Sign issue with the matrix of rotated elements (stiffness matrix)

Question

Good day All I have a doubt regarding the derivation of the following matrix

according to my basic understanding we want to go from the basis u1, v1, u2, v2, to the basis u'1, v'1,u'2 ,v'2, and for doing so we use the rotation matrix, the rotation matrix is the following and the angle theta is positive

but i still can 't understand why the signs are inverted in the matrix of the first image ( the sign are inverted)? any help would be highly appreciated Many thanks in advance!

Answer 1

Whoah, there's actually so much algebra behind this. It's all about covariant and contravariant indices.

I'll try to make it easy: to transform a basis, you use a certain matrix $R$, so that $u_2= R\cdot u_1$, which means "each vector of the new basis is obtained applying $R$ to the corresponding vector of the old basis". In keywords, new basis = R · old basis.

However, given a vector, the coordinates transform inversely, that is, they use the inverse matrix $R^{-1}$.

$v_2 = R^{-1} v_1$, new coordinates = R^-1 old coordinates.

This said, the particular case of rotations is that a rotation is a unitary orthogonal matrix, so that

$$ R^{-1}=R^t $$

Which you can easily check: since $\sin(-\alpha)=-\sin (\alpha)$ (and cosines are the same), putting the minus sign in the other place means turning a "negative angle", or rotating it backwards, so it is the inverse matrix. If you multiply both matrices, you'll see that you get the identity matrix.

Thus we can conclude that

• The minus sign is upside if you are transforming the basis vectors.
• The minus sign is below if you are transforming coordinates of vectors.

I'm assuming this is rotation around the z axis. It works the same for the x axis, but ratoations around the y axis have the sign flipped, so be careful.

Answer 2

The Rotation Matrix is a skew symmetric Matrix.

$R_\alpha$ = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix}

When we transpose it, we get:

$R_\alpha^T$ = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{bmatrix}

Therefore we see that when we transpose a skew symmetric matrix the signs get inverted.

By properties of matrices we can either write

$A'$ = $A$ * $R_\alpha$

or,

$A'$ = $R_\alpha^T$ * $A$