# Positive-definite non-symmetric matrix contraction operator?

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).

• define what a "positive definite" not symmetric matrix is, please Aug 18, 2021 at 10:05
• It's a wild guess, but probably a matrix $A$, such that $x^TAx>0$ for all $x\neq 0$. But maybe i am false. Aug 18, 2021 at 10:13
• Yes "Confusingly, the discussion of positive definite matrices is often restricted to only Hermitian matrices, or symmetric matrices in the case of real matrices". mathworld.wolfram.com/PositiveDefiniteMatrix.html Aug 18, 2021 at 10:16
• The result is false. Consider $\mathbb{R}^2$ with the $\infty$-norm and $A=\begin{pmatrix}1-\epsilon&1\\0&1-\epsilon\end{pmatrix}$, for example. Aug 18, 2021 at 12:49
• $(I-\alpha A)(1,-1)=(1+\alpha\epsilon,-(1-\alpha(1-\epsilon)))$ so $\lVert(1-\alpha A)(1,-1)\rVert_\infty=1+\alpha\epsilon>1=\lVert(1,-1)\rVert_\infty$ for all $\alpha>0$. Aug 18, 2021 at 15:00

The eigenvalues of $$I-\alpha A$$ are in the form $$1-\alpha\lambda$$ for $$\lambda\in\Lambda(A)$$. From the definition of positive definite matrix, $$Re(\lambda)>0$$ and $$|1-\alpha\lambda|^2 = (1-\alpha Re(\lambda))^2 + \alpha^2Im(\lambda)^2 = 1 - 2\alpha Re(\lambda) + \alpha^2|\lambda|^2$$ that is less than 1 whenever $$\alpha < 2 Re(\lambda)/|\lambda|^2$$ so you just need to impose $$\alpha < 2 \inf_\lambda Re(\lambda)/|\lambda|^2.$$ Notice that $$Re(\lambda)/|\lambda|^2$$ is always positive thanks to $$Re(\lambda)>0$$.