Show that the function $||u|| = sqrt^2+5$ is a norm on $V$.Show that the norm of the derivitive of a $C^1$ function over a vector space is non-negative, homogeneous and satisfies the triangle ineqIf $u_1,ldots,u_n$ is a complete orthonormal set for the inner product induced by $G$, then $sum_j=1^n u_ju_j^T=G^-1$Prove that a norm that satisfies the parallelogram inequality defines an inner productIf a norm is induced from an inner product, then that inner product is uniqueIs it true that there exists a scalar $t$ such that $langle u_2-tu_1, u_1rangle = 0$?Isometry group of a norm is always contained in some Isometry group of an inner product?Prove that there is an inner product on $mathbbR^2$, given that the associated norm is a p-norm only if p = 2Triangle inequality for specific vector distance functionfor $T:Uto W$ and $A=u_1,u_2,…,u_ksubseteq U$ and $T(u_1),T(u_2),…,T(u_k)$ linearly independent set in W. prove A is linear independentCloser in the sense of inner product
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Show that the function $||u|| = sqrtu_2$ is a norm on $V$.
Show that the norm of the derivitive of a $C^1$ function over a vector space is non-negative, homogeneous and satisfies the triangle ineqIf $u_1,ldots,u_n$ is a complete orthonormal set for the inner product induced by $G$, then $sum_j=1^n u_ju_j^T=G^-1$Prove that a norm that satisfies the parallelogram inequality defines an inner productIf a norm is induced from an inner product, then that inner product is uniqueIs it true that there exists a scalar $t$ such that $langle u_2-tu_1, u_1rangle = 0$?Isometry group of a norm is always contained in some Isometry group of an inner product?Prove that there is an inner product on $mathbbR^2$, given that the associated norm is a p-norm only if p = 2Triangle inequality for specific vector distance functionfor $T:Uto W$ and $A=u_1,u_2,…,u_ksubseteq U$ and $T(u_1),T(u_2),…,T(u_k)$ linearly independent set in W. prove A is linear independentCloser in the sense of inner product
$begingroup$
I'm given: Let $V$ be the real vector space $mathbbR^2$, and $u = [u_1 , u_2]^T in V$. Show that the function $||u|| = sqrtu_2$ is a norm on $V$. Then, determine whether this norm is derived from an inner product. If so, what is the inner product?
After many attempts, I'm having trouble (a) proving that this function satisfies the Triangle Inequality property of norms (i.e. that $||u+v|| leq ||u|| + ||v||$), and (b) determining the inner product this norm is derived from. Any help getting unstuck would be much appreciated!
linear-algebra
asked Mar 21 at 7:08
user636164user636164
455
$endgroup$
add a comment |
$begingroup$
I'm given: Let $V$ be the real vector space $mathbbR^2$, and $u = [u_1 , u_2]^T in V$. Show that the function $||u|| = sqrtu_2$ is a norm on $V$. Then, determine whether this norm is derived from an inner product. If so, what is the inner product?
After many attempts, I'm having trouble (a) proving that this function satisfies the Triangle Inequality property of norms (i.e. that $||u+v|| leq ||u|| + ||v||$), and (b) determining the inner product this norm is derived from. Any help getting unstuck would be much appreciated!
linear-algebra
asked Mar 21 at 7:08
user636164user636164
455
$endgroup$
add a comment |
$begingroup$
I'm given: Let $V$ be the real vector space $mathbbR^2$, and $u = [u_1 , u_2]^T in V$. Show that the function $||u|| = sqrtu_2$ is a norm on $V$. Then, determine whether this norm is derived from an inner product. If so, what is the inner product?
After many attempts, I'm having trouble (a) proving that this function satisfies the Triangle Inequality property of norms (i.e. that $||u+v|| leq ||u|| + ||v||$), and (b) determining the inner product this norm is derived from. Any help getting unstuck would be much appreciated!
linear-algebra
asked Mar 21 at 7:08
user636164user636164
455
$endgroup$
I'm given: Let $V$ be the real vector space $mathbbR^2$, and $u = [u_1 , u_2]^T in V$. Show that the function $||u|| = sqrtu_2$ is a norm on $V$. Then, determine whether this norm is derived from an inner product. If so, what is the inner product?
