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I have two deformation gradient tensors, corresponding to the simple shear of a cube; one at time t (say $\textbf{F0}$) and another at time $t+\Delta t$ (say $\textbf{F1}$). The basis of $\textbf{F1}$ is a mixed basis, $ e^{t+\Delta t}_i \otimes E^0_j $. The $e^{t+\Delta t}_i$ basis is attached to the material point/Gauss point/integration point and keeps rotating with it. The $E^0_j $ basis is that at the initial state (t=0) or in the undeformed configuration. Similarly, $\textbf{F0}$ is in the mixed basis $ e^{t}_i \otimes E^0_j $.

Next, I do a polar decomposition of the two and extract the orthogonal rotation tensors, $\textbf{R0}$ and $\textbf{R1}$, corresponding to $\textbf{F0}$ and $\textbf{F1}$, respectively. The basis corresponding to $\textbf{R0}$ and $\textbf{R1}$ are $ e^{t}_i \otimes E^0_j $ and $ e^{t+\Delta t}_i \otimes E^0_j $.

To obtain $\textbf{F0}$ and $\textbf{F1}$ in the completely Lagrangian basis, i.e., in $ E^0_i \otimes E^0_j $ basis, I pre-multiply $\textbf{F0}$ and $\textbf{F1}$ with $\textbf{R0}^T$ and $\textbf{R1}^T$, respectively.

$$\textbf{F0}_{\rm{Lagrangian}} = \textbf{R0}^T.\textbf{F0}$$ $$\textbf{F1}_{\rm{Lagrangian}} = \textbf{R1}^T.\textbf{F1}$$

Now if I were to perform and eigenvector extraction on $\textbf{F0}_{\rm{Lagrangian}}$ and $\textbf{F1}_{\rm{Lagrangian}}$, shouldn't their eigenvectors be the same? If not could you please explain why? My end goal is to ensure that both deformation gradients are described/projected onto the same basis system.

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The eigenvectors should not be the same.

Two different deformations, hence different stretches and stretch directions, independent on whether material or spatial.

First of all, the convective description is

$$ \mathbf F=\mathbf e_i \otimes \mathbf E_i = \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right] \mathbf e_i \otimes \mathbf E_j$$

does not say much. Without specifying the $\mathbf e_i$ in terms of the $\mathbf E_i$ we can't do anything. A simple shear deformation with the shear direction $\mathbf E_1$ and the shear plane normal $\mathbf E_2$ is

$$ \mathbf e_1 = \mathbf E_1 \\ \mathbf e_2 = \mathbf E_2 + \gamma \mathbf E_1 \\ \mathbf e_3 = \mathbf E_3 \\ $$

where you can insert different values for $\gamma$. To compare the deformation gradients for different $\gamma$, just write them down w.r.t. the same basis, where the choice $\mathbf E_i$ is most reasonable:

$$ \mathbf F=\mathbf e_i \otimes \mathbf E_i \\ = \mathbf E_1 \otimes \mathbf E_1 + (\gamma \mathbf E_1 + \mathbf E_2 ) \otimes\mathbf E_2 + \mathbf E_3 \otimes \mathbf E_3 \\ =\left[\begin{matrix} 1 & \gamma & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right] \mathbf E_i \otimes \mathbf E_j$$

Then, for the polar decomposition you can operate on this matrix, using the same base vectors for all quantities. This is the easiest way to compare $\mathbf F_0$ and $\mathbf F_1$. Doing the polar decomposition gives you different rotation angles $\phi_{0/1}$ and different stretches and stretch directions of $\mathbf U_{0/1}$ and $\mathbf V_{0/1}$, all depending on $\gamma$.

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