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Eigen vector problem

[Deleted User] · 2009-08-25 21:26:45

Right, this is driving me up the walls... 0.981 , 0.019 0.019 , 0.981 is a transition matrix, I want to get to the eigenvalues and vectors. Now I'm getting that 1 and 0.962 are the eigenvalues and that -0.70711 , 0.70711 0.70711 , 0.70711 are the eigenvalues, but those don't make sense economically (at least I don't think…

25-08-2009 10:26PM

#1

[Deleted User]
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Right, this is driving me up the walls...

0.981 , 0.019
0.019 , 0.981

is a transition matrix, I want to get to the eigenvalues and vectors.
Now I'm getting that 1 and 0.962 are the eigenvalues and that
-0.70711 , 0.70711
0.70711 , 0.70711
are the eigenvalues, but those don't make sense economically (at least I don't think they do)
as

(p(a)) = c1 (-0.70711).(1) + c2 (0.70711).962
(p(b)) = ....(.70711).........+.....(0.70711)

leads to a p(a) = p(b) = .5, which is not correct

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Comments

#2 25-08-2009 11:47PM

[Deleted User]
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Its grand.
Transition matrix was wrong. Stupid decimal places...

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#3 26-08-2009 12:59AM

alan4cult

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If you don't mind me asking what were you using this matrix for? What are p(a) and p(b)? Is it economics?

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#4 26-08-2009 09:32AM

[Deleted User]
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Yeah, its to calculate the expected probabilitiy of a future state.

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