Let $\bf A$ be an $n \times n$ matrix with rank $r$ where $r<n$. How can I get a full-rank approximation for $\bf A$? In other words, I want to find the rank-$n$ $\bf X$ that minimizes the Frobenius norm $\|\bf A-\bf X\|_{\text{F}}$.
Full-rank approximation to a square matrix
constrained optimizationmatrixoptimizationranks
Best Answer
As the simplest example, if we choose $B = t I_n$, then $X = A + B$ and we can make $\| A - (A + B) \|_F = \| B \|_F$ arbitrarily small according to the choice of $t$ such that $|t| > 0$. (But we can't minimize $t$ because the minimum value of $\| B \|_F$ occurs at $t=0$, for which $B$ is not full-rank.)
A wealth of interesting solutions exist for $B$ that are not scalar multiples of the identity matrix.