The Invertible Tensor Theorem

Every chapter of part 2 adds a way of saying “$\mathbf{A}$ can be undone”. They are collected here. For $\mathbf{A}\in \mathrm{Lin}$ the following are equivalent:

1. $\mathbf{A}$ is invertible: there is ${\mathbf{A}}^{-1}$ with $\mathbf{A}{\mathbf{A}}^{-1}={\mathbf{A}}^{-1}\mathbf{A}=\mathbf{I}$ (Theorem 2.5.2).
2. Only $0$ is sent to $0$: $\mathbf{Au}=0⟹\mathbf{u}=0$.
3. $\mathbf{A}$ sends every basis to a basis.
4. $\mathbf{A}$ sends independent pairs to independent pairs: $\mathbf{u}\times \mathbf{v}\ne 0⟹\mathbf{Au}\times \mathbf{Av}\ne 0$.
5. $\det \mathbf{A}\ne 0$: $\mathbf{A}$ does not crush volumes to zero (Definition 2.5.5).
6. The columns of $[\mathbf{A}]$ are a basis.
7. The rows of $[\mathbf{A}]$ are a basis, since $\det {\mathbf{A}}^{\top }=\det \mathbf{A}$ (Theorem 2.5.8).
8. ${\mathbf{A}}^{\top }$ is invertible, with $({\mathbf{A}}^{\top }{)}^{-1}=({\mathbf{A}}^{-1}{)}^{\top }$ (Theorem 2.5.4).
9. $0$ is not an eigenvalue of $\mathbf{A}$: $\det (\mathbf{A}-0\mathbf{I})\ne 0$ (Recipe 2.7.2).
10. ${\mathbf{A}}^{\top }\mathbf{A}$ is positive definite (Exercise 2.7.2).

And when it fails, the picture is always the same: something nonzero is crushed to $0$, the unit cube goes flat, and no tensor can bring it back (Figure 2.5.2).

In kinematics

The admissibility condition $\det \mathbf{F}>0$ (Definition 3.2.4) is statement 5 with a sign. Through this theorem it gives, at every point of the body:

• ${\mathbf{F}}^{-1}$ exists, and is the gradient of the inverse deformation;
• no material fibre is crushed to zero length (statement 2): $|\mathbf{Ft}|>0$;
• $\mathbf{C}={\mathbf{F}}^{\top }\mathbf{F}$ is positive definite (statement 10), so it has a square root $\mathbf{U}$ and the polar decomposition exists (Theorem 3.6.5);
• the cofactor ${\mathbf{F}}^{\ast }$ is defined, and Nanson’s formula makes sense (Theorem 3.4.2).