# Convergence of powers of matrix given convergence of the powers of its absolute value.

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?

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

## 1 Answer

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 $$B$$.)

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