What is the Maximum Value of $x^T Ax$?











up vote
3
down vote

favorite
2












Let $A$ be the matrix $begin{bmatrix}3 &1 \ 1&2end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?




  1. $5$

  2. $frac{(5 + sqrt{5})}{2}$

  3. $3$

  4. $frac{(5 - sqrt{5})}{2}$



Eigenvalues of $A$ are $frac{(5 pm sqrt{5})}{2}$




My question is: what is $x^T Ax$ ? Can you explain little bit please?







linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question




















  • 1




    $x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
    – Kushal Bhuyan
    Nov 30 '15 at 12:14















up vote
3
down vote

favorite
2












Let $A$ be the matrix $begin{bmatrix}3 &1 \ 1&2end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?




  1. $5$

  2. $frac{(5 + sqrt{5})}{2}$

  3. $3$

  4. $frac{(5 - sqrt{5})}{2}$



Eigenvalues of $A$ are $frac{(5 pm sqrt{5})}{2}$




My question is: what is $x^T Ax$ ? Can you explain little bit please?







linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question




















  • 1




    $x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
    – Kushal Bhuyan
    Nov 30 '15 at 12:14













up vote
3
down vote

favorite
2









up vote
3
down vote

favorite
2






2





Let $A$ be the matrix $begin{bmatrix}3 &1 \ 1&2end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?




  1. $5$

  2. $frac{(5 + sqrt{5})}{2}$

  3. $3$

  4. $frac{(5 - sqrt{5})}{2}$



Eigenvalues of $A$ are $frac{(5 pm sqrt{5})}{2}$




My question is: what is $x^T Ax$ ? Can you explain little bit please?







linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question















Let $A$ be the matrix $begin{bmatrix}3 &1 \ 1&2end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?




  1. $5$

  2. $frac{(5 + sqrt{5})}{2}$

  3. $3$

  4. $frac{(5 - sqrt{5})}{2}$



Eigenvalues of $A$ are $frac{(5 pm sqrt{5})}{2}$




My question is: what is $x^T Ax$ ? Can you explain little bit please?







linear-algebra matrices eigenvalues-eigenvectors




linear-algebra matrices eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '15 at 12:56









H. R.

9,31093261




9,31093261










asked Nov 30 '15 at 12:12









Mithlesh Upadhyay

2,88882661




2,88882661








  • 1




    $x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
    – Kushal Bhuyan
    Nov 30 '15 at 12:14














  • 1




    $x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
    – Kushal Bhuyan
    Nov 30 '15 at 12:14








1




1




$x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
– Kushal Bhuyan
Nov 30 '15 at 12:14




$x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
– Kushal Bhuyan
Nov 30 '15 at 12:14










5 Answers
5






active

oldest

votes

















up vote
3
down vote



accepted










If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.



As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
$$R(x)=frac{x^TMx}{x^Tx}$$
which is called the Rayleigh quotient associated to $M$.






share|cite|improve this answer






























    up vote
    3
    down vote













    This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient






    share|cite|improve this answer





















    • Thanks for nice article/answer.
      – Mithlesh Upadhyay
      Nov 30 '15 at 12:49




















    up vote
    2
    down vote













    OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence



    $$Ax_{lambda}=lambda x_{lambda}$$



    and observe that



    $$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$



    so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.






    share|cite|improve this answer



















    • 1




      Thanks for beautiful explanation.
      – Mithlesh Upadhyay
      Nov 30 '15 at 12:46








    • 1




      @MithleshUpadhyay: You are welcome! :)
      – H. R.
      Nov 30 '15 at 13:50


















    up vote
    1
    down vote













    $$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$






    share|cite|improve this answer





















    • I need to find maximum value, from this function, rt?
      – Mithlesh Upadhyay
      Nov 30 '15 at 12:24












    • Using the (unit) eigenvectors, yes.
      – Nigel Overmars
      Nov 30 '15 at 12:25


















    up vote
    0
    down vote













    Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$



    For finding the eigen value:



    Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.



