If I have all the elements of an $N\times N $ complex valued matrix except for the diagonal elements (e.g. I have $N^2-N$ elements out of the $N^2$ elements), is there anyway I could complete those diagonal elements if I know the rank of the matrix $r$ ?
How will the solution (if any) be different if the matrix was noisy (i.e. the matrix is the sum of a rank-$r$ matrix + noise)?
Thanks.
All matrices in the following are $N \times N$.
Let us give the name $V$ to the set of matrices having all their diagonal entries equal to $0$. Let $W$ be the set of diagonal matrices.
The question can be reformulated so:
The answer is no; there exist even some $A \in V$, such that, for any $B \in W$, rank$(A+B) \geq N-1.$ Here is an example of such an $A$ :
$$A=\begin{pmatrix}0&1&0&0&\cdots&0&0\\0&0&1&0&\cdots&0&0\\0&0&0&1&\cdots&0&0\\\cdots&&&&&&\cdots\\\cdots&&&&&&\cdots\\0&0&0&0&\cdots&0&1\\0&0&0&0&\cdots&0&0 \end{pmatrix}$$
(Entries $A_{ij}=0$ but the $A_{i,i+1}, i=1,... N-1$ equal to $1$).
Whatever $B=diag(a_1,a_2,\cdots a_n) \in W$, rank$(A+B)\geq N-1$.
The reason is that the $(N-1) \times (N-1)$ upperright block of $A+B$ is a triangular matrix with ones on its diagonal, thus with determinant 1, therefore invertible in all cases.