Is there a special relationship between a norm on a vector space V, and the operator norm $ mathcal{L}(V,...












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$begingroup$


Let $T$ be a linear operator in $mathcal{L}(V)$. An operator norm is denoted as $||T||$, where it is the smallest $M$, such that $||T(v)||$ $le$ $M||v||$ for any $v in V$.



A norm on the vector space $V$ is simply the square root of an inner-product.



Do these two types of norms have some relationship with each-other, and if so, what is it?






linear-algebra vector-spaces norm normed-spaces







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $T$ be a linear operator in $mathcal{L}(V)$. An operator norm is denoted as $||T||$, where it is the smallest $M$, such that $||T(v)||$ $le$ $M||v||$ for any $v in V$.



    A norm on the vector space $V$ is simply the square root of an inner-product.



    Do these two types of norms have some relationship with each-other, and if so, what is it?






    linear-algebra vector-spaces norm normed-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $T$ be a linear operator in $mathcal{L}(V)$. An operator norm is denoted as $||T||$, where it is the smallest $M$, such that $||T(v)||$ $le$ $M||v||$ for any $v in V$.



      A norm on the vector space $V$ is simply the square root of an inner-product.



      Do these two types of norms have some relationship with each-other, and if so, what is it?






      linear-algebra vector-spaces norm normed-spaces







      share|cite|improve this question









      $endgroup$




      Let $T$ be a linear operator in $mathcal{L}(V)$. An operator norm is denoted as $||T||$, where it is the smallest $M$, such that $||T(v)||$ $le$ $M||v||$ for any $v in V$.



      A norm on the vector space $V$ is simply the square root of an inner-product.



      Do these two types of norms have some relationship with each-other, and if so, what is it?






      linear-algebra vector-spaces norm normed-spaces




      linear-algebra vector-spaces norm normed-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 7:29









      JaigusJaigus

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          $begingroup$

          You are wrong in your assumption that “A norm on the vector space $V$ is simply the square root of an inner-product.” Most norms are not induced by inner-products. For instance, of all norms $lVertcdotrVert_p$ in $mathbb{R}^n$ defined by$$bigllVert(x_1,ldots,x_n)bigrrVert_p=bigl(lvert x_1rvert^p+cdots+lvert x_nrvert^pbigr)^{frac1p}$$($pin[1,infty)$), only the norm $lVertcdotrVert_2$ is induced from an inner-product.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for clearing this up for me. I'm in an introductory class as you might have guessed, and this is how it was introduced to us at this point. I'm reading Axler's Linear Algebra Done Right.
            $endgroup$
            – Jaigus
            Dec 10 '18 at 7:36













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          $begingroup$

          You are wrong in your assumption that “A norm on the vector space $V$ is simply the square root of an inner-product.” Most norms are not induced by inner-products. For instance, of all norms $lVertcdotrVert_p$ in $mathbb{R}^n$ defined by$$bigllVert(x_1,ldots,x_n)bigrrVert_p=bigl(lvert x_1rvert^p+cdots+lvert x_nrvert^pbigr)^{frac1p}$$($pin[1,infty)$), only the norm $lVertcdotrVert_2$ is induced from an inner-product.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for clearing this up for me. I'm in an introductory class as you might have guessed, and this is how it was introduced to us at this point. I'm reading Axler's Linear Algebra Done Right.
            $endgroup$
            – Jaigus
            Dec 10 '18 at 7:36


















          1












          $begingroup$

          You are wrong in your assumption that “A norm on the vector space $V$ is simply the square root of an inner-product.” Most norms are not induced by inner-products. For instance, of all norms $lVertcdotrVert_p$ in $mathbb{R}^n$ defined by$$bigllVert(x_1,ldots,x_n)bigrrVert_p=bigl(lvert x_1rvert^p+cdots+lvert x_nrvert^pbigr)^{frac1p}$$($pin[1,infty)$), only the norm $lVertcdotrVert_2$ is induced from an inner-product.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for clearing this up for me. I'm in an introductory class as you might have guessed, and this is how it was introduced to us at this point. I'm reading Axler's Linear Algebra Done Right.
            $endgroup$
            – Jaigus
            Dec 10 '18 at 7:36
















          1












          1








          1





          $begingroup$

          You are wrong in your assumption that “A norm on the vector space $V$ is simply the square root of an inner-product.” Most norms are not induced by inner-products. For instance, of all norms $lVertcdotrVert_p$ in $mathbb{R}^n$ defined by$$bigllVert(x_1,ldots,x_n)bigrrVert_p=bigl(lvert x_1rvert^p+cdots+lvert x_nrvert^pbigr)^{frac1p}$$($pin[1,infty)$), only the norm $lVertcdotrVert_2$ is induced from an inner-product.






          share|cite|improve this answer









          $endgroup$



          You are wrong in your assumption that “A norm on the vector space $V$ is simply the square root of an inner-product.” Most norms are not induced by inner-products. For instance, of all norms $lVertcdotrVert_p$ in $mathbb{R}^n$ defined by$$bigllVert(x_1,ldots,x_n)bigrrVert_p=bigl(lvert x_1rvert^p+cdots+lvert x_nrvert^pbigr)^{frac1p}$$($pin[1,infty)$), only the norm $lVertcdotrVert_2$ is induced from an inner-product.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 7:34









          José Carlos SantosJosé Carlos Santos

          159k22126231




          159k22126231












          • $begingroup$
            Thanks for clearing this up for me. I'm in an introductory class as you might have guessed, and this is how it was introduced to us at this point. I'm reading Axler's Linear Algebra Done Right.
            $endgroup$
            – Jaigus
            Dec 10 '18 at 7:36




















          • $begingroup$
            Thanks for clearing this up for me. I'm in an introductory class as you might have guessed, and this is how it was introduced to us at this point. I'm reading Axler's Linear Algebra Done Right.
            $endgroup$
            – Jaigus
            Dec 10 '18 at 7:36


















          $begingroup$
          Thanks for clearing this up for me. I'm in an introductory class as you might have guessed, and this is how it was introduced to us at this point. I'm reading Axler's Linear Algebra Done Right.
          $endgroup$
          – Jaigus
          Dec 10 '18 at 7:36






          $begingroup$
          Thanks for clearing this up for me. I'm in an introductory class as you might have guessed, and this is how it was introduced to us at this point. I'm reading Axler's Linear Algebra Done Right.
          $endgroup$
          – Jaigus
          Dec 10 '18 at 7:36




















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