Wang–Chin partial order for comparing block matrix capacities

Question

In polar code design over binary erasure channels (generated by some subsets of rows of a Kronecker power of $\bbox[5px, #F0FFF0]{X= \begin{bmatrix} 1 & 1\\ 1 & 0\\ \end{bmatrix}}$), the Wang–Chin partial order [arXiv:2304.07667] compares binary strings $a, b\in\left\{ 0, 1 \right\}^{n}$ via: $$I_{0}\left ( x \right )= x^{2}, \quad I_{1}\left ( x \right )= 1- \left ( 1- x \right )^{2}, \quad I_{a_{1}a_{2}\ldots a_{n}}= I_{a_{2}\ldots a_{n}}\left ( I_{a_{1}}\left ( x \right ) \right ),$$ where $a\succcurlyeq b$ if $I_{a}\left ( x \right )\geq I_{b}\left ( x \right )$ for $x\in\left [ 0, 1 \right ]$.

Consider two $6\times 6$ block matrices $A$ and $B$, with $\ell\times\ell$ blocks, forming $6\ell\times 6\ell$ matrices: $$A= \begin{bmatrix} U & L & L & L & U & L\\ L & U & L & L & U & L\\ L & L & U & L & L & U\\ L & L & L & U & L & U\\ U & U & L & L & U & L\\ L & L & U & U & L & U\\ \end{bmatrix}, \quad B= \begin{bmatrix} U & L & L & L & U & U\\ L & U & L & L & U & U\\ L & L & U & L & U & L\\ L & L & L & U & L & U\\ U & U & U & L & U & U\\ U & U & L & U & U & U\\ \end{bmatrix},$$ where $U$ is the nilpotent matrix ($1$s on superdiagonal, $0$s elsewhere) and $L$ is idempotent ($1$s in first column, $0$s elsewhere): $$U= \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 & 0\\ 0 & 0 & 1 & 0 & \cdots & 0\\ 0 & \ddots & 0 & \ddots & 0 & 0\\ 0 & 0 & \cdots & 0 & 1 & 0\\ 0 & 0 & 0 & \ddots & 0 & 1\\ 0 & 0 & 0 & 0 & \cdots & 0\\ \end{bmatrix}_{\ell\times\ell}, \quad L= \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 & 0\\ 1 & 0 & \ddots & 0 & 0 & 0\\ \vdots & 0 & 0 & \ddots & 0 & 0\\ 1 & 0 & 0 & 0 & \cdots & 0\\ 1 & 0 & 0 & 0 & 0 & \cdots\\ 1 & 0 & 0 & \cdots & 0 & 0\\ \end{bmatrix}_{\ell\times\ell}.$$ Goal

Prove that $A$'s capacity (largest real eigenvalue) exceeds $B$'s for all $\ell> 1$, for all $\ell$, using Wang–Chin partial order and matrix decomposition, despite differing minimal polynomial degrees.

Matrix structure:

• "Pattern of $U$" matrices: $$K_{A}= \begin{bmatrix} 1 & 0 & 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0 & 1\\ 1 & 1 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 1 & 0 & 1\\ \end{bmatrix}, K_{B}= \begin{bmatrix} 1 & 0 & 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 0 & 1 & 1\\ 0 & 0 & 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 1\\ 1 & 1 & 1 & 0 & 1 & 1\\ 1 & 1 & 0 & 1 & 1 & 1\\ \end{bmatrix}.$$ We have $A= K_{A}\otimes U+ \left ( J- K_{A} \right )\otimes L, B= K_{B}\otimes U+ \left ( J- K_{B} \right )\otimes L$, where $J$ is the all-ones matrix.
• Decomposition for $B$ (via dharr's solution): Similarity transformation yields $A_{2}= {K}'\otimes\left ( U- L \right )+ {J}'\otimes L$, with characteristic polynomial factored into:

1. First factor: Degree $\ell$, from $U- L$.
2. Second factor: Degree $2\ell$, from $X\otimes\left ( U- L \right ), \bbox[5px, #F0FFF0]{X= \begin{bmatrix} 1 & 1\\ 1 & 0\\ \end{bmatrix}}$.
3. Third factor: Degree $3\ell$.

• For $A$, the minimal polynomial containing the largest eigenvalue has degree $\ell$.

Challenge

$A$’s minimal polynomial (degree $\ell$) vs. $B$'s third factor (degree $3\ell$) suggests different eigenvalue structures. Wang–Chin compares equal-length strings recursively, not polynomials of differing degrees.

Questions

Can we compare $A$ and $B$'s capacities by mapping submatrices (e.g., $U- L, L$ blocks) to synthetic channels with string capacities $I_{a_{i}}\left ( x \right )$, then adapt Wang–Chin’s partial order to prove $a\succcurlyeq b$ supporting $A$'s higher capacity?

Answer 1

The following didn't work out, but maybe someone can finish the argument.

