Determine the $|A|_2$ vector-induced norm












2












$begingroup$


This seems to be a very easy exercise, but as I never had previous contact with vector induced norm of a matrix, I am getting stuck...



Given the matrix:
$$A=
begin{bmatrix}
2 & 0\
0 & 1\
end{bmatrix}
$$



I am supposed to compute its vector-induced $|A|_2$ norm, defined as:



$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}$$



My attempt so far was:



Define a vector $v=(v_1,v_2)$ so that I end up with:



$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}=sup_{vneq 0}sqrt{frac{4v_{1}^{2}+v_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$$



It seems reasonable to choose a vector $v$ such that $|v|_2=1$, then this would result in a problem where I want to find:



$$|A|_2=sup_{|v|_2=1}sqrt{4v_{1}^{2}+v_{2}^{2}}$$



Which kind of seems like an optimization problem where the constraint is:



$$sqrt{v_{1}^{2}+v_{2}^{2}}=1$$



And this is where I get stuck!



Is this a correct way to solve this? If yes, could you help me finish this line of thought? Would Lagrange Multiplier do the job?



I really would like to solve it "the hard way" before using other properties/definitions such as:



$$|A|_2=sigma_{max}(A)$$



Thanks in advance!










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    This seems to be a very easy exercise, but as I never had previous contact with vector induced norm of a matrix, I am getting stuck...



    Given the matrix:
    $$A=
    begin{bmatrix}
    2 & 0\
    0 & 1\
    end{bmatrix}
    $$



    I am supposed to compute its vector-induced $|A|_2$ norm, defined as:



    $$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}$$



    My attempt so far was:



    Define a vector $v=(v_1,v_2)$ so that I end up with:



    $$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}=sup_{vneq 0}sqrt{frac{4v_{1}^{2}+v_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$$



    It seems reasonable to choose a vector $v$ such that $|v|_2=1$, then this would result in a problem where I want to find:



    $$|A|_2=sup_{|v|_2=1}sqrt{4v_{1}^{2}+v_{2}^{2}}$$



    Which kind of seems like an optimization problem where the constraint is:



    $$sqrt{v_{1}^{2}+v_{2}^{2}}=1$$



    And this is where I get stuck!



    Is this a correct way to solve this? If yes, could you help me finish this line of thought? Would Lagrange Multiplier do the job?



    I really would like to solve it "the hard way" before using other properties/definitions such as:



    $$|A|_2=sigma_{max}(A)$$



    Thanks in advance!










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      This seems to be a very easy exercise, but as I never had previous contact with vector induced norm of a matrix, I am getting stuck...



      Given the matrix:
      $$A=
      begin{bmatrix}
      2 & 0\
      0 & 1\
      end{bmatrix}
      $$



      I am supposed to compute its vector-induced $|A|_2$ norm, defined as:



      $$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}$$



      My attempt so far was:



      Define a vector $v=(v_1,v_2)$ so that I end up with:



      $$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}=sup_{vneq 0}sqrt{frac{4v_{1}^{2}+v_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$$



      It seems reasonable to choose a vector $v$ such that $|v|_2=1$, then this would result in a problem where I want to find:



      $$|A|_2=sup_{|v|_2=1}sqrt{4v_{1}^{2}+v_{2}^{2}}$$



      Which kind of seems like an optimization problem where the constraint is:



      $$sqrt{v_{1}^{2}+v_{2}^{2}}=1$$



      And this is where I get stuck!



      Is this a correct way to solve this? If yes, could you help me finish this line of thought? Would Lagrange Multiplier do the job?



      I really would like to solve it "the hard way" before using other properties/definitions such as:



      $$|A|_2=sigma_{max}(A)$$



      Thanks in advance!










      share|cite|improve this question











      $endgroup$




      This seems to be a very easy exercise, but as I never had previous contact with vector induced norm of a matrix, I am getting stuck...



