Say $C$ is a real, symmetric and positive definite matrix. Then it has a square root $F$ such that $F^2=C$. However, I wonder if one can construct $C$ in any way such that a real (optional: positive definite) $F$ with $F^{\sf T}F=C$ does not exist.
For googlers: this is a problem of finite strain theory where $C$ is the right Cauchy-Green tensor and $F$ is the deformation gradient.
Every symmetric and positive definite real matrix $C$ has a Cholesky factorization
$C=R^{T}R$
where $R$ is a non-singular upper triangular matrix. The Cholesky factor is your desired $F$. There are many algorithms for computing $R$.
Furthermore, every symmetric and positive definite real matrix $C$ has a real and symmetric and positive definite square root matrix $D$ with
$C=DD=D^{T}D$.
$D$ can be computed from the orthogonal diagonalization of $C$. If
$C=Q\Lambda Q^{T}$
where $Q$ is orthogonal and $\Lambda$ is a diagonal matrix of the eigenvalues of $C$, then
$D=Q\sqrt{\Lambda} Q^{T}$.