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Example of finding the principal values
We will find the principal values and principal axes for the tensor below using the method outlined above.
First let's rearrange the equation:
The system has a solution other than x1 = x2 = x3 = 0 only if ...
Example of finding the principal values
... the determinant of the matrix is zero. That is ...
= (5 - λ) [(3 - λ)2 - 1] + [1 - (3 - λ)] + [1 - (3 - λ)]

= 0;

This gives us after a short amount of arrangement (be careful here) ...
λ3 − 11λ2 + 36 λ - 36 = 0

Now we know that for symmetric tensors the principal values are all real, and so there are several easy ways to find them
Example of finding the principal values
The principal values can be obtained from the cubic by:
1. Drawing the graph and reading off the intercepts
2. Using the general cubic formula for cubics with
three real roots
3. make a guess and if correct, factor out the root

The graph of the equation is to the right
We can see that the roots are around
λ = 2,3 or 6

Using the general cubic formula for 3 real roots
we find the roots are at λ = 2.00, 3.00 and 6.00

If we guessed λ = 2 to begin with, we find

λ3 - 11λ2 + 36λ - 36 = (λ - 2)(λ2 - 9λ + 18 = (λ - 2)(λ - 3)(λ - 6) = 0

And so λ= 2, 3 or 6