                                         $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$



                                         $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$



                                         $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$



                                         $Rightarrow(3-lambda)(2-lambda)-1=0$



                                          $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$



                                           $Rightarrow lambda^{2}-5lambda+5=0$



    Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$



     $lambda =frac{5pm sqrt{5}}{2}$



     So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$



    Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$



    Now find the Eigen Vectors:



               $AX=lambda X$



           $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$



           $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$



      $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$



               Put $lambda=frac{5+sqrt{5}}{2}$



      $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



      $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



      $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



    Perform some row operation



    $R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$



    $Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



    $Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



    $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



    Again do some operation $R_{2}rightarrow R_{2}+R_{1}$



    $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



    Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$



    clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]



    So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$



                                      $ -x_{1}=-(frac{1+sqrt{5}}{2})k$



                                       $ x_{1}=(frac{1+sqrt{5}}{2})k$



    Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$



                   $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$



    Now, find the unit vectors of $A:$



    $hat{u}=frac{vec{X}}{|vec{X}|}$



    $|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$



    $hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$



    $hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$



    here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$



    Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$



    $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$



    and similarly do for $lambda=frac{5-sqrt{5}}{2}$



     $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$



    Now, find the unit vectors of $A:$



    $hat{u}=frac{vec{X}}{|vec{X}|}$



    $|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$



    $hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$



    $hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$



    here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$



    Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$



    $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$



    So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$



    and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$






    share|cite|improve this answer





















      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',
      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%2f1553046%2fwhat-is-the-maximum-value-of-xt-ax%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote



      accepted










      If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.



      As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
      $$R(x)=frac{x^TMx}{x^Tx}$$
      which is called the Rayleigh quotient associated to $M$.






      share|cite|improve this answer



























        up vote
        3
        down vote



        accepted










        If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.



        As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
        $$R(x)=frac{x^TMx}{x^Tx}$$
        which is called the Rayleigh quotient associated to $M$.






        share|cite|improve this answer

























          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.



          As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
          $$R(x)=frac{x^TMx}{x^Tx}$$
          which is called the Rayleigh quotient associated to $M$.






          share|cite|improve this answer














          If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.



          As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
          $$R(x)=frac{x^TMx}{x^Tx}$$
          which is called the Rayleigh quotient associated to $M$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 '15 at 14:51

























          answered Nov 30 '15 at 12:41









          Surb

          37k94375




          37k94375






















              up vote
              3
              down vote













              This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient






              share|cite|improve this answer





















              • Thanks for nice article/answer.
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:49

















              up vote
              3
              down vote













              This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient






              share|cite|improve this answer





















              • Thanks for nice article/answer.
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:49















              up vote
              3
              down vote










              up vote
              3
              down vote









              This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient






              share|cite|improve this answer












              This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 30 '15 at 12:33









              Ant

              17.3k22873




              17.3k22873












              • Thanks for nice article/answer.
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:49




















              • Thanks for nice article/answer.
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:49


















              Thanks for nice article/answer.
              – Mithlesh Upadhyay
              Nov 30 '15 at 12:49






              Thanks for nice article/answer.
              – Mithlesh Upadhyay
              Nov 30 '15 at 12:49












              up vote
              2
              down vote













              OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence



              $$Ax_{lambda}=lambda x_{lambda}$$



              and observe that



              $$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$



              so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.






              share|cite|improve this answer



















              • 1




                Thanks for beautiful explanation.
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:46








              • 1




                @MithleshUpadhyay: You are welcome! :)
                – H. R.
                Nov 30 '15 at 13:50















              up vote
              2
              down vote













              OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence



              $$Ax_{lambda}=lambda x_{lambda}$$



              and observe that



              $$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$



              so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.






              share|cite|improve this answer



















              • 1




                Thanks for beautiful explanation.
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:46








              • 1




                @MithleshUpadhyay: You are welcome! :)
                – H. R.
                Nov 30 '15 at 13:50













              up vote
              2
              down vote










              up vote
              2
              down vote









              OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence



              $$Ax_{lambda}=lambda x_{lambda}$$



              and observe that



              $$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$



              so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.






              share|cite|improve this answer














              OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence



              $$Ax_{lambda}=lambda x_{lambda}$$



              and observe that



              $$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$



              so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 30 '15 at 13:50

























              answered Nov 30 '15 at 12:41









              H. R.

              9,31093261




              9,31093261








              • 1




                Thanks for beautiful explanation.
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:46








              • 1




                @MithleshUpadhyay: You are welcome! :)
                – H. R.
                Nov 30 '15 at 13:50














              • 1




                Thanks for beautiful explanation.
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:46








