Is there a special relationship between a norm on a vector space V, and the operator norm $ mathcal{L}(V,...
$begingroup$
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
asked Dec 10 '18 at 7:29
JaigusJaigus
2218
$endgroup$
add a comment |
$begingroup$
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
asked Dec 10 '18 at 7:29
JaigusJaigus
2218
$endgroup$
add a comment |
$begingroup$
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
asked Dec 10 '18 at 7:29
JaigusJaigus
2218
$endgroup$
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
asked Dec 10 '18 at 7:29
JaigusJaigus
2218
asked Dec 10 '18 at 7:29
JaigusJaigus
2218
asked Dec 10 '18 at 7:29
JaigusJaigus
2218
asked Dec 10 '18 at 7:29
JaigusJaigus
2218
asked Dec 10 '18 at 7:29
JaigusJaigus
2218
2218
add a comment |
add a comment |
1 Answer
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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.
answered Dec 10 '18 at 7:34
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
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.
answered Dec 10 '18 at 7:34
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 10 '18 at 7:34
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 10 '18 at 7:34
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
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.
answered Dec 10 '18 at 7:34
José Carlos SantosJosé Carlos Santos
159k22126231
answered Dec 10 '18 at 7:34
José Carlos SantosJosé Carlos Santos
159k22126231
answered Dec 10 '18 at 7:34
José Carlos SantosJosé Carlos Santos
159k22126231
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
add a comment |
$begingroup$
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
add a comment |
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