Re: The problem of derivative on scalar scailed matrix


"Peter Spellucci" <spellucci@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in
message news:e92gc9$kcv$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

In article <e927mg$tmt$1@xxxxxxxxxxxxxxxxxx>,
"shna" <nsh1979@xxxxxxxxxxxxx> writes:

"Peter Spellucci" <spellucci@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in
message news:e90qj8$d9f$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

In article <e904d7$tgj$1@xxxxxxxxxxxxxxxxxx>,
"shna" <nsh1979@xxxxxxxxxxxxx> 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.


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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