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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?

David Z
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Shawn Wang
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  • 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. – David Z Feb 05 '15 at 01:54

2 Answers2

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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 $\theta\in S^1$.

Phoenix87
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  • 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! – Shawn Wang Feb 04 '15 at 20:38
  • @ShawnWang Sorry I didn't have the time to reply to this but meanwhile I see that enzotib has written what I wanted to write in reply to your comment, so I refer you to his/her answer. – Phoenix87 Feb 04 '15 at 23:32
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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 $3, 3, 1$.

enzotib
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    This is certainly true from a mathematical perspective. But is it true in continuum mechanics (which may or may not impose additional constraints on the matrix)? I'm not saying it is or isn't, but to correctly answer the question, you would also need to show that such a matrix can arise in continuum mechanics. – tpg2114 Feb 04 '15 at 21:22
  • This is likely to be the case – Phoenix87 Feb 04 '15 at 22:10
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    @tpg2114: there is not other restriction that I know of. Also, the difference from this matrix and the diagonal one is a simple change of basis, and one can always choose the reference frame at will. – enzotib Feb 05 '15 at 07:54