X.B.6 Eigenvectors of symmetric tensors and orthogonal matrices

Most of the tensors corresponding to finite element results are order 2 symmetric tensors. The three eigenvectors of these tensors are mutually orthogonal, and it is possible to find a coordinate system with respect to which tensor $T$ has a diagonal form. More precisely, this is done as follows:

• The eigenvalue problem is solved:

  $$\left(T-\lambda I\right)v=0.$$

  This leads to the calculation of three eigenvalues ${\lambda }_i$ and associated vectors $v_i$.

• The three eigenvectors are defined up to a multiplicative constant. To avoid ambiguity regarding their definition we impose unit length to these vectors:

  $$v_i\cdot v_i=1.$$

  If tensor $T$ is symmetric, the three vectors are mutually orthogonal and we have:

  $$v_i\cdot v_j={\delta }_{ij},$$

• An orthogonal matrix or tensor $Q$ can then be created by assembling the three eigenvectors, each corresponding to a column of $Q$:

  $$Q=\left(\begin{array}{ccc}\hfill c_{11}\hfill & \hfill c_{12}\hfill & \hfill c_{13}\hfill \\ \hfill c_{21}\hfill & \hfill c_{22}\hfill & \hfill c_{23}\hfill \\ \hfill c_{31}\hfill & \hfill c_{32}\hfill & \hfill c_{33}\hfill \end{array}\right)$$

  in which the components are related to the eigenvectors as follows:

  $$v_1=\left(\begin{array}{c}\hfill c_{11}\hfill \\ \hfill c_{21}\hfill \\ \hfill c_{31}\hfill \end{array}\right)\text{ , }{v}_2=\left(\begin{array}{c}\hfill c_{12}\hfill \\ \hfill c_{22}\hfill \\ \hfill c_{32}\hfill \end{array}\right)\text{ and }{v}_3=\left(\begin{array}{c}\hfill c_{13}\hfill \\ \hfill c_{23}\hfill \\ \hfill c_{33}\hfill \end{array}\right).$$

• Matrix or tensor $Q$ can be considered as a rotation tensor. allowing diagonalization of tensor $T$ as follows:

  $$T^{\prime }=\left(\begin{array}{ccc}\hfill {\lambda }_1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {\lambda }_2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {\lambda }_3\hfill \end{array}\right)=QT\cdot T\cdot Q,$$

  and the reverse operation is done as follows:

  $$T=Q\cdot T^{\prime }\cdot QT,$$

• The rotation tensor $Q$ can also be used to transform vectors:

  $$v^{\prime }=QT\cdot v,$$

  $$v=Q\cdot v^{\prime }.$$