id]^S_B$ (where S denotes the standart-Basis) then $[T]^B_B=[id]^B_S T [id]^S_B$



          hint:
          $[id]^B_S$ the transformation matrix from S to B is:
          $$[id]^B_S=begin{bmatrix}
          -2 & 1 & 0 & 0 \
          1 & 0 & 0 & 0 \
          0 & 0 & 1& 0 \
          0 & 0 & 0 & 1
          end{bmatrix} $$

          to understand this look at $[id]^B_S e_j $.



          Also note $[id]^B_S=([id]^S_B)^{-1}$






          share|cite|improve this answer











          $endgroup$



          Up till now all correct :)
          For exercise D) you'll need The differentiation Matrix $$D=
          begin{bmatrix}
          0 & 1 & 0 & 0 \
          0 & 0 & 2 & 0 \
          0 & 0 & 0 & 3 \
          0 & 0 & 0 & 0
          end{bmatrix}$$

          $$T:=4D+3I=begin{bmatrix}
          3 & 4 & 0 & 0 \
          0 & 3 & 8 & 0 \
          0 & 0& 3 & 12 \
          0 & 0 & 0 & 3
          end{bmatrix} $$

          T is bijective because $det(T)=3^4neq 0$ If you have not learned about determinants jet you can simply use the Gauss-algorithm to Calculate $A^{-1}$.
          Then have to write T with as a Matrix with Respect to the Basis B. For this you will need the Transformation Matricies $[id]^B_S$ and $[id]^S_B$ (where S denotes the standart-Basis) then $[T]^B_B=[id]^B_S T [id]^S_B$



          hint:
          $[id]^B_S$ the transformation matrix from S to B is:
          $$[id]^B_S=begin{bmatrix}
          -2 & 1 & 0 & 0 \
          1 & 0 & 0 & 0 \
          0 & 0 & 1& 0 \
          0 & 0 & 0 & 1
          end{bmatrix} $$

          to understand this look at $[id]^B_S e_j $.



          Also note $[id]^B_S=([id]^S_B)^{-1}$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 17:56

























          answered Dec 31 '18 at 0:28









          A. PA. P

          1186




          1186












          • $begingroup$
            I think I'm doing something wrong when trying to calculate the $[id]^S_B$ matrix. I get a matrix with a zero in the diagonal such that it isn't invertible.
            $endgroup$
            – Jens Kramer
            Jan 2 at 15:54










          • $begingroup$
            I have misunderstood some things, but correct me if I'm wrong. I just calculated the matrix in the hint, but i thought that it was $[id]^S_B$ not $[id]^B_S$. When you write $e_j$, do you refer to the j'th orthonormal basisvector. I'm not really sure what that should clarify. Last I calculate the inverse of $[id]^B_S$ to be $begin{bmatrix}0 & 1 & 0 & 0\ 1 &2 &0&0\0&0&1&0\0&0&0&1end{bmatrix}$. My previous comment said that I didn't think this was invertible because one of the diagonals had a zero, which seems to be wrong. Is it right to say it isn't diagonizable then?
            $endgroup$
            – Jens Kramer
            Jan 2 at 16:58












          • $begingroup$
            the matrix seems right, it is the inverse of the matrix I've written (i'm not quite sure about my notation i might have done it the wrong way round) , $e_j$ is the j-basisvektor $e_1=(1,0,0,0),e_2=(0,1,0,0),e_3=(0,0,1,0),e_4=(0,0,0,1)$
            $endgroup$
            – A. P
            Jan 2 at 17:07












          • $begingroup$
            You might be right about it, its just I don't use the same notation, it took a minute to figure the id meant the identity :p So multiplying $[id]^B_Se_j$ for each $e_j$ would yield a basis for $B$, right? After finding $[id]^B_S$ and its inverse I determine $[T]^B_B=begin{bmatrix}3&4&0&0\0&3&0&0\8&16&3&0\0&0&12&3end{bmatrix}$ To summarize to find the matrix of $T$, $M(T)=[T]^B_B$, I first compute $T$ from the assumptions, then find an invertible matrix $[id]^B_S$ wrt the basis B, finding its inverse, $[id]^S_B$ then I compute $M(T)=[id]^B_ST_S[id]^S_B$?
            $endgroup$
            – Jens Kramer
            Jan 2 at 17:24








          • 1




            $begingroup$
            👍Wolfram alpha gives me the same answer,
            $endgroup$
            – A. P
            Jan 2 at 18:11


















          • $begingroup$
            I think I'm doing something wrong when trying to calculate the $[id]^S_B$ matrix. I get a matrix with a zero in the diagonal such that it isn't invertible.
            $endgroup$
            – Jens Kramer
            Jan 2 at 15:54










          • $begingroup$
            I have misunderstood some things, but correct me if I'm wrong. I just calculated the matrix in the hint, but i thought that it was $[id]^S_B$ not $[id]^B_S$. When you write $e_j$, do you refer to the j'th orthonormal basisvector. I'm not really sure what that should clarify. Last I calculate the inverse of $[id]^B_S$ to be $begin{bmatrix}0 & 1 & 0 & 0\ 1 &2 &0&0\0&0&1&0\0&0&0&1end{bmatrix}$. My previous comment said that I didn't think this was invertible because one of the diagonals had a zero, which seems to be wrong. Is it right to say it isn't diagonizable then?
            $endgroup$
            – Jens Kramer
            Jan 2 at 16:58












          • $begingroup$
            the matrix seems right, it is the inverse of the matrix I've written (i'm not quite sure about my notation i might have done it the wrong way round) , $e_j$ is the j-basisvektor $e_1=(1,0,0,0),e_2=(0,1,0,0),e_3=(0,0,1,0),e_4=(0,0,0,1)$
            $endgroup$
            – A. P
            Jan 2 at 17:07












