I'm performing an optimization where for a first order condition I wish to solve an equation in square, say $n\times n$, matrices of the following form:

$$ I \circ [\Lambda - R^{-1}\Lambda^{-1}S] = 0, $$

where $\circ$ denotes the elementwise product, i.e. I need to set the diagonal of the expression in brackets to zero. Assume $\Lambda$ is a diagonal matrix, $R$ is positive definite and $S$ is positive semi-definite (known to be of less than full rank).

I tried iteratively for $\Lambda^0 = I$ and $k =1, 2,\dots$ solving $\Lambda^{k+1} = I \circ (R^{-1}[\Lambda^{k}]^{-1}S)$ but did not get convergence.

I wonder if there is a closed form solution, or if there is a theorem about the convergence of such an iterative scheme that I can check?


I assume that the unknown is the matrix $\Lambda$. Let $\Lambda=diag(1/a_1,\cdots,1/a_n)$ where $a_i\not= 0$, $R^{-1}=[r_{i,j}],S=[s_{i,j}]$. The problem reduces to the matricial equation:

$[1/a_1,\cdots,1/a_n]^T=\sum_j r_{i,j}a_js_{j,i}=U[a_1,\cdots,a_n]^T$ where $U=[u_{i,j}]$, with $u_{i,j}=r_{i,j}s_{j,i}$ that is, $U=R^{-1}\circ S$ a symmetric $\geq 0$ matrix. There are $2^n$ solutions in $\mathbb{C}^n$; I have an example with $n=3$ which admits $8$ solutions in $\mathbb{R}^3$.

Numerically. Let $X_k=[x_{k,1},\cdots,x_{k,n}]^T,Y_k=[1/x_{k,1},\cdots,1/x_{k,n}]^T$. By the Newton's method:

$X_{k+1}=X_k-{Z_k}^{-1}(UX_k-Y_k)$ where $Z_k=U+diag({x_{k,1}}^{-2},\cdots,{x_{k,n}}^{-2})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.