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Let $\hat{\bf H}$ be a $p\hat{N}\times p \hat{N}$ sparse matrix consisting of $p\times p$ blocks, where each block is of size $\hat{N}\times\hat{N}$. The values in $\hat{\bf H}$ is illustrated below (empty places are zero):

enter image description here

The infinite norm of $\hat{\bf H}$ is obviously 1, and I know spectral radius is no larger than any natural norm. My question is how do I prove the spectral radius of this matrix is smaller than 1?

I did a simple numerical experiment, and found the claim should hold. If $p \to \infty$ and $\hat{N} \to \infty$, then the spectral radius should approach to 1.

If we fix $p = 5$ and let $\hat{N}$ go from 5 to 100, we have

enter image description here

If we fix $\hat{N} = 10$ and let $p$ go from 5 to 100, we have

enter image description here

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    $\begingroup$ $\lambda I - H$ is diagonally dominant for $|\lambda| \geq 1$. $\endgroup$ – Federico Poloni Dec 10 '18 at 8:57
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    $\begingroup$ Imagine a (complex) eigenvector with absolute value of eigenvalue equal to 1. Consider the coordinate with maximal absolute value. Get new coordinates with the same absolute value. Proceed. Also look for general statements, keywords should be Perron - Frobenius theorem and recursive Markov chain. $\endgroup$ – Fedor Petrov Dec 10 '18 at 9:01
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    $\begingroup$ @FedericoPoloni but not strictly $\endgroup$ – Fedor Petrov Dec 10 '18 at 9:02
  • $\begingroup$ @FedorPetrov Irreducibility considerations can be used to relax the strictness requirement (and the proof I know is exactly the one that you sketch). See e.g. here. $\endgroup$ – Federico Poloni Dec 10 '18 at 9:05
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    $\begingroup$ In simple terms, this matrix is conjugate to a block diagonal matrix with tridiagonal blocks (with each block having 0’s on the diagonal) so that each block is irreducible and sub-stochastic. Hence the spectral radius is the maximum of the spectral radius of the blocks, which are all strictly less than 1. $\endgroup$ – Anthony Quas Dec 10 '18 at 14:23

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