Page 56 - Introduction to Continuum Mechanics

Part B Eigenvalues and Eigenvectors of a Tensor 41

ever, there are actually infinitely many eigenvectors corresponding to the double root. In fact,
since Te 1=2e 1 and Tc 2=2e2, therefore,

i.e., aei-f/Se2 is an eigenvector with eigenvalue 2. This fact can also be obtained from
Eqs.(2B17.3c). With A=2 these equations give

Thus, 0.1 and«2 are arbitrary and«3=0 so that any vector perpendicular to e 3, i.e.,
n=a e +6Z e s an
i l 2 2 i eigenvector.

Example 2B17.4
Find the eigenvalues and eigenvectors for the tensor

Solution. The characteristic equation gives

Thus, there are three distinct eigenvalues, Aj=2, A 2=5 and A 3= -5.

Corresponding to Aj=2, Eqs. (2B17.3c) give

and Eq. (2B17.5) gives

3 0 ^ «i=±l, so that the eigenvector corresponding to Aj=2 is n 1=±ej. We
Thus, «2 =C1: = an
note that from the Example 2B17.2, this eigenvalue 2 and the corresponding eigenvector
±«i can be written down by inspection without computation.

Corresponding to A 2=5, we have