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shna science forum beginner
Joined: 29 Mar 2006
Posts: 20

Posted: Thu Jul 13, 2006 5:34 am Post subject:
Re: The problem of derivative on scalar scailed matrix



"Peter Spellucci" <spellucci@fb04373.mathematik.tudarmstadt.de> wrote in
message news:e92gc9$kcv$1@fb04373.mathematik.tudarmstadt.de...
Quote: 
In article <e927mg$tmt$1@news.kreonet.re.kr>,
"shna" <nsh1979@postech.ac.kr> writes:
"Peter Spellucci" <spellucci@fb04373.mathematik.tudarmstadt.de> wrote in
message news:e90qj8$d9f$1@fb04373.mathematik.tudarmstadt.de...
In article <e904d7$tgj$1@news.kreonet.re.kr>,
"shna" <nsh1979@postech.ac.kr> writes:
Hi, all.
I have trouble with the following problem while reading a book about
numerical analysis.
Let H be given square matrix.
H = b M (b is scalar, M is a square matrix)
Let \lambda_i be eigenvalues of H.
Then,
d \lambda_i / d b = \lambda_i / b  (1)
(d indicates 'derivative')
Why be formula (1) correct?
Could you explain in detail?
Thank you.
From SeungHoon Na
??? homework
if M has the eigenvalues mu(j) which eigenvalues does then have b*M?
name these lam(j), hence which relation is there between lam(j) and
mu(j).
now compute (d/db) lam(j) = ??
lam(j)/b = ??
hth
peter
Thank you.
Your guideline help me solve this problem.
Here is the solution.
By applying determinant to "b M = H", we obtain
b M = H
where H = lam(1) * ... lam(d)
(d = # dimension)
After applying derivative on both sides, then
M db = H / lam(i) d lam(i)
Thus,
d lam(i) / d b = lam(i) * M / H
d lam(i) / d b = lam(i) / b
example how a false proof may lead to correct results.
if
H = b*M b scalar, H,M square matrices
then
det(H)=b^d*det(M), d=rowlength(H)

You are right. It is a mistake, but, I have never recognized).
I should have used b^d instead of b.
Quote:  but what about this:
M*x=lambda*x x not= 0
=
b*M*x = (b*lambda)*x
hth
peter

All your comments are summarized into " lambda(i) = b mu(i) " !
Then, lambda(i) has linear relation with b,
so the above formula (d lambda(i) / d b = lambda(i) / b ) is naturally
correct.
It make our problem more easy to manipulate, rather when considering the
determinant.
Thank you. 

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Peter Spellucci science forum Guru
Joined: 29 Apr 2005
Posts: 702

Posted: Wed Jul 12, 2006 9:45 am Post subject:
Re: The problem of derivative on scalar scailed matrix



In article <e927mg$tmt$1@news.kreonet.re.kr>,
"shna" <nsh1979@postech.ac.kr> writes:
Quote: 
"Peter Spellucci" <spellucci@fb04373.mathematik.tudarmstadt.de> wrote in
message news:e90qj8$d9f$1@fb04373.mathematik.tudarmstadt.de...
In article <e904d7$tgj$1@news.kreonet.re.kr>,
"shna" <nsh1979@postech.ac.kr> writes:
Hi, all.
I have trouble with the following problem while reading a book about
numerical analysis.
Let H be given square matrix.
H = b M (b is scalar, M is a square matrix)
Let \lambda_i be eigenvalues of H.
Then,
d \lambda_i / d b = \lambda_i / b  (1)
(d indicates 'derivative')
Why be formula (1) correct?
Could you explain in detail?
Thank you.
From SeungHoon Na
??? homework
if M has the eigenvalues mu(j) which eigenvalues does then have b*M?
name these lam(j), hence which relation is there between lam(j) and mu(j).
now compute (d/db) lam(j) = ??
lam(j)/b = ??
hth
peter
Thank you.
Your guideline help me solve this problem.
Here is the solution.
By applying determinant to "b M = H", we obtain
b M = H
where H = lam(1) * ... lam(d)
(d = # dimension)
After applying derivative on both sides, then
M db = H / lam(i) d lam(i)
Thus,
d lam(i) / d b = lam(i) * M / H
d lam(i) / d b = lam(i) / b

example how a false proof may lead to correct results.
if
H = b*M b scalar, H,M square matrices
then
det(H)=b^d*det(M), d=rowlength(H)
but what about this:
M*x=lambda*x x not= 0
=>
b*M*x = (b*lambda)*x
hth
peter 

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shna science forum beginner
Joined: 29 Mar 2006
Posts: 20

Posted: Wed Jul 12, 2006 7:17 am Post subject:
Re: The problem of derivative on scalar scailed matrix



"Peter Spellucci" <spellucci@fb04373.mathematik.tudarmstadt.de> wrote in
message news:e90qj8$d9f$1@fb04373.mathematik.tudarmstadt.de...
Quote: 
In article <e904d7$tgj$1@news.kreonet.re.kr>,
"shna" <nsh1979@postech.ac.kr> writes:
Hi, all.
I have trouble with the following problem while reading a book about
numerical analysis.
Let H be given square matrix.
H = b M (b is scalar, M is a square matrix)
Let \lambda_i be eigenvalues of H.
Then,
d \lambda_i / d b = \lambda_i / b  (1)
(d indicates 'derivative')
Why be formula (1) correct?
Could you explain in detail?
Thank you.
From SeungHoon Na
??? homework
if M has the eigenvalues mu(j) which eigenvalues does then have b*M?
name these lam(j), hence which relation is there between lam(j) and mu(j).
now compute (d/db) lam(j) = ??
lam(j)/b = ??
hth
peter

Thank you.
Your guideline help me solve this problem.
Here is the solution.
By applying determinant to "b M = H", we obtain
b M = H
where H = lam(1) * ... lam(d)
(d = # dimension)
After applying derivative on both sides, then
M db = H / lam(i) d lam(i)
Thus,
d lam(i) / d b = lam(i) * M / H
d lam(i) / d b = lam(i) / b 

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Peter Spellucci science forum Guru
Joined: 29 Apr 2005
Posts: 702

Posted: Tue Jul 11, 2006 6:27 pm Post subject:
Re: The problem of derivative on scalar scailed matrix



In article <e904d7$tgj$1@news.kreonet.re.kr>,
"shna" <nsh1979@postech.ac.kr> writes:
Quote:  Hi, all.
I have trouble with the following problem while reading a book about
numerical analysis.
Let H be given square matrix.
H = b M (b is scalar, M is a square matrix)
Let \lambda_i be eigenvalues of H.
Then,
d \lambda_i / d b = \lambda_i / b  (1)
(d indicates 'derivative')
Why be formula (1) correct?
Could you explain in detail?
Thank you.
From SeungHoon Na

??? homework
if M has the eigenvalues mu(j) which eigenvalues does then have b*M?
name these lam(j), hence which relation is there between lam(j) and mu(j).
now compute (d/db) lam(j) = ??
lam(j)/b = ??
hth
peter 

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shna science forum beginner
Joined: 29 Mar 2006
Posts: 20

Posted: Tue Jul 11, 2006 12:09 pm Post subject:
The problem of derivative on scalar scailed matrix



Hi, all.
I have trouble with the following problem while reading a book about
numerical analysis.
Let H be given square matrix.
H = b M (b is scalar, M is a square matrix)
Let \lambda_i be eigenvalues of H.
Then,
d \lambda_i / d b = \lambda_i / b  (1)
(d indicates 'derivative')
Why be formula (1) correct?
Could you explain in detail?
Thank you.
From SeungHoon Na 

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