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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).

Robin Carter
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  • define what a "positive definite" not symmetric matrix is, please – Exodd Aug 18 '21 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. – Jochen Aug 18 '21 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". https://mathworld.wolfram.com/PositiveDefiniteMatrix.html – Robin Carter Aug 18 '21 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. – user10354138 Aug 18 '21 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 $\alpha>0$. – user10354138 Aug 18 '21 at 15:00
  • @user10354138 I misread A somehow the first time and confused the issue, apologies. A as defined does indeed seem to be positive definite and not an ∞-norm contraction. Would the result have been True if I had specified an L2-norm for the contraction definition in my question? I can see for instance that a rotation with a slight inward spiral would indeed fail ∞-norm starting from $(1,-1)$ e.g. after 45 degrees. But such an inward spiral would still be an operation that when applied repeatedly leads to a fixed point in the limit. – Robin Carter Aug 18 '21 at 16:47
  • For the 2-norm, I think you need an additional constraint, something along the line orthogonal decomposition into eigenspaces of $A$ (or the symmetric part and antisymmetric parts of $A$ commute). – user10354138 Aug 19 '21 at 10:04
  • Hmmm, the book doesn't offer any. It says that all that is needed to apply the Banach fixed-point theorem is that A is positive-definite (but I don't see why). Here's a link to the book if you are willing to look at it, jump to "Theorem 6.1" on p58. The relevant and self-contained part is the short section from "Therefore we define the function..." to "Therefore it suffices to show that A is positive definite". Also, I'm curious if can you fault @Exodd logic because I'm unable to. Thanks again. http://old.ins.sjtu.edu.cn/files/paper/20191021090916_Book%20(3).pdf – Robin Carter Aug 19 '21 at 10:27
  • @RobinCarter I supposed (and the same seems to do the book) that "contraction" means "there exists a norm in which the operator is a contraction". Since you can always find a norm close enough to the spectral radius, it is enough to impose the spectral radius less than 1. Obviously, if you choose a single norm, then it may fail, but there always exists one where it works – Exodd Aug 19 '21 at 11:37
  • @Exodd And I think I'm right in saying that a contraction under any norm is sufficient to reach a fixed point in the limit of repeated application of the operator. Just to confirm: If the spectral radius is less than 1 as you've shown, then that's sufficient to show that under the L2-norm this operator is a contraction? No additional constraint is required as user10354138 suggests? Thanks again. I'm learning a lot! – Robin Carter Aug 19 '21 at 14:53
  • @RobinCarter It may be false for the L2-norm. What's true is that there exists a norm for which it is a contraction. – Exodd Aug 19 '21 at 14:55

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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 $Re(\lambda)>0$.

Exodd
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  • Thanks but as I understand it, the existence of an eigendecomposition is only guaranteed if the positive definite matrix is symmetric. I've also edited the question to make clear that the matrix of interest is real if that helps. – Robin Carter Aug 18 '21 at 10:48
  • OK I've now read and understood that all square matrices are guaranteed to have complex eigenvalues and eigenvectors. – Robin Carter Aug 18 '21 at 12:06
  • @RobinCarter Indeed. In fact you can define a contraction just by requiring that all eigenvalues have norm less than 1 – Exodd Aug 18 '21 at 13:29