I'm currently reading the article "Full Banach Mean Values On Countable Groups" by Harry Kesten (see http://www.mscand.dk/article/view/10568/8589) for my bachelor's thesis.
For a given matrix $M$, the spectrum of $M$ is defined as the set of all complex numbers $\alpha$ such that $M - \alpha I$ does not have an inverse. ($I$ is the unit matrix of the same dimension as $M$)
One defines $\lambda(M)$ as follows: $$\lambda(M)=\text{sup}\lbrace |\alpha|:\alpha \in \text{spectrum(M)} \rbrace $$
At page 153, the infinite matrix $C^{(1)}=(c_{ij}^{(1)})$ is defined such that $c_{11}^{(1)}=2k$ (for a certain $k>0$), $c_{r1}^{(1)}=c_{1r}^{(1)}=k$ for all $r=2,..,m$, where $m$ is an integer and $m \geq 2$ and $c_{ij}^{(1)}=0$ otherwise.
It says that a trivial computation gives you the following statement: $$\lambda(C^{(1)})=k(1+m^{\frac{1}{2}})$$
I have trouble understanding this, especially since $C^{(1)}$ isn't finite-dimensional... Can someone give me some insight?
It's not finite dimensional, but writing out the infinite matrix corresponding to $C-\lambda I$ shows that it is a "block" matrix, where the upper left block is the $m\times m$ matrix given by $c-\lambda I$, the bottom right block is the infinite matrix $-\lambda I$, and the other two blocks are zero.
Simple arguments as in finite dimensional linear algebra then show that $v$ satisfies $(C-\lambda I)v=0$ if and only if it is non-zero only in the first $m$ components and the vector $u$ given by its first $m$ components satisfies $(c-\lambda I_m)u=0$.
Similarly, if $C-\lambda I$ fails to surject on some vector $v$, then $c-\lambda I_m$ must fail to surject onto the vector $u$ given by the first $m$ components of $v$; and conversely, for any $u$ upon which $c-\lambda I$ fails to surject, $C-\lambda I$ fails to surject onto the vector whose first $m$ components are given by $u$ and whose remaining components are zero.
Thus the problem is reduced to finding the eigenvalues of the $m\times m$ matrix given by $c_{ij}$. I calculated that $det(c-\lambda I)=(2k-\lambda)(-\lambda)^{m-1}-k^2(-\lambda)^{m-2}(m-1)$. The roots of this are $0$ and $k(1\pm m^{\frac{1}{2}})$.