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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:
0 1 3 x1 x2 x3 5 = λ to 5 - λ 3 - λ 3 - λ =
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 ...
0 1 3 x1 x2 x3 5 - λ 3 - λ 3 - λ = (5 - λ) 3 - λ 3 - λ + 1 × 3 - λ + 1 × 3 - λ
= (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
Layer 2 2 4 6 8 40 30 20 10 0 -10 -20 -30 -40
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