              • 1




                @MithleshUpadhyay: You are welcome! :)
                – H. R.
                Nov 30 '15 at 13:50








              1




              1




              Thanks for beautiful explanation.
              – Mithlesh Upadhyay
              Nov 30 '15 at 12:46






              Thanks for beautiful explanation.
              – Mithlesh Upadhyay
              Nov 30 '15 at 12:46






              1




              1




              @MithleshUpadhyay: You are welcome! :)
              – H. R.
              Nov 30 '15 at 13:50




              @MithleshUpadhyay: You are welcome! :)
              – H. R.
              Nov 30 '15 at 13:50










              up vote
              1
              down vote













              $$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$






              share|cite|improve this answer





















              • I need to find maximum value, from this function, rt?
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:24












              • Using the (unit) eigenvectors, yes.
                – Nigel Overmars
                Nov 30 '15 at 12:25















              up vote
              1
              down vote













              $$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$






              share|cite|improve this answer





















              • I need to find maximum value, from this function, rt?
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:24












              • Using the (unit) eigenvectors, yes.
                – Nigel Overmars
                Nov 30 '15 at 12:25













              up vote
              1
              down vote










              up vote
              1
              down vote









              $$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$






              share|cite|improve this answer












              $$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 30 '15 at 12:20









              Nigel Overmars

              2,59911225




              2,59911225












              • I need to find maximum value, from this function, rt?
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:24












              • Using the (unit) eigenvectors, yes.
                – Nigel Overmars
                Nov 30 '15 at 12:25


















              • I need to find maximum value, from this function, rt?
                – Mithlesh Upadhyay
                Nov 30 '15 at 12:24












              • Using the (unit) eigenvectors, yes.
                – Nigel Overmars
                Nov 30 '15 at 12:25
















              I need to find maximum value, from this function, rt?
              – Mithlesh Upadhyay
              Nov 30 '15 at 12:24






              I need to find maximum value, from this function, rt?
              – Mithlesh Upadhyay
              Nov 30 '15 at 12:24














              Using the (unit) eigenvectors, yes.
              – Nigel Overmars
              Nov 30 '15 at 12:25




              Using the (unit) eigenvectors, yes.
              – Nigel Overmars
              Nov 30 '15 at 12:25










              up vote
              0
              down vote













              Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$



              For finding the eigen value:



              Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.



                                                   $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$



                                                   $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$



                                                   $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$



                                                   $Rightarrow(3-lambda)(2-lambda)-1=0$



                                                    $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$



                                                     $Rightarrow lambda^{2}-5lambda+5=0$



              Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$



               $lambda =frac{5pm sqrt{5}}{2}$



               So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$



              Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$



              Now find the Eigen Vectors:



                         $AX=lambda X$



                     $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$



                     $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$



                $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$



                         Put $lambda=frac{5+sqrt{5}}{2}$



                $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



              Perform some row operation



              $R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$



              $Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



              $Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



              $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



              Again do some operation $R_{2}rightarrow R_{2}+R_{1}$



              $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



              Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$



              clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]



              So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$



                                                $ -x_{1}=-(frac{1+sqrt{5}}{2})k$



                                                 $ x_{1}=(frac{1+sqrt{5}}{2})k$



              Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$



                             $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$



              Now, find the unit vectors of $A:$



              $hat{u}=frac{vec{X}}{|vec{X}|}$



              $|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$



              $hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$



              $hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$



              here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$



              Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$



              $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$



              and similarly do for $lambda=frac{5-sqrt{5}}{2}$



               $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$



              Now, find the unit vectors of $A:$



              $hat{u}=frac{vec{X}}{|vec{X}|}$



              $|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$



              $hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$



              $hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$



              here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$



              Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$



              $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$



              So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$



              and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$






              share|cite|improve this answer

























                up vote
                0
                down vote













                Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$



                For finding the eigen value:



                Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.



