# 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?

• For the future, Shawn: don't include a description of your edit in the post itself. When you make an edit, there's a separate text box for that. Feb 5, 2015 at 1:54

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 $\theta\in S^1$.
• Thank you for your answer! I really appreciate it. However, I think I was trying to ask about $|\mathbf F|$ in your notation when I was referring to $\mathbf U$. I tried to make it clearer by adding some definitions and a reference link in the OP. My apologies on the ambiguity in the OP! Feb 4, 2015 at 20:38
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 $3, 3, 1$.