Can $U_{ij}$ or $v_{ij}$ in continuum mechanics be negative? In continuum mechanics, we have the deformation gradient $\mathbf F$ to be:
$$d\mathbf x = \mathbf F d \mathbf X$$
And then, we do a polar decomposition (A good reference here would be http://www.continuummechanics.org/cm/polardecomposition.html), we may get:
$$\mathbf F = \mathbf{RU} = \mathbf{vR}$$
where $\mathbf R$ is the rotation tensor, and is real, proper orthogonal; $\mathbf U$ and $\mathbf v$ are right and left stretch tensors, and they are both real, symmetric, positive-definite matrices.
And my question is: I know that $\mathbf U$ and $\mathbf v$ are real, symmetric, positive-definite matrices. My actual question is - can components of $\mathbf U$ and $\mathbf v$, i.e., $U_{ij}$ or $v_{ij}$ ever be negative? They seem to me that they are always non-negative, because a "stretch" can only be from 0 to infinity, and cannot have negative length in reality.
Would this be right or wrong?
 A: I struggle to understand the notation in this question, since 1. I'm not familiar with the fine details of the theory of continuum mechanics and 2. the OP hasn't defined all of the symbols.
Nonetheless, the second equation suggests a polar decomposition, which usually of the form $\mathbf F = \mathbf U|\mathbf F|$, where $\mathbf U$ is a unitary (orthogonal matrix if everything is real) and $|\mathbf F|$ is a positive matrix (i.e. a symmetric matrix with eigenvalues in the non-negative real line). As an example for the matrix $\mathbf U$, consider
$$\begin{bmatrix}\cos\theta & \sin\theta & 0\\-\sin\theta & \cos\theta & 0\\0&0&1\end{bmatrix}.$$
This is a valid possibility for the matrix $\mathbf U$, which clearly has at least one negative component for any angle $\theta\in S^1$.
A: A positive definite matrix has all eigenvalues positive, but one or more of the matrix elements can be negative, for example
\begin{pmatrix}
 2 & -1 & 0 \\
-1 &  2 & 0 \\
 0 &  0 & 3
\end{pmatrix}
has eigenvalues $3, 3, 1$.
