Page 55 - Introduction to Continuum Mechanics

40 Tensors

use Eq. (2B17.3) to find the eigenvalues and Eqs. (2B17.4) to find the eigenvectors. Indeed,
Eq. (2B17.3) gives, for this tensor the following characteristic equation:

So we have a triple root A=2. Substituting A=2 in Eqs. (2B17.3c), we obtain

an
Thus, all three equations are automatically satisfied for arbitrary values of a^ «2» ^ «3, so
that vectors in all directions are eigenvectors. We can choose any three directions as the three
independent eigenvectors. In particular, we can choose the basis {e,-} as a set of linearly
independent eigenvectors.

Example 2B17.2

Show that if 72i=73 1=0, then ±ej is an eigenvector of T with eigenvalue T\\.
Solution. From Tej = T^ei* 72162+73163, we have
Te 1 = ri 1e 1andT(-e 1) = r 1i(-e 1)

Thus, by definition, Eq. (2B17.1), ±61 are eigenvectors with TH as its eigenvalue. Similarly, if
7*12=732=0, then ±62 are eigenvectors with corresponding eigenvalue T22 and if
7i3=7*23=0, then ±63 are eigenvectors with corresponding eigenvalue T33.

Example 2B17.3

Given that

\J \J +S
Find the eigenvalues and their corresponding eigenvectors.
Solution. The characteristic equation is

Thus, Ai=3, A2=A 3=2. (note the ordering of the eigenvalues is arbitrary). These results are
obvious in view of Example 2B17.2. In fact, that example also tells us that the eigenvector
corresponding to Ai=3 is 63 and eigenvectors corresponding to A2=A 3=2 are 61 and 62- How-