The approach is to show directly that the largest eigenvalue of $A$ is larger than the largest eigenvalue of $B$. It is based on a theorem from ref. [1], concerning non-negative matrices $M$ and positive vectors $z$. $$ \min_{1\leq i\leq n}\frac{1}{z_{i}}\sum_{j=1}^{n}m_{ij}z_{j}\leq \rho _{M}\leq \max_{1\leq i\leq n}\frac{1}{z_{i}}\sum_{j=1}^{n}m_{ij}z_{j} $$ where $\rho _{M}$ is the spectral radius, in this case the largest positive eigenvalue. $A$ is an irreducible nonnegative matrix, which by Perron-Frobenius has a positive eigenvector $u$ corresponding to $\rho _{A}$. Numerically, we find that using the eigenvector of $A$ as the positive vector in the theorem applied to $B$ gives the required result, e.g., for $ l=4$ we have $\rho _{A}=5.9025$, and surprisingly find $\max_{1\leq i\leq n} \frac{1}{u_{i}}\sum_{j=1}^{n}b_{ij}u_{j}=5.9025,$ showing that $\rho _{B}\leq 5.9025=\rho _{A}$. (Actually $\rho _{B}=5.5807$.).

I propose that $u$ may be written in the form $[1,1,1,1,0,0]^{T}\otimes x+$ $ [0,0,0,0,1,1]^{T}\otimes y$ where $x$ and $y$ are $\ell $-vectors, i.e., the components of $u$ are 4 repeats of $x_{1},\ldots ,x_{\ell }$ and then 2 repeats of $ y_{1},\ldots ,y_{\ell }$. Directly calculating $Au$ and setting it equal to $ \lambda u$ componentwise gives $6\ell $ equations of which only $2\ell $ are distinct. The first 4 sets of $\ell $ equations are identical sets and the last two sets are identical. Taking the first set and the last set, we have $ 2\ell $ linear equations in the $2\ell $ unknowns $v=[x_{1},\ldots ,x_{\ell },y_{1},\ldots ,y_{\ell }]^{T}$ of the form $Qv=0,$ which has a solution when the determinant is zero; the determinant is the expected degree $2\ell $ polynomial in $\lambda $ that has $\rho _{A}$ as a root.

Let $B[i]$ denote the $i$th row of $B$. Compare $A[2\ell +1]$ ("LLULLU") with $B[2\ell +1]$ ("LLULUL") $$ A[2\ell +1] =[1,0,\ldots ;1,0,\ldots ;0,1,\ldots ;1,0,\ldots ;1,0,\ldots ;0,1,\ldots ;] \\ B[2\ell +1] =[1,0,\ldots ;1,0,\ldots ;0,1,\ldots ;1,0,\ldots ;0,1,\ldots ;1,0,\ldots ;] $$ where the dots ($\ldots $) indicate zeros and the semicolons serve only to better visualize the separation between the blocks of size $\ell $. Consider the dot product of $A[2\ell +1]$ with $u,$ which is $ x_{1}+x_{1}+x_{2}+x_{1}+y_{1}+y_{2}$ and we know this is equal to $\rho _{A}u_{2\ell +1}=\rho _{A}x_{1}.$ The dot product of $B[2\ell +1]$ with $u$ is $x_{1}+x_{1}+x_{2}+x_{1}+y_{2}+y_{1}$, which it the same as for $A[2\ell +1]$ and so is also equal to $\rho _{A}u_{2\ell +1}$. That is, $\frac{1}{ u_{i}}\sum_{j=1}^{n}b_{ij}u_{j}=\rho _{A}$ for $i = 2\ell +1$. The cases for $ i=2\ell +2,\ldots ,4\ell $ also give $\rho _{A}$ by the same argument. It remains to show that for other $i$ the weighted sums are smaller, so $\rho _{A}$ is the maximum.

Edit:

Example for $\ell =2$.

$A$ has largest eigenvalue $\rho _{A}=5.2644$ and corresponding eigenvector $[x_{1},x_{2},x_{1},x_{2},x_{1},x_{2},x_{1},x_{2},y_{1},y_{2},y_{1},y_{2}]^{T} $ with $x_{1}=0.339045$, $x_{2}=0.254566$, $y_{1}=0.322998$, $y_{2}=0.190163$ (normalized so sum of squares is 1).

$$ B = \begin{bmatrix} 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1\\ 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0\\ 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0\\ 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1\\ 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$

with $\rho _{B}=4.52099$.

The values of $\frac{1}{u_{i}}\sum_{j=1}^{n}b_{ij}u_{j}$ for $i=1,\ldots ,12$ are $$ (3x_{1}+x_{2}+2y_{2})/x_{1},3x_{1}/x_{2},(3x_{1}+x_{2}+2y_{2})/x_{1},3x_{1}/x_{2},(3x_{1}+x_{2}+y_{1}+y_{2})/x_{1},\\ (3x_{1}+y_{1})/x_{2}, (3x_{1}+x_{2}+y_{1}+y_{2})/x_{1},(3x_{1}+y_{1})/x_{2},(3x_{2}+x_{1}+2y_{1})/y_{1},x_{1}/y_{2},(3x_{2}+x_{1}+2y_{1})/y_{1},x_{1}/y_{2} $$ which for the $x_{i}$ and $y_{i}$ values above are $$ 4.8726,3.9956,4.8726,3.9956,5.2644,5.2644,5.2644,5.2644,4.5916,1.7829,4.5916,1.7829 $$ for which the maximum is $5.2644=\rho _{A}$.

[1]: R.A.Horn and C.R. Johnson, Matrix Analysis, Cambridge University Press, 1985, Thm 8.1.26, p. 493.