Upper and lower bound of infinity norm of a matrix-vector product $lVert A x rVert_{infty}$?
Multi tool use
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Do we have an upper and lower bound of an infinity norm of a matrix-vector product $lVert A x rVert_{infty}$, where $A in M_{m,n}(mathbb{C})$ and $x in M_{n,1}(mathbb{C})$?
I can see a lower bound with infinity norm assuming real valued, e.g., here.
So, do we have an upper and lower bound with complex matrix and vector? thank you.
linear-algebra complex-analysis
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add a comment |
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Do we have an upper and lower bound of an infinity norm of a matrix-vector product $lVert A x rVert_{infty}$, where $A in M_{m,n}(mathbb{C})$ and $x in M_{n,1}(mathbb{C})$?
I can see a lower bound with infinity norm assuming real valued, e.g., here.
So, do we have an upper and lower bound with complex matrix and vector? thank you.
linear-algebra complex-analysis
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1
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The lower bound in the link still applies in the complex case. And an obvious upper bound is $|A|_infty|x|_infty$.
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– user1551
Nov 30 '18 at 9:21
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Thank you. So, if I summarize, ${1 over sqrt{n}} sigma_min(A) |x|_infty leq | A x |_infty leq | A |_infty | x |_infty $. Is it correct?
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– learning
Nov 30 '18 at 9:48
1
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Yes.$phantom{}$
$endgroup$
– user1551
Nov 30 '18 at 9:54
add a comment |
$begingroup$
Do we have an upper and lower bound of an infinity norm of a matrix-vector product $lVert A x rVert_{infty}$, where $A in M_{m,n}(mathbb{C})$ and $x in M_{n,1}(mathbb{C})$?
I can see a lower bound with infinity norm assuming real valued, e.g., here.
So, do we have an upper and lower bound with complex matrix and vector? thank you.
linear-algebra complex-analysis
$endgroup$
Do we have an upper and lower bound of an infinity norm of a matrix-vector product $lVert A x rVert_{infty}$, where $A in M_{m,n}(mathbb{C})$ and $x in M_{n,1}(mathbb{C})$?
I can see a lower bound with infinity norm assuming real valued, e.g., here.
So, do we have an upper and lower bound with complex matrix and vector? thank you.
linear-algebra complex-analysis
linear-algebra complex-analysis
asked Nov 30 '18 at 8:42
learninglearning
346
346
1
$begingroup$
The lower bound in the link still applies in the complex case. And an obvious upper bound is $|A|_infty|x|_infty$.
$endgroup$
– user1551
Nov 30 '18 at 9:21
$begingroup$
Thank you. So, if I summarize, ${1 over sqrt{n}} sigma_min(A) |x|_infty leq | A x |_infty leq | A |_infty | x |_infty $. Is it correct?
$endgroup$
– learning
Nov 30 '18 at 9:48
1
$begingroup$
Yes.$phantom{}$
$endgroup$
– user1551
Nov 30 '18 at 9:54
add a comment |
1
$begingroup$
The lower bound in the link still applies in the complex case. And an obvious upper bound is $|A|_infty|x|_infty$.
$endgroup$
– user1551
Nov 30 '18 at 9:21
$begingroup$
Thank you. So, if I summarize, ${1 over sqrt{n}} sigma_min(A) |x|_infty leq | A x |_infty leq | A |_infty | x |_infty $. Is it correct?
$endgroup$
– learning
Nov 30 '18 at 9:48
1
$begingroup$
Yes.$phantom{}$
$endgroup$
– user1551
Nov 30 '18 at 9:54
1
1
$begingroup$
The lower bound in the link still applies in the complex case. And an obvious upper bound is $|A|_infty|x|_infty$.
$endgroup$
– user1551
Nov 30 '18 at 9:21
$begingroup$
The lower bound in the link still applies in the complex case. And an obvious upper bound is $|A|_infty|x|_infty$.
$endgroup$
– user1551
Nov 30 '18 at 9:21
$begingroup$
Thank you. So, if I summarize, ${1 over sqrt{n}} sigma_min(A) |x|_infty leq | A x |_infty leq | A |_infty | x |_infty $. Is it correct?
$endgroup$
– learning
Nov 30 '18 at 9:48
$begingroup$
Thank you. So, if I summarize, ${1 over sqrt{n}} sigma_min(A) |x|_infty leq | A x |_infty leq | A |_infty | x |_infty $. Is it correct?
$endgroup$
– learning
Nov 30 '18 at 9:48
1
1
$begingroup$
Yes.$phantom{}$
$endgroup$
– user1551
Nov 30 '18 at 9:54
$begingroup$
Yes.$phantom{}$
$endgroup$
– user1551
Nov 30 '18 at 9:54
add a comment |
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1
$begingroup$
The lower bound in the link still applies in the complex case. And an obvious upper bound is $|A|_infty|x|_infty$.
$endgroup$
– user1551
Nov 30 '18 at 9:21
$begingroup$
Thank you. So, if I summarize, ${1 over sqrt{n}} sigma_min(A) |x|_infty leq | A x |_infty leq | A |_infty | x |_infty $. Is it correct?
$endgroup$
– learning
Nov 30 '18 at 9:48
1
$begingroup$
Yes.$phantom{}$
$endgroup$
– user1551
Nov 30 '18 at 9:54