It's not possible to do that, since no matrix of that form is positive definite.
Claim: Given a 1xn (real, n>1) matrix A, the symmetric matrix M = A'A is not positive definite:
Proof: By definition, M is positive definite iff x'Mx > 0 for all non zero x. That is, iff x'A'Ax = (Ax)'Ax = (Ax)^2 = (A_1 x_1 + ... + A_n x_n) > 0 for all non zero x.
Since the real values A_i are linearly dependent, there exists x_i, not all zero, such that A_1 x_1 + ... + A_n x_n = 0. We found a non zero vector x such that x'Mx = 0, so M is not positive definite.
A different proof, that can be applied directly to the complex case is this: Let A be an 1xn (complex, n>1) matrix. Positive definiteness implies invertibility, so M = A*A must have full rank to be positive definite. It clearly has rank 1, so it's not invertible and thus not positive definite.