Let $T$ be a mapping from $\mathbb{R}^n \to \mathbb{R}^n$.
Fix $x^\star \in \mathbb{R}^n$, and suppose that the Jacobian matrix of $T(x) $ at $x = x^\star$is symmetric.
Then, I know that if all the eigenvales of the Jacobain of $T$ at $x = x^\star$ have absolute values less than 1, then $T$ is a (local) contraction at $x = x^\star$.
But, is the converse true? Is it necessary for $T$ to have all the eigenvalues of its Jacobain less than 1 in absolute value for $T$ to be a local contraction?
It is certainly true, if $T'$ is continuous at $x^*$.
Assume $T$ admits a Lipschitz constant $L<1$ w.r.t the norm $\|\cdot\|$ on a neighborhood $U$ of $x^*$. Let $h$ such that $\|h\|=1$ and $\| T'(x^*)h \| = \| T'(x^*) \|$. Then, for $t > 0$ and sufficiently small, by mean value theorem, we have some $\tau\in(0,t)$ such that $$ 1 > L \ge \frac{\|T(x^* + th) - T(x^*)\|}{\|th\|} = \left\| T'(x^* + \tau h)h \right\| \to \| T'(x^*)h \| $$ for $t\to 0$.
Now, as the spectral radius is a lower bound for any operator norm, we proved the claim.