# Show the spectral radius of a matrix is smaller than 1

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):

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

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

• I don't understand the pattern of the matrix $\hat H$, could you give and example with $N=1$ and $p=6$ ? and then with $N=2$ and $p=2$ ? – P. Quinton Dec 10 '18 at 7:15
• Thanks. If $\hat{N}=1$ and $p=6$, then $\hat{\mathbf{H}} = \left( {\begin{array}{*{20}{c}} {}&{\rm{1}}&{}&{}&{}&{}\\ {\rm{1}}&{}&{\rm{1}}&{}&{}&{}\\ {}&{\rm{1}}&{}&{\rm{1}}&{}&{}\\ {}&{}&{\rm{1}}&{}&{\rm{1}}&{}\\ {}&{}&{}&{\rm{1}}&{}&{\rm{1}}\\ {}&{}&{}&{}&{\rm{1}}&{} \end{array}} \right)$. If $\hat{N}=2$ and $p=2$, I think the matrix is $\left( {\begin{array}{*{20}{c}} {}&{}&{\rm{1}}&{}\\ {}&{}&{\rm{2}}&{}\\ {}&{\rm{2}}&{}&{}\\ {}&{\rm{1}}&{}&{} \end{array}} \right)$ – Tony Dec 10 '18 at 7:21
• Ok, perfect for the p=6 example, for the other one, I'm still confused, could you place all the zeroes ? Are the blocs diagonal ? – P. Quinton Dec 10 '18 at 7:25
• Sure. I think it is $\left( {\begin{array}{*{20}{c}} {\rm{0}}&{\rm{0}}&{\rm{1}}&{\rm{0}}\\ {\rm{0}}&{\rm{0}}&{\rm{2}}&{\rm{0}}\\ {\rm{0}}&{\rm{2}}&{\rm{0}}&{\rm{0}}\\ {\rm{0}}&{\rm{1}}&{\rm{0}}&{\rm{0}} \end{array}} \right)$ with zeros. Looks not exactly block diagonal. – Tony Dec 10 '18 at 7:26