                                                     $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$



                                                     $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$



                                                     $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$



                                                     $Rightarrow(3-lambda)(2-lambda)-1=0$



                                                      $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$



                                                       $Rightarrow lambda^{2}-5lambda+5=0$



                Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$



                 $lambda =frac{5pm sqrt{5}}{2}$



                 So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$



                Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$



                Now find the Eigen Vectors:



                           $AX=lambda X$



                       $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$



                       $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$



                  $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$



                           Put $lambda=frac{5+sqrt{5}}{2}$



                  $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                Perform some row operation



                $R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$



                $Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                $Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                Again do some operation $R_{2}rightarrow R_{2}+R_{1}$



                $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$



                clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]



                So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$



                                                  $ -x_{1}=-(frac{1+sqrt{5}}{2})k$



                                                   $ x_{1}=(frac{1+sqrt{5}}{2})k$



                Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$



                               $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$



                Now, find the unit vectors of $A:$



                $hat{u}=frac{vec{X}}{|vec{X}|}$



                $|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$



                $hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$



                $hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$



                here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$



                Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$



                $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$



                and similarly do for $lambda=frac{5-sqrt{5}}{2}$



                 $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$



                Now, find the unit vectors of $A:$



                $hat{u}=frac{vec{X}}{|vec{X}|}$



                $|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$



                $hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$



                $hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$



                here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$



                Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$



                $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$



                So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$



                and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$



                  For finding the eigen value:



                  Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.



                                                       $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$



                                                       $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$



                                                       $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$



                                                       $Rightarrow(3-lambda)(2-lambda)-1=0$



                                                        $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$



                                                         $Rightarrow lambda^{2}-5lambda+5=0$



                  Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$



                   $lambda =frac{5pm sqrt{5}}{2}$



                   So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$



                  Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$



                  Now find the Eigen Vectors:



                             $AX=lambda X$



                         $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$



                         $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$



                    $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$



                             Put $lambda=frac{5+sqrt{5}}{2}$



                    $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                    $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                    $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  Perform some row operation



                  $R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$



                  $Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  $Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  Again do some operation $R_{2}rightarrow R_{2}+R_{1}$



                  $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$



                  clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]



                  So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$



                                                    $ -x_{1}=-(frac{1+sqrt{5}}{2})k$



                                                     $ x_{1}=(frac{1+sqrt{5}}{2})k$



                  Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$



                                 $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$



                  Now, find the unit vectors of $A:$



                  $hat{u}=frac{vec{X}}{|vec{X}|}$



                  $|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$



                  $hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$



                  $hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$



                  here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$



                  Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$



                  $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$



                  and similarly do for $lambda=frac{5-sqrt{5}}{2}$



                   $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$



                  Now, find the unit vectors of $A:$



                  $hat{u}=frac{vec{X}}{|vec{X}|}$



                  $|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$



                  $hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$



                  $hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$



                  here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$



                  Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$



                  $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$



                  So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$



                  and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$






                  share|cite|improve this answer












                  Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$



                  For finding the eigen value:



                  Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.



                                                       $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$



                                                       $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$



                                                       $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$



                                                       $Rightarrow(3-lambda)(2-lambda)-1=0$



                                                        $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$



                                                         $Rightarrow lambda^{2}-5lambda+5=0$



                  Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$



                   $lambda =frac{5pm sqrt{5}}{2}$



                   So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$



                  Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$



                  Now find the Eigen Vectors:



                             $AX=lambda X$



                         $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$



                         $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$



                    $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$



                             Put $lambda=frac{5+sqrt{5}}{2}$



                    $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                    $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                    $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  Perform some row operation



                  $R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$



                  $Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  $Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  Again do some operation $R_{2}rightarrow R_{2}+R_{1}$



                  $Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$



                  Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$



                  clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]



                  So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$



                                                    $ -x_{1}=-(frac{1+sqrt{5}}{2})k$



                                                     $ x_{1}=(frac{1+sqrt{5}}{2})k$



                  Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$



                                 $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$



                  Now, find the unit vectors of $A:$



                  $hat{u}=frac{vec{X}}{|vec{X}|}$



                  $|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$



                  $hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$



                  $hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$



                  here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$



                  Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$



                  $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$



                  and similarly do for $lambda=frac{5-sqrt{5}}{2}$



                   $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$



                  Now, find the unit vectors of $A:$



                  $hat{u}=frac{vec{X}}{|vec{X}|}$



                  $|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$



                  $hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$



                  $hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$



                  here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$



                  Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$



                  $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$



                  So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$



                  and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Lakshman Patel

                  11




                  11






























                       

                      draft saved


                      draft discarded



















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1553046%2fwhat-is-the-maximum-value-of-xt-ax%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

                      Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                      Image
                      28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                      Read more

                      Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                      Image
                      up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                      Read more

                      A Topological Invariant for $pi_3(U(n))$

                      Image
                      3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                      Read more