I have a matrix A and a matrix B such that $B_{i, j} = |A_{i, j}|$. I am given that all of the eigenvalues of B have magnitude less than 1, and therefore: $ \displaystyle \lim_{k \to \infty} B^k = 0$ through analysis of the Jordan normal form.

What I want to prove now is that $\displaystyle \lim_{k \to \infty} A^k = 0$ and that, therefore, all the eigenvalues of A also have magnitude less than 1. At first, this seems intuitively true, but I'm struggling to find a formal proof for the convergence of $A^k$. Can someone offer a formal explanation?

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    $\begingroup$ in general $|\sum a_k b_k| \le \sum_k |a_k| |b_k|$. Compare the $ij$ entry of $A^2$ with $B^2$. $\endgroup$ – copper.hat Feb 23 at 2:46

Hint: Try and show (e.g. by induction on $k$) that for all $i,j$, we have $\left|A^{k}_{i,j}\right| \leq B^{k}_{i,j}$ for any $k\in \mathbb{Z}^{+}$. You can use the triangle inequality and definition of matrix multiplication. ($A^k_{i,j}$ refers to the $i,j$ element of $A^{k}$, and similarly for $B$.)

  • $\begingroup$ Thanks! That did it. $\endgroup$ – Jordan Field Feb 23 at 3:06
  • $\begingroup$ You're welcome! $\endgroup$ – Minus One-Twelfth Feb 23 at 3:06

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