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I have an iterative map $X^{n+1} = T(X)$ which maps a k-tuple of vectors in $\mathbb{R}^n$, that is, $X = (\bf {x_1,x_2,x_3,\dots,x_k})$, where $\bf x_i\in\mathbb{R}^n$, into another $k$-tuple, i.e. $T(X) = ({y_1,y_2,y_3,\dots,y_k})$, where $\bf y_i\in\mathbb{R}^n$.

I am interested in fixed points of this map $T$, and the paper that I am reading says that:

...the eigenvalues of $\dfrac{\partial T}{\partial Z}(Z)$ at a fixed point $Z^*$ are real and and less than 1, a necessary condition for $T$ to be a local contraction

But, shouldn't this say a sufficient condition for $T$ to be a local contraction?

What I know is that:

$$ \text {T is a local contraction at a fixed point Z^* if and only if} \\ \text {the eigenvalues of the Jacobian of T have modulus strictly less than 1} \tag 1$$

So, I do not understand why the eigenvalues being real is a necesary condition. Or, is my condition (1) wrong?

user74261
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1 Answers1

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What you write in (1) looks correct to me.

What's in the yellow box is wrong. For a counterexample, the linear contraction $$T(x_1,x_2) = \biggl( -\frac{x_2}{2}, \frac{x_1}{2} \biggr) $$ has complex eigenvalues $\pm \frac{1}{2} \, i$.

Lee Mosher
  • 120,280
  • By the way, I am still not certain about what I write in (1). Sufficient condition is clear to me, but on the other hand, is it necessary for a local contraction to have the modulus of all its eigenvalues strictly less than 1? What if one of the eigenvalues has modulus exactly equal to 1? Wouldn't it not be a local contraction? – user74261 May 23 '15 at 00:37