After many attempts, I'm having trouble (a) proving that this function satisfies the Triangle Inequality property of norms (i.e. that $||u+v|| leq ||u|| + ||v||$), and (b) determining the inner product this norm is derived from. Any help getting unstuck would be much appreciated!
linear-algebra
linear-algebra
asked Mar 21 at 7:08
user636164user636164
455
asked Mar 21 at 7:08
user636164user636164
455
asked Mar 21 at 7:08
user636164user636164
455
asked Mar 21 at 7:08
user636164user636164
455
asked Mar 21 at 7:08
user636164user636164
455
455
add a comment |
add a comment |
1 Answer
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$begingroup$
If $|.|'$ is the usual norm then $|(u,v)|=|(sqrt 2, sqrt 5v)|'$. it should be easy to see from this triangle inequality for $|(u,v)|'$ gives triangle inequality for $|(u,v)|$. This inner norm is derived form the inner product $ langle (u,v) (u',v') rangle= langle (sqrt 2 u, sqrt 5v), (sqrt 2 u, sqrt 5v) rangle'$ where the prime is used to denote the usual inner product.
answered Mar 21 at 7:27
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
$begingroup$
Thank you, that helped me see it!
$endgroup$
– user636164
Mar 21 at 17:13
add a comment |
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$begingroup$
If $|.|'$ is the usual norm then $|(u,v)|=|(sqrt 2, sqrt 5v)|'$. it should be easy to see from this triangle inequality for $|(u,v)|'$ gives triangle inequality for $|(u,v)|$. This inner norm is derived form the inner product $ langle (u,v) (u',v') rangle= langle (sqrt 2 u, sqrt 5v), (sqrt 2 u, sqrt 5v) rangle'$ where the prime is used to denote the usual inner product.
answered Mar 21 at 7:27
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
$begingroup$
Thank you, that helped me see it!
$endgroup$
– user636164
Mar 21 at 17:13
add a comment |
$begingroup$
If $|.|'$ is the usual norm then $|(u,v)|=|(sqrt 2, sqrt 5v)|'$. it should be easy to see from this triangle inequality for $|(u,v)|'$ gives triangle inequality for $|(u,v)|$. This inner norm is derived form the inner product $ langle (u,v) (u',v') rangle= langle (sqrt 2 u, sqrt 5v), (sqrt 2 u, sqrt 5v) rangle'$ where the prime is used to denote the usual inner product.
answered Mar 21 at 7:27
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
$begingroup$
Thank you, that helped me see it!
$endgroup$
– user636164
Mar 21 at 17:13
add a comment |
$begingroup$
If $|.|'$ is the usual norm then $|(u,v)|=|(sqrt 2, sqrt 5v)|'$. it should be easy to see from this triangle inequality for $|(u,v)|'$ gives triangle inequality for $|(u,v)|$. This inner norm is derived form the inner product $ langle (u,v) (u',v') rangle= langle (sqrt 2 u, sqrt 5v), (sqrt 2 u, sqrt 5v) rangle'$ where the prime is used to denote the usual inner product.
answered Mar 21 at 7:27
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
If $|.|'$ is the usual norm then $|(u,v)|=|(sqrt 2, sqrt 5v)|'$. it should be easy to see from this triangle inequality for $|(u,v)|'$ gives triangle inequality for $|(u,v)|$. This inner norm is derived form the inner product $ langle (u,v) (u',v') rangle= langle (sqrt 2 u, sqrt 5v), (sqrt 2 u, sqrt 5v) rangle'$ where the prime is used to denote the usual inner product.
answered Mar 21 at 7:27
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
edited Mar 21 at 7:32
answered Mar 21 at 7:27
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Mar 21 at 7:27
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Mar 21 at 7:27
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
71.9k53170
$begingroup$
Thank you, that helped me see it!
$endgroup$
– user636164
Mar 21 at 17:13
add a comment |
$begingroup$
Thank you, that helped me see it!
$endgroup$
– user636164
Mar 21 at 17:13
$begingroup$
Thank you, that helped me see it!
$endgroup$
– user636164
Mar 21 at 17:13
$begingroup$
Thank you, that helped me see it!
$endgroup$
– user636164
Mar 21 at 17:13
add a comment |
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