Eigenvectors in discrete biological process

[garcijon]

Hi All,

I am new to the forums. I was away from school for like 3-4 yrs and just went back to do a grad program. I've taken linear algebra before but am having a hard time wrapping my head around 1 proof they are asking for. I apologize in advance for the long post and thank you for any help

They originally defined M as a matrix:

 a[SUB]11[/SUB] a[SUB]12[/SUB]
 a[SUB]21[/SUB] a[SUB]22
[/SUB]

I know that an eigenvector is one such that Mv = λV where λ is the eigenvalue.

Also (M - λI) v = 0 where M is the matrix, I is the identity matrix, x is the eigenvalue and v is the eigenvector. They asked us to proof that for M an eigenvector is

1
(λi- a11) / a12

If I try to apply (M - λiI) I get the below matrix

a11-λi      a12
a21      a22 -λi
[FONT=Verdana]
[/FONT]

mutliplication by v yields

(a11-λi)+ a12 ((λi-a11)/a12) = 0

which does yield 0
and

a21 + (a22 -λi) *((λi- a11) / a12)[COLOR=#333333][FONT=PT Sans Caption][SIZE=3] = 0 
[/SIZE][/FONT][/COLOR]

which does not yield zero

What I was thinking is that since the det(M) has already been determined to be zero then these two equation represent the same thing. If λi can solve one linear equation than a linear combo of it will solve the second equation?

 

[garcijon]

Was able to solve it.

 

[DrPhil]

Hi All,

I am new to the forums. I was away from school for like 3-4 yrs and just went back to do a grad program. I've taken linear algebra before but am having a hard time wrapping my head around 1 proof they are asking for. I apologize in advance for the long post and thank you for any help

They originally defined M as a matrix:

 a[SUB]11[/SUB] a[SUB]12[/SUB]
 a[SUB]21[/SUB] a[SUB]22
[/SUB]

I know that an eigenvector is one such that Mv = λV where λ is the eigenvalue.

Also (M - λI) v = 0 where M is the matrix, I is the identity matrix, x is the eigenvalue and v is the eigenvector. They asked us to proof that for M one of the eigenvectors is

[B][COLOR=#ff0000]v1[/COLOR][/B] =         1
        (λ[B][COLOR=#ff0000]1[/COLOR][/B] - a11) / a12

If I try to apply (M - λiI) I get the below matrix

a11-λi      a12
a21      a22 -λi
[FONT=Verdana]
[/FONT]

mutliplication by v1 yields

(a11-λ[B][COLOR=#ff0000]1[/COLOR][/B])+ a12 ((λ[B][COLOR=#ff0000]1[/COLOR][/B]-a11)/a12) = 0

which does yield 0
and

a21 + (a22 -λ[B]1[/B]) *((λ[B]1[/B]- a11) / a12)[COLOR=#333333][FONT=PT Sans Caption][SIZE=3] = 0 
[/SIZE][/FONT][/COLOR]

which does not yield zero Solve for λ1

What I was thinking is that since the det(M - λi) has already been determined to be zero then these two equation represent the same thing. If λi can solve one linear equation than a linear combo of it will solve the second equation?

The other optiion is to solve det(M - λi) = 0 for lambda. Since they gave you an eigenvector you can use that to find the corresponding eigenvalue.