Find real s.p.d. matrix $C$ to which there is no real $F$ with $F^{\sf T}F=C$

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.

• Cholesky decomposition – Will Jagy Aug 23 '18 at 19:04
• Every real, symmetric, positive definite matrix $C$ has a real, symmetric, positive definite square root $F$. Since $F$ is symmetric, $F^T F = F^2 = C$. – Robert Israel Aug 23 '18 at 19:05

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

$D=Q\sqrt{\Lambda} Q^{T}$.