I asked this question on math.stackexchange and haven't received an answer in two weeks, so I'm repeating it here.

Let

$$ H=\left(\begin{array}{cccc} 0 & 1/2 & 0 & 1/2 \cr 1/2 & 0 & 1/2 & 0 \cr 1/2 & 0 & 0 & 1/2\cr 0 & 1/2 & 1/2 & 0 \end{array}\right), $$

$K_1(\alpha)=\left(\begin{array}{c}1 \\\\ \alpha\end{array}\right)$ and consider the sequence of matrices defined by $$ K_L(\alpha) = \left[H\otimes I_{2^{L-2}}\right]\left[I_2 \otimes K_{L-1}(\alpha)\right], $$ where $\otimes$ denotes the Kronecker product, and $I_n$ is the $n\times n$ identity matrix.

I am interested in the limiting behaviour of the singular values of $K_L(\alpha)$ -- in particular, $K_L(0)$ -- as $L$ tends to infinity. Some calculation indicate that the $2^L\times 2^{L-1}$-matrix $K_L$ has $L$ non-zero singular values and that, for any positive integer $k$, the $k$ largest singular values converges to some limit.

Question: Can this limit be described in terms of the matrix $H$?

I did some experiments and it seems that the limiting behaviour of the singular values of $K_L$ does not only depend on the matrix $H$, but also on the initial value $K_1(\alpha)$. This makes it unlikely for fixed-point arguments to work in this setting.

I also tried to obtain combinatorical expressions for the coefficients in the characteristic polynomial $\chi_L^\alpha(\lambda)$ of $K_L(\alpha)K_L(\alpha)^T$ but was successful only for the three highest non-trivial powers of $\lambda$.

Edit:

The analysis of $\Sigma(\alpha):=\lim_{L\to\infty}\sigma_1(K_L(\alpha))$ as a function of $\alpha$, as suggested by Suvrit, seems to be a good idea. Numerical calculations indicate that, asymptotically, $$ \Sigma(\alpha)\sim \Sigma(0)\left(.3540+\alpha\right),\quad \alpha\to\infty,\quad \Sigma(0)\approx .8254, $$ and that $\Sigma(\alpha)$ has a minimum at $\approx(-.2936,.7696)$.

I do not see yet, however, if this can be used to compute $\Sigma(0)$ more precisely.

Edit:

Using the improved bound $\sigma_1(K_L)\leq \frac{1}{2}\sqrt{3+2\alpha +3 \alpha ^2}$, which is sharp for $\alpha=\pm 1$, we can deduce that $d/d\alpha \Sigma(-1)=-1$, and $d/d\alpha \Sigma(1)=2$.