Page 54 - Introduction to Continuum Mechanics

Part B Eigenvalues and Eigenvectors of a Tensor 39

Let n = a/e/, then in component form

In long form, we have

Equations (2B17.3c) are a system of linear homogeneous equations in a lt a 2, and #3.
Obviously, regardless of the values of A, a solution for this system is a 1=a 2=a3=0. This is know
as the trivial solution. This solution simply states the obvious fact that a = 0 satisfies the
equation Ta = Aa, independent of the value of A. To find the nontrivial eigenvectors for T, we
note that a homogeneous system of equations admits nontrivial solution only if the determinant
of its coefficients vanishes. That is

i.e.,

For a given T, the above equation is a cubic equation in A. It is called the characteristic equation
of T. The roots of this characteristic equation are the eigenvalues of T.
Equations (2B173), together with the equation

allow us to obtain eigenvectors of unit length. The following examples illustrate how eigen-
vectors and eigenvalues of a tensor can be obtained.

Example 2B 17.1
If, with respect to some basis {e/}, the matrix of T is
"2 0 0"
[T]= 02 0
[00 2
find the eigenvalues and eigenvectors for this tensor.

Solution. We note that this tensor is 21, so that Ta = 2Ia = 2a, for any vector a. Therefore,
by the definition of eigenvector,(see Eq. (2B17.1)), any direction is a direction for an eigen-
vector. The eigenvalues for all the directions are the same, which is 2. However, we can also