# Eigenvalues of matrix and stability

Suppose that $x(t)$ represents outputs of all sectors and parts of the whole economy - represented as matrix. How outputs evolve to $x(t+1)$ is determined by the matrix $A$ - $A$ representes how previous outputs are used as inputs to produce new ouputs - so $A$ can be said as table of outputs produced from inputs. $$x(t+1) = A \cdot x(t)$$ Assume that all sectors/parts of the economy satisfy the equal growth rate in ouput: $$x(t+1) = (1+\alpha) \cdot x(t)$$

Also, $$p = (1+\pi) \cdot p \cdot A$$ where $p$ is equilibrium price, and $\pi$ is the uniform rate of profit. (The equation is saying that the whole cost of input plus the profit is the total market price of outputs.)

The first and second equation show that $$(A-(1+\alpha) \cdot I) \cdot x(t) = 0$$ where $I$ is identity matrix, and the third equation implies $$p \cdot (A^{-1} - (1+\pi) \cdot I) = 0$$

The question is, then John Blatt and Steve Keen say,

for stability, biggest eigenvalue (root) of $A$ and $A^{-1}$ must be less than 1.

and I am not getting this. Why do $A$ and $A^{-1}$ need to be like this? Is this related to the stability of equilibrium? If so, can anyone explain this using linear algebra stuffs?

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There's something strange here. If the biggest eigenvalue of $A$ is $c$, then the biggest eigenvalue of $A^{-1}$ is $1/c$, and the numbers $c$ and $1/c$ can't both be less than $1$. Maybe you can point us to the exact statement? –  Gerry Myerson Nov 20 '12 at 3:43
@GerryMyerson That's how Blatt and Keen say that general equilibrium approach is flawed. So it's strange, but it must be strange :) –  analysit Nov 20 '12 at 4:18

I've expressed some misgivings in a comment, so I'll only take up a part of the question here. You have $$x(t+1)=Ax(t)$$ from which it follows for $n=0,1,2,\dots$ that $$x(t+n)=A^nx(t)$$ If the biggest eigenvalue of $A$ has (modulus) $c\gt1$, then (it can be proved that) the entries of $A^n$ grow like $c^n$, that is, exponentially --- they blow up. It follows that $x(t)$ blows up as $t\to\infty$ (small print: provided $x(0)$ has a component in the direction of the eigenvector of $A$ corresponding to that biggest eigenvector). That, presumably, is considered unstable.