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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:

First let's rearrange the equation:

The system has a solution other than x_{1} = x_{2} = x_{3} = 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) ...

= 0;

This gives us after a short amount of arrangement (be careful here) ...

λ^{3} − 11λ^{2} +
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

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

λ^{3} - 11λ^{2} + 36λ - 36 = (λ
- 2)(λ^{2} - 9λ + 18 = (λ - 2)(λ
- 3)(λ - 6) = 0

And so λ= 2, 3 or 6

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

λ

And so λ= 2, 3 or 6