Prove operator norm cannot be longer than len(basis) times max(norm(basis))operator norm
Let $V$ be a finite dimensional normed linear space and let
$T in mathscr L (V)$. Define the operator norm of $T$ to be the smallest number $M$ such
that $||T v|| ≤ M||v||$ for any $v in V$ . We will write $||T||$ to mean that smallest
number $M$, the operator norm.
Let $B = e_1, ..., e_n$ be an orthonormal basis for V, a normed linear
space of dimension $n$. Let $T in mathscr L (V )$. Let
$m = Max{||T e_1||, ||T e_2||, ..., ||T e_n||}$. That is, $m$ is the length of the longest
vector in the list $T e_1, ..., T e_n$. Prove that for any vector $v in V , ||Tv|| ≤ mn$.
linear-algebra abstract-algebra
asked Dec 8 at 23:23
jmars
22
|
show 1 more comment
Let $V$ be a finite dimensional normed linear space and let
$T in mathscr L (V)$. Define the operator norm of $T$ to be the smallest number $M$ such
that $||T v|| ≤ M||v||$ for any $v in V$ . We will write $||T||$ to mean that smallest
number $M$, the operator norm.
Let $B = e_1, ..., e_n$ be an orthonormal basis for V, a normed linear
space of dimension $n$. Let $T in mathscr L (V )$. Let
$m = Max{||T e_1||, ||T e_2||, ..., ||T e_n||}$. That is, $m$ is the length of the longest
vector in the list $T e_1, ..., T e_n$. Prove that for any vector $v in V , ||Tv|| ≤ mn$.
linear-algebra abstract-algebra
asked Dec 8 at 23:23
jmars
22
Welcome to MSE. What are your thoughts on the problem?
– MisterRiemann
Dec 8 at 23:25
What is $A$? Does $A = T$?
– Robert Lewis
Dec 8 at 23:28
I was confused about that too, may be a typo from the professor, let's treat A=T
– jmars
Dec 8 at 23:29
As far as my thoughts go, I'm confused as to how why it's not simply $||Tv||leq m$. Shouldn't the longest ||Tv|| be m since m is the max norm of the images of the basis?
– jmars
Dec 8 at 23:34
2
I think you might need $Vert Tv Vert le mn Vert v Vert$. since $Vert Tv Vert$ must depend on $v$.
– Robert Lewis
Dec 8 at 23:39
|
show 1 more comment
Let $V$ be a finite dimensional normed linear space and let
$T in mathscr L (V)$. Define the operator norm of $T$ to be the smallest number $M$ such
that $||T v|| ≤ M||v||$ for any $v in V$ . We will write $||T||$ to mean that smallest
number $M$, the operator norm.
Let $B = e_1, ..., e_n$ be an orthonormal basis for V, a normed linear
space of dimension $n$. Let $T in mathscr L (V )$. Let
$m = Max{||T e_1||, ||T e_2||, ..., ||T e_n||}$. That is, $m$ is the length of the longest
vector in the list $T e_1, ..., T e_n$. Prove that for any vector $v in V , ||Tv|| ≤ mn$.
linear-algebra abstract-algebra
asked Dec 8 at 23:23
jmars
22
Let $V$ be a finite dimensional normed linear space and let
$T in mathscr L (V)$. Define the operator norm of $T$ to be the smallest number $M$ such
that $||T v|| ≤ M||v||$ for any $v in V$ . We will write $||T||$ to mean that smallest
number $M$, the operator norm.
Let $B = e_1, ..., e_n$ be an orthonormal basis for V, a normed linear
space of dimension $n$. Let $T in mathscr L (V )$. Let
$m = Max{||T e_1||, ||T e_2||, ..., ||T e_n||}$. That is, $m$ is the length of the longest
vector in the list $T e_1, ..., T e_n$. Prove that for any vector $v in V , ||Tv|| ≤ mn$.
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 8 at 23:23
jmars
22
asked Dec 8 at 23:23
jmars
22
edited Dec 9 at 5:29
asked Dec 8 at 23:23
jmars
22
asked Dec 8 at 23:23
jmars
22
asked Dec 8 at 23:23
jmars
22
22
Welcome to MSE. What are your thoughts on the problem?
– MisterRiemann
Dec 8 at 23:25
What is $A$? Does $A = T$?
– Robert Lewis
Dec 8 at 23:28
I was confused about that too, may be a typo from the professor, let's treat A=T
– jmars
Dec 8 at 23:29
As far as my thoughts go, I'm confused as to how why it's not simply $||Tv||leq m$. Shouldn't the longest ||Tv|| be m since m is the max norm of the images of the basis?
– jmars
Dec 8 at 23:34
2
I think you might need $Vert Tv Vert le mn Vert v Vert$. since $Vert Tv Vert$ must depend on $v$.
– Robert Lewis
Dec 8 at 23:39
|
show 1 more comment
Welcome to MSE. What are your thoughts on the problem?
– MisterRiemann
Dec 8 at 23:25
What is $A$? Does $A = T$?
– Robert Lewis
Dec 8 at 23:28
I was confused about that too, may be a typo from the professor, let's treat A=T
– jmars
Dec 8 at 23:29
As far as my thoughts go, I'm confused as to how why it's not simply $||Tv||leq m$. Shouldn't the longest ||Tv|| be m since m is the max norm of the images of the basis?
