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Part B Principal Values and Principal Directions of Real Symmetric tensors 43

Weobtainaj=0, a 2=0, «3=±1, i.e., n=±e 3, which, of course, is parallel to the axis of rotation.

2B18 Principal Values and Principal Directions of Real Symmetric tensors

In the following chapters, we shall encounter several tensors (stress tensor, strain tensor,
rate of deformation tensor, etc.) which are symmetric, for which the following theorem, stated
without proof, is important: "the eigenvalues of any real symmetric tensor are all real." Thus,
for a real symmetric tensor, there always exist at least three real eigenvectors which we shall
also call the principal directions. The corresponding eigenvalues are called the principal
values. We now prove that there always exist three principal directions which are mutually
perpendicular.
Let RI and n 2 be two eigenvectors corresponding to the eigenvalues A i and A 2 respectively
of a tensor T. Then

and

Thus,

The definition of the transpose of T gives nj/Tn 2 = n 2 -T n 1? thus for a symmetric tensor
T
T, T=T , so that n^T^ = n 2-Tn 1. Thus, from Eqs. (iii) and (iv), we have

It follows that if Aj is not equal to A 2, then nj -n 2 = 0, i.e., nj and n 2 are perpendicular to each
other. We have thus proven that if the eigenvalues are all distinct, then the three principal
directions are mutually perpendicular.
Next, let us suppose that n^ and n 2 are two eigenvectors corresponding to the same eigen-
value A. Then, by definition, Tnj = An^ and Tn 2 = An 2 so that for any a, and ft,
T(an 1+/3n 2)=aTn 1+/TTn 2=A(ani+/?n 2). That is ctn 1-»-j8n 2 is also an eigenvector with the same
eigenvalue A . In other words, if there are two distinct eigenvectors with the same eigenvalue,
then, there are infinitely many eigenvectors (which forms a plane) with the same eigenvalue.
This situation arises when the characteristic equation has a repeated root. Suppose the
characteristic equation has roots A! and A 2=A 3=A (Aj distinct from A). Let nj^ be the eigenvec-
tor corresponding to Aj, then nj is perpendicular to any eigenvector of A. Now, corresponding
to A, the equations