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Example of finding the principal axes

We will now find the principal axes of the tensor below
using the method outlined above.

We will start off with the equation

We will start off with the equation

We want to find the principal axis

By applying each principal value to the equation above and expanding out the equations we will find a relationship between the components. We can then scale the vector to be unit length.

By applying each principal value to the equation above and expanding out the equations we will find a relationship between the components. We can then scale the vector to be unit length.

Example of finding the principal axes

Try λ = 2 first. Substituting in

These three equations can be manipulated to give

2x_{1} = 0 and x_{1} + x_{2} + x_{3}
= 0

giving x_{1} = 0 and x_{2} = −x_{3}. The
means our principal axis is of the form

2x

giving x

By setting the scalar product with itself equal to 1,
we set the vector to unit length and find x_{2} = 1/√2
giving us the principal axis

x = 1/√2

x = 1/√2

Example of finding the principal axes

Now do λ = 3. Substituting in gives

The top line is the some of the other two lines, so the
euqations we need are

x_{1} + x_{3} = 0 and x_{1} + x_{2}
= 0

giving x_{3} = x_{2} = −x_{1}. The means
our principal axis is of the form

x

giving x

Again by setting the scalar product with itself equal
to 1, we set the vector to unit length and find x_{1} = 1/√3
giving us the principal axis

x = 1/√3

x = 1/√3

Example of finding the principal axes

To find the third basis vector we could take the vector
product of the other two basis vectors or do the same thing again. Following
the previous pattern using the principal value of 6 we find the normalised
basis vector is

x = 1/√6

The problem with doing it this way is that we do not
know if the basis is right- or left-handed. We almost always work in
a right handed basis, so to check this we check the determinent of the
rotation matrix built up from these basis vectors.

For the diagonal matrix

the rotation matrix is

1/√6

with respect to the original tensor

Note the principal axes line up with the principal values

Example of finding the principal axes

For a transformation matrix, a determinent of 1 means
it is a pure rotation and the right-handed basis is preserved. If the
determinent is −1 an inversion has occured and we are left with
a left-handed basis. To switch from left-handed to right-handed, simply
multiply one of the basis vectors by −1

The determinent of 1/√6

1/√6^{3}(−√2x(√3 × 1 − (−√3
x 1)) + 2 × (√3 × −√2 − (-√3
× −√2)))

= 1/√6^{3}(−2√6 − 4√6) = −1

1/√6

= 1/√6

is

Choosing a vector to invert, we finally obtain the principal
vectors as

x = 1/√2

x = 1/√3

x = 1/√6