Let A be positive definite and not symmetric (edit: and real).
Why is $I - \alpha A$ a contraction for sufficiently small $\alpha$?
I see why this is the case if A is symmetric since it will have an eigendecomposition and give:
$Q(I - \alpha \Lambda)Q^T$
Where $\Lambda$ is diagonal with positive eigenvalues on the diagonal. But what can be said if A if not symmetric? Is it valid to use a singular value decomposition instead and say that the singular values must be positive and somehow argue that way?
The question comes from reading Reinforcement Learning by Sutton & Barto and the boxed text Proof of Convergence of Linear TD(0) and also Reinforcement Learning: Algorithms and Convergence by Heitzinger (Theorem 6.1).