Operator norm $||A||=sup_{x neq 0} frac{||A(x)||}{||x||}$












0












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My teacher defined the operator norm as $||A||=sup_{x neq 0} frac{||A(x)||}{||x||}$. I'm not really sure if I understand this definition, so, can anyone explain it to me, for example, let $A$ be $begin{pmatrix}1&3\ 5&7end{pmatrix}$. Or if it has to be with functions, let $A$ be $begin{pmatrix}x&3x\ 5x&7xend{pmatrix}$.



Thank you.






matrices eigenvalues-eigenvectors norm







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  • 2




    $begingroup$
    $A(x)$ would be a vector, $begin{bmatrix} x_1 + 3x_2 \ 5x_1 + 7x_2 end{bmatrix}$ in your example, so $frac{||A(x)||}{||x||} = frac{sqrt{(x_1 + 3x_2)^2 + (5x_1 + 7x_2)^2}{sqrt{x_1^2 + x_2^2}}$. Taking the supremum of this gives the operator norm.
    $endgroup$
    – Anthony Ter
    Jan 22 at 0:27


















0












$begingroup$


My teacher defined the operator norm as $||A||=sup_{x neq 0} frac{||A(x)||}{||x||}$. I'm not really sure if I understand this definition, so, can anyone explain it to me, for example, let $A$ be $begin{pmatrix}1&3\ 5&7end{pmatrix}$. Or if it has to be with functions, let $A$ be $begin{pmatrix}x&3x\ 5x&7xend{pmatrix}$.



Thank you.






matrices eigenvalues-eigenvectors norm







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $A(x)$ would be a vector, $begin{bmatrix} x_1 + 3x_2 \ 5x_1 + 7x_2 end{bmatrix}$ in your example, so $frac{||A(x)||}{||x||} = frac{sqrt{(x_1 + 3x_2)^2 + (5x_1 + 7x_2)^2}{sqrt{x_1^2 + x_2^2}}$. Taking the supremum of this gives the operator norm.
    $endgroup$
    – Anthony Ter
    Jan 22 at 0:27
















0












0








0





$begingroup$


My teacher defined the operator norm as $||A||=sup_{x neq 0} frac{||A(x)||}{||x||}$. I'm not really sure if I understand this definition, so, can anyone explain it to me, for example, let $A$ be $begin{pmatrix}1&3\ 5&7end{pmatrix}$. Or if it has to be with functions, let $A$ be $begin{pmatrix}x&3x\ 5x&7xend{pmatrix}$.



Thank you.






matrices eigenvalues-eigenvectors norm







share|cite|improve this question









$endgroup$




My teacher defined the operator norm as $||A||=sup_{x neq 0} frac{||A(x)||}{||x||}$. I'm not really sure if I understand this definition, so, can anyone explain it to me, for example, let $A$ be $begin{pmatrix}1&3\ 5&7end{pmatrix}$. Or if it has to be with functions, let $A$ be $begin{pmatrix}x&3x\ 5x&7xend{pmatrix}$.



Thank you.






matrices eigenvalues-eigenvectors norm




matrices eigenvalues-eigenvectors norm






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asked Jan 22 at 0:18









user25568user25568

1138




1138








  • 2




    $begingroup$
    $A(x)$ would be a vector, $begin{bmatrix} x_1 + 3x_2 \ 5x_1 + 7x_2 end{bmatrix}$ in your example, so $frac{||A(x)||}{||x||} = frac{sqrt{(x_1 + 3x_2)^2 + (5x_1 + 7x_2)^2}{sqrt{x_1^2 + x_2^2}}$. Taking the supremum of this gives the operator norm.
    $endgroup$
    – Anthony Ter
    Jan 22 at 0:27
















  • 2




    $begingroup$
    $A(x)$ would be a vector, $begin{bmatrix} x_1 + 3x_2 \ 5x_1 + 7x_2 end{bmatrix}$ in your example, so $frac{||A(x)||}{||x||} = frac{sqrt{(x_1 + 3x_2)^2 + (5x_1 + 7x_2)^2}{sqrt{x_1^2 + x_2^2}}$. Taking the supremum of this gives the operator norm.
    $endgroup$
    – Anthony Ter
    Jan 22 at 0:27










2




2




$begingroup$
$A(x)$ would be a vector, $begin{bmatrix} x_1 + 3x_2 \ 5x_1 + 7x_2 end{bmatrix}$ in your example, so $frac{||A(x)||}{||x||} = frac{sqrt{(x_1 + 3x_2)^2 + (5x_1 + 7x_2)^2}{sqrt{x_1^2 + x_2^2}}$. Taking the supremum of this gives the operator norm.
$endgroup$
– Anthony Ter
Jan 22 at 0:27