– jmars
Dec 8 at 23:34
2
I think you might need $Vert Tv Vert le mn Vert v Vert$. since $Vert Tv Vert$ must depend on $v$.
– Robert Lewis
Dec 8 at 23:39
Welcome to MSE. What are your thoughts on the problem?
– MisterRiemann
Dec 8 at 23:25
Welcome to MSE. What are your thoughts on the problem?
– MisterRiemann
Dec 8 at 23:25
What is $A$? Does $A = T$?
– Robert Lewis
Dec 8 at 23:28
What is $A$? Does $A = T$?
– Robert Lewis
Dec 8 at 23:28
I was confused about that too, may be a typo from the professor, let's treat A=T
– jmars
Dec 8 at 23:29
I was confused about that too, may be a typo from the professor, let's treat A=T
– jmars
Dec 8 at 23:29
As far as my thoughts go, I'm confused as to how why it's not simply $||Tv||leq m$. Shouldn't the longest ||Tv|| be m since m is the max norm of the images of the basis?
– jmars
Dec 8 at 23:34
As far as my thoughts go, I'm confused as to how why it's not simply $||Tv||leq m$. Shouldn't the longest ||Tv|| be m since m is the max norm of the images of the basis?
– jmars
Dec 8 at 23:34
2
2
I think you might need $Vert Tv Vert le mn Vert v Vert$. since $Vert Tv Vert$ must depend on $v$.
– Robert Lewis
Dec 8 at 23:39
I think you might need $Vert Tv Vert le mn Vert v Vert$. since $Vert Tv Vert$ must depend on $v$.
– Robert Lewis
Dec 8 at 23:39
|
show 1 more comment
1 Answer
1
active
oldest
votes
This may be what you need. Write $v = sum_{i=1}^n v_i e_i $ where $v_i = langle v,e_irangle$. Then we have
$$
||Tv||^2 = ||sum_{i=1}^n v_i T(e_i)||^2 leq sum_{i=1}^n |v_i|^2sum_{i=1}^n ||T(e_i)||^2leq ||v||^2nm^2,
$$ by Cauchy-Schwarz inequality. Hence, it follows that
$$
||T||leq msqrt{n}.
$$
answered Dec 9 at 7:36
Song
4,200316
add a comment |
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1 Answer
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1 Answer
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This may be what you need. Write $v = sum_{i=1}^n v_i e_i $ where $v_i = langle v,e_irangle$. Then we have
$$
||Tv||^2 = ||sum_{i=1}^n v_i T(e_i)||^2 leq sum_{i=1}^n |v_i|^2sum_{i=1}^n ||T(e_i)||^2leq ||v||^2nm^2,
$$ by Cauchy-Schwarz inequality. Hence, it follows that
$$
||T||leq msqrt{n}.
$$
answered Dec 9 at 7:36
Song
4,200316
add a comment |
This may be what you need. Write $v = sum_{i=1}^n v_i e_i $ where $v_i = langle v,e_irangle$. Then we have
$$
||Tv||^2 = ||sum_{i=1}^n v_i T(e_i)||^2 leq sum_{i=1}^n |v_i|^2sum_{i=1}^n ||T(e_i)||^2leq ||v||^2nm^2,
$$ by Cauchy-Schwarz inequality. Hence, it follows that
$$
||T||leq msqrt{n}.
$$
answered Dec 9 at 7:36
Song
4,200316
add a comment |
This may be what you need. Write $v = sum_{i=1}^n v_i e_i $ where $v_i = langle v,e_irangle$. Then we have
$$
||Tv||^2 = ||sum_{i=1}^n v_i T(e_i)||^2 leq sum_{i=1}^n |v_i|^2sum_{i=1}^n ||T(e_i)||^2leq ||v||^2nm^2,
$$ by Cauchy-Schwarz inequality. Hence, it follows that
$$
||T||leq msqrt{n}.
$$
answered Dec 9 at 7:36
Song
4,200316
This may be what you need. Write $v = sum_{i=1}^n v_i e_i $ where $v_i = langle v,e_irangle$. Then we have
$$
||Tv||^2 = ||sum_{i=1}^n v_i T(e_i)||^2 leq sum_{i=1}^n |v_i|^2sum_{i=1}^n ||T(e_i)||^2leq ||v||^2nm^2,
$$ by Cauchy-Schwarz inequality. Hence, it follows that
$$
||T||leq msqrt{n}.
$$
answered Dec 9 at 7:36
Song
4,200316
answered Dec 9 at 7:36
Song
4,200316
answered Dec 9 at 7:36
Song
4,200316
answered Dec 9 at 7:36
Song
4,200316
4,200316
add a comment |
add a comment |
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Welcome to MSE. What are your thoughts on the problem?
– MisterRiemann
Dec 8 at 23:25
What is $A$? Does $A = T$?
– Robert Lewis
Dec 8 at 23:28
I was confused about that too, may be a typo from the professor, let's treat A=T
– jmars
Dec 8 at 23:29
As far as my thoughts go, I'm confused as to how why it's not simply $||Tv||leq m$. Shouldn't the longest ||Tv|| be m since m is the max norm of the images of the basis?
– jmars
Dec 8 at 23:34
2
I think you might need $Vert Tv Vert le mn Vert v Vert$. since $Vert Tv Vert$ must depend on $v$.
– Robert Lewis
Dec 8 at 23:39