Hello,


If I have a matrix A, is it possible to construct a positive definite matrix M with the same range as range(A')? I am trying to use the property x'Mx > 0 to remove an absolute value constraint in an optimization problem.


Thanks



constructing a positive definite basis
If $M$ is a symmetric and positive definite matrix of size $n$ by $n$, then the range of $M$ is $R^{n}$. Unless $A^{T}$ also has $R^{n}$ as its range, you can't make this happen. 


You'd have a much better chance of getting a useful answer if you asked your original question about dealing with the absolute value constraint. 