      Given the matrix:
      $$A=
      begin{bmatrix}
      2 & 0\
      0 & 1\
      end{bmatrix}
      $$



      I am supposed to compute its vector-induced $|A|_2$ norm, defined as:



      $$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}$$



      My attempt so far was:



      Define a vector $v=(v_1,v_2)$ so that I end up with:



      $$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}=sup_{vneq 0}sqrt{frac{4v_{1}^{2}+v_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$$



      It seems reasonable to choose a vector $v$ such that $|v|_2=1$, then this would result in a problem where I want to find:



      $$|A|_2=sup_{|v|_2=1}sqrt{4v_{1}^{2}+v_{2}^{2}}$$



      Which kind of seems like an optimization problem where the constraint is:



      $$sqrt{v_{1}^{2}+v_{2}^{2}}=1$$



      And this is where I get stuck!



      Is this a correct way to solve this? If yes, could you help me finish this line of thought? Would Lagrange Multiplier do the job?



      I really would like to solve it "the hard way" before using other properties/definitions such as:



      $$|A|_2=sigma_{max}(A)$$



      Thanks in advance!







      linear-algebra optimization vectors norm






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 2:02







      bertozzijr

















      asked Jan 8 at 1:52









      bertozzijrbertozzijr

      6331720




      6331720






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          Alternative:



          You want to solve $$max 4v_1^2 + v_2^2$$



          subject to $$v_1^2+v_2^2=1$$



          Let $y_i = v_i^2$, we have



          $$max 4y_1+y_2$$



          subject to $$y_1+y_2=1$$
          $$y_i ge 0 ,forall i in {1,2}$$



          which is just a linear programming problem which has optimal solution at the corners of the polyhedral.



          We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.



          Taking square root gives us $2$ as the optimal solution






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
            $$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
            Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.






              share|cite|improve this answer









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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                Alternative:



                You want to solve $$max 4v_1^2 + v_2^2$$



                subject to $$v_1^2+v_2^2=1$$



                Let $y_i = v_i^2$, we have



                $$max 4y_1+y_2$$



                subject to $$y_1+y_2=1$$
                $$y_i ge 0 ,forall i in {1,2}$$



                which is just a linear programming problem which has optimal solution at the corners of the polyhedral.



                We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.



                Taking square root gives us $2$ as the optimal solution






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Alternative:



                  You want to solve $$max 4v_1^2 + v_2^2$$



                  subject to $$v_1^2+v_2^2=1$$



                  Let $y_i = v_i^2$, we have



                  $$max 4y_1+y_2$$



                  subject to $$y_1+y_2=1$$
                  $$y_i ge 0 ,forall i in {1,2}$$



                  which is just a linear programming problem which has optimal solution at the corners of the polyhedral.



                  We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.



                  Taking square root gives us $2$ as the optimal solution






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Alternative:



                    You want to solve $$max 4v_1^2 + v_2^2$$



                    subject to $$v_1^2+v_2^2=1$$



                    Let $y_i = v_i^2$, we have



                    $$max 4y_1+y_2$$



                    subject to $$y_1+y_2=1$$
                    $$y_i ge 0 ,forall i in {1,2}$$



                    which is just a linear programming problem which has optimal solution at the corners of the polyhedral.



                    We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.



                    Taking square root gives us $2$ as the optimal solution






                    share|cite|improve this answer









                    $endgroup$



                    Alternative:



                    You want to solve $$max 4v_1^2 + v_2^2$$



                    subject to $$v_1^2+v_2^2=1$$



                    Let $y_i = v_i^2$, we have



                    $$max 4y_1+y_2$$



                    subject to $$y_1+y_2=1$$
                    $$y_i ge 0 ,forall i in {1,2}$$



                    which is just a linear programming problem which has optimal solution at the corners of the polyhedral.



                    We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.



                    Taking square root gives us $2$ as the optimal solution







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 8 at 2:12









                    Siong Thye GohSiong Thye Goh

                    104k1468120




                    104k1468120























                        1












                        $begingroup$

                        Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
                        $$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
                        Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
                          $$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
                          Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
                            $$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
                            Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….






                            share|cite|improve this answer









                            $endgroup$



                            Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
                            $$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
                            Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 8 at 2:04









                            zipirovichzipirovich

                            11.4k11731




                            11.4k11731























                                0












                                $begingroup$

                                Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 8 at 2:15









                                    Mustafa SaidMustafa Said

                                    3,0611913




                                    3,0611913






























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