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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
0 1 3 x1 x2 x3 5 − λ 3 − λ 3 − λ =
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.
Example of finding the principal axes
Try λ = 2 first. Substituting in
0 1 3 x1 x2 x3 = 3x1 + + + + + + =
These three equations can be manipulated to give

2x1 = 0 and x1 + x2 + x3 = 0

giving x1 = 0 and x2 = −x3. The means our principal axis is of the form
By setting the scalar product with itself equal to 1, we set the vector to unit length and find x2 = 1/√2 giving us the principal axis


   x = 1/√2
Example of finding the principal axes
Now do λ = 3. Substituting in gives
0 1 3 x1 x2 x3 2 = 2x1 + + + + =
The top line is the some of the other two lines, so the euqations we need are

x1 + x3 = 0 and x1 + x2 = 0

giving x3 = x2 = −x1. The means our principal axis is of the form
Again by setting the scalar product with itself equal to 1, we set the vector to unit length and find x1 = 1/√3 giving us the principal axis


   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
0 1 2 3 6 x1 x2 x3 2 1 1
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
0 1 2 3 6 x1 x2 x3 2
The determinent of 1/√6


1/√63(−√2x(√3 × 1 − (−√3 x 1)) + 2 × (√3 × −√2 − (-√3 × −√2)))
= 1/√63(−2√6 − 4√6) = −1
is
Choosing a vector to invert, we finally obtain the principal vectors as
x = 1/√2
x = 1/√3
x = 1/√6