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











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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







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    $x=(x_1,x_2)^T$ 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

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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













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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






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






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






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      5 Answers
      5






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      5 Answers
      5






      active

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      active

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      active

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







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                  answered yesterday









                  Lakshman Patel

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