$begingroup$
$A(x)$ would be a vector, $begin{bmatrix} x_1 + 3x_2 \ 5x_1 + 7x_2 end{bmatrix}$ in your example, so $frac{||A(x)||}{||x||} = frac{sqrt{(x_1 + 3x_2)^2 + (5x_1 + 7x_2)^2}{sqrt{x_1^2 + x_2^2}}$. Taking the supremum of this gives the operator norm.
$endgroup$
– Anthony Ter
Jan 22 at 0:27












2 Answers
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oldest

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$begingroup$

Using the example provided, the operator norm of $A$ is the quantity,
$$sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} 1 & 3 \ 5 & 7end{pmatrix} begin{pmatrix} x \ y end{pmatrix}right|}{left|begin{pmatrix} x \ y end{pmatrix}right|} = sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} x + 3y \ 5x + 7y end{pmatrix}right|}{left|begin{pmatrix} x \ yend{pmatrix}right|} = sup_{(x, y) neq (0, 0)} sqrt{frac{(x + 3y)^2 + (5x + 7y)^2 }{x^2 + y^2}}.$$






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$endgroup$





















    1












    $begingroup$

    Let's look at an example: take
    $$
    A = pmatrix{3&1\1&3}
    $$
    we have
    $$
    A(x) = pmatrix{3&1\1&3}pmatrix{x_1\x_2} = pmatrix{3x_1+x_2\x_1+3x_2}
    $$
    So, we have $|Ax| = sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}$. Our goal is to compute
    $$
    sup_{xneq 0} frac{|Ax|}{|x|}
    = sup_{(x_1,x_2) neq (0,0)}frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}}
    $$
    For this operator, we can make things easier to compute with the substitution $u_1 = x_1 + x_2$ and $u_2 = x_1 - x_2$. With this we can write
    $$
    frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} = frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
    $$
    So that
    $$
    sup_{(x_1,x_2)neq (0,0)} frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} =
    sup_{(u_1,u_2) neq (0,0)}frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
    $$
    so that in this case, $|A| = 5$.






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      2 Answers
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      2 Answers
      2






      active

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      active

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      1












      $begingroup$

      Using the example provided, the operator norm of $A$ is the quantity,
      $$sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} 1 & 3 \ 5 & 7end{pmatrix} begin{pmatrix} x \ y end{pmatrix}right|}{left|begin{pmatrix} x \ y end{pmatrix}right|} = sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} x + 3y \ 5x + 7y end{pmatrix}right|}{left|begin{pmatrix} x \ yend{pmatrix}right|} = sup_{(x, y) neq (0, 0)} sqrt{frac{(x + 3y)^2 + (5x + 7y)^2 }{x^2 + y^2}}.$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Using the example provided, the operator norm of $A$ is the quantity,
        $$sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} 1 & 3 \ 5 & 7end{pmatrix} begin{pmatrix} x \ y end{pmatrix}right|}{left|begin{pmatrix} x \ y end{pmatrix}right|} = sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} x + 3y \ 5x + 7y end{pmatrix}right|}{left|begin{pmatrix} x \ yend{pmatrix}right|} = sup_{(x, y) neq (0, 0)} sqrt{frac{(x + 3y)^2 + (5x + 7y)^2 }{x^2 + y^2}}.$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Using the example provided, the operator norm of $A$ is the quantity,
          $$sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} 1 & 3 \ 5 & 7end{pmatrix} begin{pmatrix} x \ y end{pmatrix}right|}{left|begin{pmatrix} x \ y end{pmatrix}right|} = sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} x + 3y \ 5x + 7y end{pmatrix}right|}{left|begin{pmatrix} x \ yend{pmatrix}right|} = sup_{(x, y) neq (0, 0)} sqrt{frac{(x + 3y)^2 + (5x + 7y)^2 }{x^2 + y^2}}.$$






          share|cite|improve this answer









          $endgroup$



          Using the example provided, the operator norm of $A$ is the quantity,
          $$sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} 1 & 3 \ 5 & 7end{pmatrix} begin{pmatrix} x \ y end{pmatrix}right|}{left|begin{pmatrix} x \ y end{pmatrix}right|} = sup_{(x, y) neq (0, 0)} frac{left|begin{pmatrix} x + 3y \ 5x + 7y end{pmatrix}right|}{left|begin{pmatrix} x \ yend{pmatrix}right|} = sup_{(x, y) neq (0, 0)} sqrt{frac{(x + 3y)^2 + (5x + 7y)^2 }{x^2 + y^2}}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 22 at 0:30









