I am trying to prove the following using basic eigenvalue/eigenvector results but getting stuck. Any help will be greatly appreciated.
Suppose we have a matrix $A_{n \times n}$ with all eigenvalues in $(0,1)$, i.e. $0 < \lambda_{(1)} \le \ldots \le \lambda_{(n)} < 1$.
Is it true that for any vector $x \in \mathbb{R}^n$, multiplying by $A$ will shrink $x$ elementwise in absolute value, i.e. will $|(Ax)_i| \le |x|_i$ for all $1 \le i \le n$?
(I can show that the entire vector will be shrunk $\rVert Ax \lVert < \rVert x \lVert$ in some norm. Also, having the spectral radius / maximum eigenvalue $\rho(A) < 1$ will imply $\lim_{k \to \infty} A^k x \to 0$.)
Best Answer
The answer to the question is no. While you have correctly assessed that the norm of the resulting vector is decreasing, the individual entries can increase.
Take e.g. $$ A = \begin{pmatrix} 1/2 & 1/5 \\ 1/5 & 1/2 \end{pmatrix}, \qquad x = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$ and you will find $(Ax)_2 = 1/5 > 0 = x_2$.