# Diagonal entries of a Cholesky factorization

Let $I$ denote an identity matrix, $E$ denote the all-one matrix of dimension $k\times k$ and $c$ some positive real number. Define $X=B(I-cE)B^T$ where $B$ is given by

$B:=\begin{pmatrix} 1 &\beta_1 &\beta_2 &\ldots, &\beta_{k-1}\\ 0 &1 &0 &\ldots &0\\ 0 &0 &1 &\ldots &0\\ \vdots &\vdots &\vdots &\ddots &\vdots\\ 0 &0 &0 &0 &1 \end{pmatrix}$

i.e., $B$ is the identity matrix with its first row filled up with $k-1$ positive real numbers $\beta_1,\ldots,\beta_{k-1}$.

I am interested in the cholesky factorization of $X$, $X=LL^T$ where $L$ is a lower triangular matrix. In particular, I am interested in the case when all the diagonal entries of $L$ are equal.

So the question is: under what condition of $c$, there exist $k-1$ reals $\beta_1,\ldots,\beta_{k-1}$ such that $L$ has equal diagonal entries?

Here are some observations I have made:

1. It's not true that for any $c$, $L$ can be made having equal diagonals by adjusting $\beta$'s. Numerical results suggest that for $c$ larger than a certain value, this is always possible.
2. Clearly $\det(L)=\det(X)^{1/2}=(\det(B)\det(I-cE)\det(B^T))^{1/2}=\det(I-cE)^{1/2}$ which does not depend on $\beta$. Hence asking $L$ to have equal diagonals is equivalent to asking $L$ to have minimum trace possible because of the inequality $\operatorname{trace}(L)\geq K\det(L)^{1/K}$. We want the above inequality to be tight: this happens when all diagonals (also eigenvalues) of $L$ are equal. I do not know if this observation could help.
3. Of course we could calculate the diagonals of $L$ out using a brut force calculation. I think each entry $L_{ii}$ is a quadratic expression of $\beta$'s. However the expressions are too complicated to make any useful statement.
• Have you tried computing an upper triangular Cholesky decomposition $X=UU^T$ (which is easier because $U=BU_0$ for $1-cE = U_0 U_0^T$) and using that knowledge? – Johannes Hahn Apr 29 '14 at 15:52
• Thanks @JohannesHahn, it's a good idea. I tried the upper triangular Cholesky on $X$ (surprisingly I did not find many references on the upper triangular Cholesky decomposition), but I think there is no simple relationship between $L_{ii}$ (diagonals of the lower triangular matrix) and $U_{ii}$ (diagonals of the upper triangular matrix) with $X=LL^T=UU^T$ unless $X$ is very special. In my case, the upper triangular decomposition is even weirder, it can be seen that the diagonals of $U=BU_0$ do not contain the parameters $\beta$... – EEStudent Apr 30 '14 at 13:56