          Theo BenditTheo Bendit

          19.3k12353




          19.3k12353























              1












              $begingroup$

              Let's look at an example: take
              $$
              A = pmatrix{3&1\1&3}
              $$
              we have
              $$
              A(x) = pmatrix{3&1\1&3}pmatrix{x_1\x_2} = pmatrix{3x_1+x_2\x_1+3x_2}
              $$
              So, we have $|Ax| = sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}$. Our goal is to compute
              $$
              sup_{xneq 0} frac{|Ax|}{|x|}
              = sup_{(x_1,x_2) neq (0,0)}frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}}
              $$
              For this operator, we can make things easier to compute with the substitution $u_1 = x_1 + x_2$ and $u_2 = x_1 - x_2$. With this we can write
              $$
              frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} = frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
              $$
              So that
              $$
              sup_{(x_1,x_2)neq (0,0)} frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} =
              sup_{(u_1,u_2) neq (0,0)}frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
              $$
              so that in this case, $|A| = 5$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let's look at an example: take
                $$
                A = pmatrix{3&1\1&3}
                $$
                we have
                $$
                A(x) = pmatrix{3&1\1&3}pmatrix{x_1\x_2} = pmatrix{3x_1+x_2\x_1+3x_2}
                $$
                So, we have $|Ax| = sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}$. Our goal is to compute
                $$
                sup_{xneq 0} frac{|Ax|}{|x|}
                = sup_{(x_1,x_2) neq (0,0)}frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}}
                $$
                For this operator, we can make things easier to compute with the substitution $u_1 = x_1 + x_2$ and $u_2 = x_1 - x_2$. With this we can write
                $$
                frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} = frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
                $$
                So that
                $$
                sup_{(x_1,x_2)neq (0,0)} frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} =
                sup_{(u_1,u_2) neq (0,0)}frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
                $$
                so that in this case, $|A| = 5$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let's look at an example: take
                  $$
                  A = pmatrix{3&1\1&3}
                  $$
                  we have
                  $$
                  A(x) = pmatrix{3&1\1&3}pmatrix{x_1\x_2} = pmatrix{3x_1+x_2\x_1+3x_2}
                  $$
                  So, we have $|Ax| = sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}$. Our goal is to compute
                  $$
                  sup_{xneq 0} frac{|Ax|}{|x|}
                  = sup_{(x_1,x_2) neq (0,0)}frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}}
                  $$
                  For this operator, we can make things easier to compute with the substitution $u_1 = x_1 + x_2$ and $u_2 = x_1 - x_2$. With this we can write
                  $$
                  frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} = frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
                  $$
                  So that
                  $$
                  sup_{(x_1,x_2)neq (0,0)} frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} =
                  sup_{(u_1,u_2) neq (0,0)}frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
                  $$
                  so that in this case, $|A| = 5$.






                  share|cite|improve this answer









                  $endgroup$



                  Let's look at an example: take
                  $$
                  A = pmatrix{3&1\1&3}
                  $$
                  we have
                  $$
                  A(x) = pmatrix{3&1\1&3}pmatrix{x_1\x_2} = pmatrix{3x_1+x_2\x_1+3x_2}
                  $$
                  So, we have $|Ax| = sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}$. Our goal is to compute
                  $$
                  sup_{xneq 0} frac{|Ax|}{|x|}
                  = sup_{(x_1,x_2) neq (0,0)}frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}}
                  $$
                  For this operator, we can make things easier to compute with the substitution $u_1 = x_1 + x_2$ and $u_2 = x_1 - x_2$. With this we can write
                  $$
                  frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} = frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
                  $$
                  So that
                  $$
                  sup_{(x_1,x_2)neq (0,0)} frac{sqrt{(3x_1+x_2)^2 + (x_1+3x_2)^2}}{sqrt{x_1^2 + x_2^2}} =
                  sup_{(u_1,u_2) neq (0,0)}frac{sqrt{25u_1^2 + u_2^2}}{sqrt{u_1^2+u_2^2}}
                  $$
                  so that in this case, $|A| = 5$.







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Jan 22 at 0:42









                  OmnomnomnomOmnomnomnom

                  128k791186




                  128k791186






























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