Let $H$ be a $m\times n$ matrix that is not full rank. Hence $H^T H \neq \mathbb{1}_{m\times m}$. Can we say anything about the diagonal entries, or the trace, of matrix products of the form:
$P_0 = H^T H$
$P_1 = H^T(H^T H)H$
$P_2 = H^T(H^T(H^T H)H)H$
$\vdots$
$P_n = H^T P_{n-1} H$
Can we estimate the trace of $P_n$ as a function of $n$ ?
Context: This is unrelated as such to the question, but If anyone is curious:
The matrix $H$ to which I want to apply the above result is the parity check matrix for a linear code. The adjacency matrix of the tanner graph which describes the linear code is simply:
$A = \begin{pmatrix} 0_{m\times m} & H \\ H^T & 0_{n\times n} \end{pmatrix}$
and $A^2$ can be expressed as
$A^2 = \begin{pmatrix} H H^T & 0_{n\times n} \\ 0_{n\times n} & H^T H \end{pmatrix} = \begin{pmatrix} P^T_0 & 0_{n\times n} \\ 0_{n\times n} & P_0 \end{pmatrix}$
and more generally
$A^{2^{(n+1)}} = \begin{pmatrix} P^T_n & 0_{n\times n} \\ 0_{n\times n} & P_n \end{pmatrix}$
If the trace of the above matrix is $\text{Tr}(A^{2^{(n+1)}})$, then it is equal to the number of cycles in the Tanner graph of length $2^{(n+1)}$. It is this quantity that I want to comment about.