Some matrices can be transformed to diagonal matrices, that is, a matrix where the terms not on the leading diagonal are zero.
For a symmetrical matrix we can rotate it to get a diagonal matrix, do some operation, then rotate it back to its original coordinates. This rotation matrix is the eigen matrix or the orthonormal basis of [A], in other words:
[D] = [Q]-1 [A] [Q]
where:
- [D] = Diagonal matrix, diagonal terms are eigenvectors of A
- [A] = Symmetrical Matrix
- [Q] = Orthogonal matrix, columns are eigenvectors of A
- [Q]-1 = inverse of [Q]
Derivation
The length of a vector squared is given by:
|V|² = Vt * V
where
- V = vector
- Vt = transpose of vector
- |V| = length of vector
This length will be unchanged if the coordinates are rotated by a matrix [R]. In this case the vector V is replaced by [R]V and the transposed vector Vt is replaced by Vt[R]t (transposing both operands reverses the order) and the unchanged length is therefore:
|V|² = Vt * V = Vt[R]t[R]V
therefore:
[R]t[R] = [1]
where
- [1] = identity matrix
- [R]t = transpose of rotation matrix
See quadratic form.
Inertia Tensor
An example of diagonalisation is an inertia tensor.
- Find the eigenvalues a by solving 0 = det{[A] - a[1]) for a. The values of a are the principal moments of inertia.
- Find the eigenvectors v of A by solving A v = a v for v.
- Normalize the eigenvectors.
- Form the matrix C whose whose columns consist of the normalized
- D = Ct A C is the diagonal matrix of principal moments of inertia.
In principle, you can write down D directly after (1), however, completing (1) to (5) gives a check on your work.
Note: Ct is the transpose of C.
For this case where the only off diagonal terms are 12 and 21, you
know it only needs a rotation about axis 3 to diagonalize it. Use a
similarity transformation:
A'JA where A is the 3x3 rotation matrix about z.
Solve to find
j12 = 0 = j11cos²(a) -j22sin²(a) solve for a
j22 = j22cos²(a) - j11sin²(a)
j11 = j11cos²2(a) - j22sin²(a)
I don't know if that's easier.
