Dear Sir,
We start from the leslie matrix 12x12:
a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12
b1 0 0 0 0 0 0 0 0 0 0 0
0 b2 0 0 0 0 0 0 0 0 0 0
0 0 b3 0 0 0 0 0 0 0 0 0
0 0 0 b4 0 0 0 0 0 0 0 0
0 0 0 0 b5 0 0 0 0 0 0 0
0 0 0 0 0 b6 0 0 0 0 0 0
0 0 0 0 0 0 b7 0 0 0 0 0
0 0 0 0 0 0 0 b8 0 0 0 0
0 0 0 0 0 0 0 0 b9 0 0 0
0 0 0 0 0 0 0 0 0 b10 0 0
0 0 0 0 0 0 0 0 0 0 b11 0
to calculate the eigenvalues:
the determinant has to be 0
a1 - ë a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12
b1 -ë 0 0 0 0 0 0 0 0 0 0
0 b2 -ë 0 0 0 0 0 0 0 0 0
0 0 b3 -ë 0 0 0 0 0 0 0 0
0 0 0 b4 -ë 0 0 0 0 0 0 0
det( 0 0 0 0 b5 -ë 0 0 0 0 0 0 ) =0
0 0 0 0 0 b6 -ë 0 0 0 0 0
0 0 0 0 0 0 b7 -ë 0 0 0 0
0 0 0 0 0 0 0 b8 -ë 0 0 0
0 0 0 0 0 0 0 0 b9 -ë 0 0
0 0 0 0 0 0 0 0 0 b10 -ë 0
0 0 0 0 0 0 0 0 0 0 b11 -ë
we look to the first column and we "develop" to it:
(a1-ë)* det( -ë 0 0 0 0 0 0 0 0 0 0 )
b2 -ë 0 0 0 0 0 0 0 0 0
0 b3 -ë 0 0 0 0 0 0 0 0
0 0 b4 -ë 0 0 0 0 0 0 0
0 0 0 b5 -ë 0 0 0 0 0 0
0 0 0 0 b6 -ë 0 0 0 0 0
0 0 0 0 0 b7 -ë 0 0 0 0
0 0 0 0 0 0 b8 -ë 0 0 0
0 0 0 0 0 0 0 b9 -ë 0 0
0 0 0 0 0 0 0 0 b10 -ë 0
0 0 0 0 0 0 0 0 0 b11 -ë
-b1* det( a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 )
b2 -ë 0 0 0 0 0 0 0 0 0
0 b3 -ë 0 0 0 0 0 0 0 0
0 0 b4 -ë 0 0 0 0 0 0 0
0 0 0 b5 -ë 0 0 0 0 0 0
0 0 0 0 b6 -ë 0 0 0 0 0
0 0 0 0 0 b7 -ë 0 0 0 0
0 0 0 0 0 0 b8 -ë 0 0 0
0 0 0 0 0 0 0 b9 -ë 0 0
0 0 0 0 0 0 0 0 b10 -ë 0
0 0 0 0 0 0 0 0 0 b11 -ë
= 0
the first determinant, is a "beneath triangle" det = the
multiplication of the diagonal elements = -ë^11
we have:
(a1 - ë)* (-ë^11)= ë^12 - a1*ë^11 (this is the first term)
we have to develop the second determinant further(again to the first
column): (RECURSIVE)
we notice that these determinant is from the same shape as the first;
-b1* det( a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 )
b2 -ë 0 0 0 0 0 0 0 0 0
0 b3 -ë 0 0 0 0 0 0 0 0
0 0 b4 -ë 0 0 0 0 0 0 0
0 0 0 b5 -ë 0 0 0 0 0 0
0 0 0 0 b6 -ë 0 0 0 0 0
0 0 0 0 0 b7 -ë 0 0 0 0
0 0 0 0 0 0 b8 -ë 0 0 0
0 0 0 0 0 0 0 b9 -ë 0 0
0 0 0 0 0 0 0 0 b10 -ë 0
0 0 0 0 0 0 0 0 0 b11 -ë
= -b1*a2 det( -ë 0 0 0 0 0 0 0 0 0 )
b3 -ë 0 0 0 0 0 0 0 0
0 b4 -ë 0 0 0 0 0 0 0
0 0 b5 -ë 0 0 0 0 0 0
0 0 0 b6 -ë 0 0 0 0 0
0 0 0 0 b7 -ë 0 0 0 0
0 0 0 0 0 b8 -ë 0 0 0
0 0 0 0 0 0 b9 -ë 0 0
0 0 0 0 0 0 0 b10 -ë 0
0 0 0 0 0 0 0 0 b11 -ë
-b1 * -b2 det(b3 -ë 0 0 0 0 0 0 0 0 )
0 b4 -ë 0 0 0 0 0 0 0
0 0 b5 -ë 0 0 0 0 0 0
0 0 0 b6 -ë 0 0 0 0 0
0 0 0 0 b7 -ë 0 0 0 0
0 0 0 0 0 b8 -ë 0 0 0
0 0 0 0 0 0 b9 -ë 0 0
0 0 0 0 0 0 0 b10 -ë 0
0 0 0 0 0 0 0 0 b11 -ë
here is the first again the multiplicant of the diagonal elements =
ë^10
we have for the first part again:
-b1*a2*ë^10 (this is thesecond term of the total equation)
if we do so on....after 12 steps, we have
ë^12 – a1*ë^11 – a2b1*ë^10 -...– an(b1*b2*..bn-1)=0
this could be transformed to:
– a1*ë^11 – a2b1*ë^10 -...– an(b1*b2*..bn-1)= -ë^12
if we change ë^12 from side
this could be transformed to:
a1/ë1 a2b1/ë^2 +...+ an(b1,b2…..bn-1)/ë^12 = 1
if we divide the both parts of the equitation by: -ë^12
If you need a clarification, just ask, i will be pleased to help you
again.
Kind Regards,
Pythagoras |
Clarification of Answer by
pythagoras-ga
on
02 Jun 2002 07:50 PDT
Dear Sir,
A little mistake in the last part, it has to be b11 in stead of bn-1,
and a12 instead of an:
if we do so on....after 12 steps, we have
ë^12 – a1*ë^11 – a2b1*ë^10 -...– a12(b1*b2*..b11)=0
this could be transformed to:
– a1*ë^11 – a2b1*ë^10 -...– a12(b1*b2*..b11)= -ë^12
if we change ë^12 from side
this could be transformed to:
a1/ë1 a2b1/ë^2 +...+ a12(b1,b2…..b11)/ë^12 = 1
if we divide the both parts of the equitation by: -ë^12
Kind Regards,
Pythagoras
|
Request for Answer Clarification by
virus-ga
on
02 Jun 2002 20:11 PDT
DOes this men that from q(ë) we can show that A has at least one real
root ë=ë1, which is simple (distinct) and positive?
Can you explain please
Thanks
|
Clarification of Answer by
pythagoras-ga
on
03 Jun 2002 09:00 PDT
Dear Sir,
The original question doesn't include the clarification you are asking for.
Maybe you should repost the question, and then a researcher can look for the proof.
Excuse me, I can't help you directly.
Kind Regards,
Pythagoras
|
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