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.