I have an orthonormal vector $C$ of size ($r*n$). I want to pre-multiply the Hermitian transpose of the vector C, by a vector D (size ($r*n$)). The aim is to find vector D such as the result is an identity matrix (size ($n*n$)) or close to an identity matrix.
$D * C^{H} \approx I$.
Please note that $r < n$. Is it possible that the resultant $n*n$ matrix has a higher rank than $r$?
Thank you.
The product $AB$ has a rank that's no greater than the smaller of rank $A$ and rank $B$. Hence in your example, the rank of $D*C^H$ will be at most $r < n$. So it'll never be the identity.
When you say "orthonormal vector", I assume you mean "orthonormal matrix"; if we assume that it's the rows of your matrix that are orthogonal and are unit vectors, then your best choice for $D$ is simply $D = C$, for then $$ D * C^H = \pmatrix{I_r & 0 \\ 0 & 0} $$ where the $0$s denote block matrices of size $r \times n-r$, $n-r \times r$, and $n-r \times n -r $, in ordinary reading order, and $I_r$ denotes the $r \times r $ identity matrix.