One way to compute the spectral radius of a matrix is that it is the square root of the spectral radius of (the star denoting the hermitian adjoint) (see correction below). If you apply that idea to the matrix then you get

This splits up as a direct sum

If I have done the computations correctly, then each of these matrices splits up into smaller ones, and you end up with a direct sum of copies of each of the one- and two-dimensional matrices

The largest eigenvalue in those matrices is So it appears that the spectral radius of each of the matrices is the golden ratio

Edit.Serious case of brain failure: the spectral radius of isnotequal to the square root of the spectral radius of That would give you the operator norm of However, the spectral radius is dominated by the norm, so (unless I have made any other silly mistakes) the spectral radius of is at most

Second edit.Numerical computations with X(2) and X(3) make it look as though X(n) has just a single positive eigenvalue, which appears to be increasing towards as n increases. Can't prove that though.