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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 Seung-Hoon Na
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 Seung-Hoon 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
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.tu-darmstadt.de> wrote in
message news:e90qj8$d9f$1@fb04373.mathematik.tu-darmstadt.de...
 Quote: In article , "shna" 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 Seung-Hoon 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
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$1@news.kreonet.re.kr>,
"shna" <nsh1979@postech.ac.kr> writes:
 Quote: "Peter Spellucci" wrote in message news:e90qj8$d9f$1@fb04373.mathematik.tu-darmstadt.de... In article , "shna" 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 Seung-Hoon 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
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.tu-darmstadt.de> wrote in
message news:e92gc9$kcv$1@fb04373.mathematik.tu-darmstadt.de...
 Quote: In article , "shna" writes: "Peter Spellucci" wrote in message news:e90qj8$d9f$1@fb04373.mathematik.tu-darmstadt.de... In article , "shna" 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 Seung-Hoon 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.
Google

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