          • $begingroup$
            You might be right about it, its just I don't use the same notation, it took a minute to figure the id meant the identity :p So multiplying $[id]^B_Se_j$ for each $e_j$ would yield a basis for $B$, right? After finding $[id]^B_S$ and its inverse I determine $[T]^B_B=begin{bmatrix}3&4&0&0\0&3&0&0\8&16&3&0\0&0&12&3end{bmatrix}$ To summarize to find the matrix of $T$, $M(T)=[T]^B_B$, I first compute $T$ from the assumptions, then find an invertible matrix $[id]^B_S$ wrt the basis B, finding its inverse, $[id]^S_B$ then I compute $M(T)=[id]^B_ST_S[id]^S_B$?
            $endgroup$
            – Jens Kramer
            Jan 2 at 17:24








          • 1




            $begingroup$
            👍Wolfram alpha gives me the same answer,
            $endgroup$
            – A. P
            Jan 2 at 18:11
















          $begingroup$
          I think I'm doing something wrong when trying to calculate the $[id]^S_B$ matrix. I get a matrix with a zero in the diagonal such that it isn't invertible.
          $endgroup$
          – Jens Kramer
          Jan 2 at 15:54




          $begingroup$
          I think I'm doing something wrong when trying to calculate the $[id]^S_B$ matrix. I get a matrix with a zero in the diagonal such that it isn't invertible.
          $endgroup$
          – Jens Kramer
          Jan 2 at 15:54












          $begingroup$
          I have misunderstood some things, but correct me if I'm wrong. I just calculated the matrix in the hint, but i thought that it was $[id]^S_B$ not $[id]^B_S$. When you write $e_j$, do you refer to the j'th orthonormal basisvector. I'm not really sure what that should clarify. Last I calculate the inverse of $[id]^B_S$ to be $begin{bmatrix}0 & 1 & 0 & 0\ 1 &2 &0&0\0&0&1&0\0&0&0&1end{bmatrix}$. My previous comment said that I didn't think this was invertible because one of the diagonals had a zero, which seems to be wrong. Is it right to say it isn't diagonizable then?
          $endgroup$
          – Jens Kramer
          Jan 2 at 16:58






          $begingroup$
          I have misunderstood some things, but correct me if I'm wrong. I just calculated the matrix in the hint, but i thought that it was $[id]^S_B$ not $[id]^B_S$. When you write $e_j$, do you refer to the j'th orthonormal basisvector. I'm not really sure what that should clarify. Last I calculate the inverse of $[id]^B_S$ to be $begin{bmatrix}0 & 1 & 0 & 0\ 1 &2 &0&0\0&0&1&0\0&0&0&1end{bmatrix}$. My previous comment said that I didn't think this was invertible because one of the diagonals had a zero, which seems to be wrong. Is it right to say it isn't diagonizable then?
          $endgroup$
          – Jens Kramer
          Jan 2 at 16:58














          $begingroup$
          the matrix seems right, it is the inverse of the matrix I've written (i'm not quite sure about my notation i might have done it the wrong way round) , $e_j$ is the j-basisvektor $e_1=(1,0,0,0),e_2=(0,1,0,0),e_3=(0,0,1,0),e_4=(0,0,0,1)$
          $endgroup$
          – A. P
          Jan 2 at 17:07






          $begingroup$
          the matrix seems right, it is the inverse of the matrix I've written (i'm not quite sure about my notation i might have done it the wrong way round) , $e_j$ is the j-basisvektor $e_1=(1,0,0,0),e_2=(0,1,0,0),e_3=(0,0,1,0),e_4=(0,0,0,1)$
          $endgroup$
          – A. P
          Jan 2 at 17:07














          $begingroup$
          You might be right about it, its just I don't use the same notation, it took a minute to figure the id meant the identity :p So multiplying $[id]^B_Se_j$ for each $e_j$ would yield a basis for $B$, right? After finding $[id]^B_S$ and its inverse I determine $[T]^B_B=begin{bmatrix}3&4&0&0\0&3&0&0\8&16&3&0\0&0&12&3end{bmatrix}$ To summarize to find the matrix of $T$, $M(T)=[T]^B_B$, I first compute $T$ from the assumptions, then find an invertible matrix $[id]^B_S$ wrt the basis B, finding its inverse, $[id]^S_B$ then I compute $M(T)=[id]^B_ST_S[id]^S_B$?
          $endgroup$
          – Jens Kramer
          Jan 2 at 17:24






          $begingroup$
          You might be right about it, its just I don't use the same notation, it took a minute to figure the id meant the identity :p So multiplying $[id]^B_Se_j$ for each $e_j$ would yield a basis for $B$, right? After finding $[id]^B_S$ and its inverse I determine $[T]^B_B=begin{bmatrix}3&4&0&0\0&3&0&0\8&16&3&0\0&0&12&3end{bmatrix}$ To summarize to find the matrix of $T$, $M(T)=[T]^B_B$, I first compute $T$ from the assumptions, then find an invertible matrix $[id]^B_S$ wrt the basis B, finding its inverse, $[id]^S_B$ then I compute $M(T)=[id]^B_ST_S[id]^S_B$?
          $endgroup$
          – Jens Kramer
          Jan 2 at 17:24






          1




          1




          $begingroup$
          👍Wolfram alpha gives me the same answer,
          $endgroup$
          – A. P
          Jan 2 at 18:11




          $begingroup$
          👍Wolfram alpha gives me the same answer,
          $endgroup$
          – A. P
          Jan 2 at 18:11


















          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%2f3057295%2fset-of-real-polynomials-organized-as-a-vectorspace-over-mathbbr%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

          Wiesbaden

          Marschland

          Dieringhausen